A duality result in locally convex cones

Abstract

In this paper, we consider a locally convex cone $({\mathcal {P}},{\mathcal {V}})$ and verify the dual of $(Conv({\mathcal {P}}),{\overline{{\mathcal {V}}}})$ the locally convex cone of the non-empty convex subsets of ${\mathcal {P}}$. Under some semilattice conditions, we characterize the dual of $Conv(\underbrace{\cdots }_{n\ times} (Conv({\mathcal {P}}))$.

1 Introduction

Duality theory is a powerfull technique to study a wide class of related problems in pure and applied mathematics. For example the Hahn-Banach extension and separation theorems studied by means of duals (see [8]). The collection of all non-empty convex subsets of a cone (or a vector space) is interesting in convexity and approximation theory (for example see [5]). This collection is a cone. We consider the non-empty convex subsets of a cone ${\mathcal {P}}$, denoted by $Conv({\mathcal {P}})$, and verify the dual of it, when ${\mathcal {P}}$ is a locally convex cone. We note that some elements of the dual of $Conv({\mathcal {P}})$ have already been introduced (see [6], I: Example 2.1(e) and Example 5.31 (b)). Firstly we review the structure of locally convex cones briefly:

A nonempty set ${\mathcal {P}}$ endowed with an addition and a scalar multiplication for nonnegative real numbers is called a cone whenever the addition is associative and commutative, there is a neutral element $0 \in {\mathcal {P}}$ and for the scalar multiplication the usual associative and distributive properties hold, that is $\alpha (\beta a) = (\alpha \beta )a$, $(\alpha + \beta )a = \alpha a + \beta a$, $\alpha (a + b) = \alpha a + \alpha b$, $1a = a$ and $0a = 0$ for all $a, b \in {\mathcal {P}}$ and nonnegative reals $\alpha$ and $\beta$.

The theory of locally convex cones as introduced and developed by K. Keimel and W. Roth in [4]. It uses an order theoretical concept or a convex quasi-uniform structure on a cone. In this paper, we use the former. For some recent researches see [1,2,3, 7].

A (preordered cone) is a cone ${\mathcal {P}}$ endowed with a preorder (reflexive transitive relation) $\le$ which is compatible with the addition and scalar multiplication, that is $x\le y$ implies $x+z\le y+z$ and $r\cdot x\le r\cdot y$ for all $x,y,z\in {\mathcal {P}}$ and $r\in {\mathbb {R}}_+=\{r\in {\mathbb {R}} \ : \ r\ge 0\}.$ Every ordered vector space is an ordered cone. The cones ${\overline{{\mathbb {R}}}}={\mathbb {R}}\cup \{+\infty \}$ and ${\overline{{\mathbb {R}}}}_+={\mathbb {R}}_+\cup \{+\infty \}$, with the usual order and algebraic operations (specially $0\cdot (+\infty )=0$), are ordered cones that are not embeddable in vector spaces.

A subset ${\mathcal {V}}$ of a preordered cone ${\mathcal {P}}$ is called an (abstract) 0-neighborhood system, if

$(v_1)$:

$0<v$ for all $v\in {\mathcal {V}}$;

$(v_2)$:

for all $u,v\in {\mathcal {V}}$ there is a $w\in {\mathcal {V}}$ with $w\le u$ and $w\le v$;

$(v_3)$:

$u+v\in {\mathcal {V}}$ and $\alpha v\in {\mathcal {V}}$ whenever $u,v\in {\mathcal {V}}$ and $\alpha >0$.

Let $a\in {\mathcal {P}}$ and $v\in {\mathcal {V}}$. We define $v(a)=\{b\in {\mathcal {P}}\ | \ b\le a+v\}$, resp. $(a)v=\{b\in {\mathcal {P}}\ | \ a\le b+v\},$ to be a neighborhood of a in the upper, resp. lower topologies on ${\mathcal {P}}$. The common refinement of the upper and lower topologies is called symmetric topology. We denote the neighborhoods of a in the symmetric topology by v(a)v. The pair $({\mathcal {P}},{\mathcal {V}})$ is called a full locally convex cone if the elements of ${\mathcal {P}}$ are bounded below, i.e. for every $a\in {\mathcal {P}}$ and $v\in {\mathcal {V}}$ we have $0\le a+\rho v$ for some $\rho >0$. Each subcone of ${\mathcal {P}}$, not necessarily containing ${\mathcal {V}}$, is called a locally convex cone.

We note that if $({\mathcal {Q}},{\mathcal {V}})$ is a locally convex cone, ${\mathcal {Q}}\oplus ({\mathcal {V}}\cup \{0\})$ with the algebraic operation

$$\begin{aligned} (a,v_1)+(b,v_2)= & {} (a+b,v_1+v_2), \\ \alpha (a,v_1)= & {} (\alpha a,\alpha v_1), \end{aligned}$$

and the preorder

$$\begin{aligned} \left\{ \begin{array}{c} (a,0)\le (b,0)\Leftrightarrow a\le b\\ (0,v_1)\le (0,v_2)\Leftrightarrow v_1\le v_2\\ (a,0)\le (b,v_1)\Leftrightarrow a\le b+v_1,\\ \end{array} \right. \end{aligned}$$

for all $a,b\in {\mathcal {Q}}$, $v_1,v_2\in {\mathcal {V}}$ and $\alpha \in {\mathbb {R}}^+$, $({\mathcal {Q}}\oplus ({\mathcal {V}}\cup \{0\}),{\mathcal {V}})$ is a full locally convex cone which ${\mathcal {Q}}$ and ${\mathcal {V}}$ can be embedded in ${\mathcal {Q}}\oplus ({\mathcal {V}}\cup \{0\})$ by the mappings $a\rightarrow (a,0)$ and $v\rightarrow (0,v)$ for all $a\in {\mathcal {Q}}$ and $v \in {\mathcal {V}}$.

For cones ${\mathcal {P}}$ and ${\mathcal {Q}}$ a mapping $t:{\mathcal {P}}\rightarrow {\mathcal {Q}}$ is called a linear operator if $t(a+b)=t(a)+t(b)$ and $t(\alpha a)=\alpha t(a)$ hold for $a,b\in {\mathcal {P}}$ and $\alpha \ge 0$.

A linear functional on a cone ${\mathcal {P}}$ is a linear mapping $\mu :{\mathcal {P}}\rightarrow {\overline{{\mathbb {R}}}}$.

Let $({\mathcal {P}}, {\mathcal {V}})$ and $({\mathcal {Q}},{\mathcal {W}})$ be two locally convex cones. The linear operator $t:({\mathcal {P}}, {\mathcal {V}})\rightarrow ({\mathcal {Q}},{\mathcal {W}})$ is called uniformly continuous or simply u-continuous if for every $w\in {\mathcal {W}}$ one can find a $v\in {\mathcal {V}}$ such that $a\le b+v$ implies $t(a)\le t(b)+w$. It is easy to see that the u-continuity implies continuity with respect to the upper, lower and symmetric topologies on ${\mathcal {P}}$ and ${\mathcal {Q}}$.

According to the definition of u-continuity, a linear functional $\mu$ on $({\mathcal {P}}, {\mathcal {V}})$ is u-continuous if there is a $v\in {\mathcal {V}}$ such that $a\le b+v$ implies $\mu (a)\le \mu (b)+1$. The u-continuous linear functionals on a locally convex cone $({\mathcal {P}},{\mathcal {V}})$ (into ${\overline{{\mathbb {R}}}}$) form a cone with the usual addition and scalar multiplication of functions. This cone is called the dual cone of ${\mathcal {P}}$ and denoted by ${\mathcal {P}}^*$.

For a locally convex cone $({\mathcal {P}},{\mathcal {V}})$, the polar $v^{\circ }$ of $v\in {\mathcal {V}}$ consists of all linear functionals $\mu$ on ${\mathcal {P}}$ satisfying $\mu (a)\le \mu (b)+1$ whenever $a\le b+v$ for $a,b\in {\mathcal {P}}$. We have $\cup \{v^{\circ }: v\in {\mathcal {V}}\}={\mathcal {P}}^*$. The cones ${\overline{{\mathbb {R}}}}$ and ${\overline{{\mathbb {R}}}}_{+}=\{a\in {\overline{{\mathbb {R}}}}:a\ge 0 \}$ with (abstract) 0-neighborhood ${\mathcal {V}}=\{\varepsilon >0:\varepsilon \in {\mathbb {R}}\}$ are locally convex cones. The dual cones of ${\overline{{\mathbb {R}}}}$ and ${\overline{{\mathbb {R}}}}_{+}$ under ${\mathcal {V}}$ consists of all nonnegative reals and the functional $0_{\infty }$ such that $0_{\infty }(a)=0$ for all $a\in {\mathbb {R}}$ and $0_{\infty } (+\infty )=+\infty$.

2 Dual of the cone of non-empty convex sets of a locally convex cone

A subset A of a cone ${\mathcal {P}}$ is said convex, if $\lambda a+(1-\lambda ) b\in A$, whenever $a,b\in {\mathcal {P}}$ and $0\le \lambda \le 1$. Let ${\mathcal {P}}$ be a preordered cone and $Conv({\mathcal {P}})$ be the cone of all non-empty convex subsets of ${\mathcal {P}}$, endowed with the usual addition and multiplication of sets by non-negative scalars, that is $\alpha A = \{\alpha a\ |\ a \in A\}$ and $A + B = \{a + b\ |\ a \in A \ and \ b \in B\}$ for $A, B \in Conv({\mathcal {P}})$ and $\alpha \ge 0$. We consider the order on $Conv({\mathcal {P}})$ by

$$\begin{aligned} A \preceq B \ \ \ if \ \ \ A \subseteq \downarrow {B}, \end{aligned}$$

where $\downarrow B = \{x \in {\mathcal {P}}| x \le b \ \text{ for } \text{ some } \ b \in B\}$ is the decreasing hull of the set B in ${\mathcal {P}}$. Note that $\downarrow B$ is again a convex subset of ${\mathcal {P}}$. The requirements for a preordered cone are easily checked. The neighborhood system in $Conv({\mathcal {P}})$ is ${\overline{{\mathcal {V}}}}:=\{ {\overline{v}}=\{v\}\ |\ v\in {\mathcal {V}}\}$, that is

$$\begin{aligned} A \preceq B + {\overline{v}} \ \ \ if \ \ \ A \subseteq \downarrow (B + \{v\} ) \end{aligned}$$

for $A, B \in Conv({\mathcal {P}})$ and ${\overline{v}} \in {\overline{{\mathcal {V}}}}$. The cone $Conv({\mathcal {P}})$ with (abstract) 0-neighborhood system ${\overline{{\mathcal {V}}}})$ is a locally convex cone. Via the embedding $x\rightarrow \{x\}: {\mathcal {P}}\rightarrow Conv({\mathcal {P}})$ the preordered cone ${\mathcal {P}}$ itself may be considered as a subcone of $Conv({\mathcal {P}})$ (see [6], I, Example 1.4 (c)).

Definition 1

We say that a preordered cone ${\mathcal {P}}$ is a $\bigvee$-semilattice cone if the order of ${\mathcal {P}}$ is antisymmetric and if

$(\bigvee 1)$ every non-empty subset $A\subseteq {\mathcal {P}}$ has a supremum $\sup A\in {\mathcal {P}}$ and $\sup (A+b)=\sup A+b$ hold for all $b\in {\mathcal {P}}$.

Moreover, if ${\mathcal {P}}$ with an abstract neighborhood system ${\mathcal {V}}$ is a locally convex cone and

$(\bigvee 2)$ for $\emptyset \ne A\subseteq {\mathcal {P}}$, $b\in {\mathcal {P}}$ and $v\in V$ such that $a\le b+v$ for all $a\in A$, we have $\sup A\le b+v$,

then $({\mathcal {P}}, {\mathcal {V}})$ is said a $\bigvee$-semilattice locally convex cone.

In particular, every $\bigvee$-semilattice cone ${\mathcal {P}}$ contains a largest element, that is $+\infty =\sup {\mathcal {P}},$ which can be adjoined as a maximal element to any $\bigvee$-semilattice cone with the convention that $a+(+\infty )=+\infty$, $\alpha \cdot (+\infty )=+\infty$, $0 \cdot (+\infty )=0$ and $a\le +\infty$ for all $a\in {\mathcal {P}}$ and $\alpha > 0$.

Remark 1

We note that the condition $(\bigvee 2)$ of definition 1 is necessary and the definition of supremum does not imply this condition in locally convex cones necessarily. We show this in the following example.

Example 1

Let ${\mathbb {R}}$ be as a cone and ${\mathcal {V}}=\{{\bar{\epsilon }}=(-\infty ,\epsilon ) \ : \ \epsilon \in {\mathbb {R}}_{>0}\}$.

Let

$$\begin{aligned} {\mathcal {P}}=\{(a,B) \ : \ a\in {\mathbb {R}} \ \text{ and } \ B\in {\mathcal {V}}\cup \{\{0\}\}\}. \end{aligned}$$

We define

$$\begin{aligned} (a,B)+(c+D)=(a+c,B+D), \end{aligned}$$

and

$$\begin{aligned} \lambda (a,B)=(\lambda a, \lambda B) \end{aligned}$$

for all $(a,B), (c,D)\in {\mathcal {P}}$. Also, we define the preorder

$$\begin{aligned} (a,B)\le (c,D)\Leftrightarrow \left\{ \begin{array}{cc} a\le c &{} \text{ if } \ B=D=\{0\} \\ a+B\subseteq c+D &{} \ \text{ if } \ D\ne \{0\}\\ \end{array} \right. , \end{aligned}$$

for all $(a,B), (c,D)\in {\mathcal {P}}$. Then $({\mathcal {P}},{\mathcal {V}})$ is a full locally convex cone. Now, we can embedded ${\mathbb {R}}$ in ${\mathcal {P}}$ by $a\rightarrow (a,\{0\})$ and we can consider ${\mathbb {R}}$ as a subcone of ${\mathcal {P}}$. We have

$$\begin{aligned} a\le b+{\bar{\epsilon }}\Leftrightarrow (a, \{0\})\le (b, (-\infty ,\epsilon ))\Leftrightarrow \{a\}\subseteq (-\infty ,b+\epsilon )\Leftrightarrow a\in (-\infty ,b+\epsilon ). \end{aligned}$$

Now, for the set $A=(0,5)\subseteq {\mathbb {R}}$, by considering the embedding, we have ${\bar{A}}=\{(a,\{0\}) \ : \ a\in (0,5)\}$. Let $b=4$ and ${\bar{1}}=(-\infty , 1)\in {\mathcal {V}}$. Then

$$\begin{aligned}&a\in (0,5) \Leftrightarrow a \in (0,4+1) \Rightarrow a \in (-\infty ,4 + 1)\\&\quad \Rightarrow (a,\{0\}) \le (4,\{0\}) + (0,(-\infty ,1)), \end{aligned}$$

for all $(a,\{0\})\in {\bar{A}}$, i.e.

$$\begin{aligned} a\le 4+{\bar{1}}, \end{aligned}$$

for all $a\in A=(0,5)$. On the other hand, $\sup A=5$ (in ${\mathbb {R}}$) and we have

$$\begin{aligned} 5\not \in (-\infty ,5)=(-\infty ,4+1)\Rightarrow (5,\{0\})\not \le (4,\{0\})+(0,(-\infty ,1)), \end{aligned}$$

i.e. $5\not \le 4+{\bar{1}}$. Although, ${\mathcal {P}}$ is not a $\bigvee$-semilattice cone, ${\mathbb {R}}$ is a $\bigvee$-semilattice cone. Also, the locally convex cone $({\mathbb {R}},{\mathcal {V}})$ is not a $\bigvee$-semilattice locally convex cone.

Remark 2

We note that definition 1 is similar to the definition of “locally convex $\bigvee$-semilattice cone" in [6], I, 5.4. In this definition, the order do not coincide with the weak preorder necessarily.

We define $Conv^n({\mathcal {P}}):=Conv(Conv^{n-1}({\mathcal {P}}))$ for $n=2,3,\ldots$ and $Conv^1({\mathcal {P}})=Conv({\mathcal {P}})$. Let

$$\begin{aligned} \{a\}^n:= \underbrace{\{\cdots \{}_{n\ times}a\underbrace{\}\cdots \}}_{n\ times} \end{aligned}$$

(1)

for all $a\in {\mathcal {P}}$. It is easy to see that $\{a\}^n \in Conv^n({\mathcal {P}})$ for all $n\in {\mathbb {N}}$. This shows that ${\mathcal {P}}$ is embedded in $Conv^n({\mathcal {P}})$ (the mapping $a\longrightarrow \{a\}^n$ is the embedding). The cone $Conv^n({\mathcal {P}})$ with the (abstract) 0-neighborhood system ${\overline{{\mathcal {V}}}}^n$ is a locally convex cone, where ${\overline{{\mathcal {V}}}}^n:=\{{\bar{v}}^n:= \{v\}^n\ |\ v\in {\mathcal {V}}\}$.

Example 2

For the cone ${\mathbb {R}}$, we have $A^1=[0,1] \in {Conv}(\overline{{\mathbb {R}}})$ , $A^2=\{ [0,a]\ |\ ,a \in [0,1] \}$ is an element of $Conv^2({\mathbb {R}})$ and $A^3=\{\{ [0,a]\ |\ ,a \in [0,b] \}\ |\ b \in [0,1] \}$ is an element of $Conv^3({\mathbb {R}})$.

For the element $A^n$ of $Conv^n({\mathcal {P}})$ we define

$$\begin{aligned} sup ^s(A^n):=\sup \{sup^s(A^{n-1}) \ | \ A^{n-1}\in A^n\} \end{aligned}$$

for $n=2,3, \ldots$ and $sup^s(A^1)=\sup A$. It is easy to see that $sup^s(A^n)\in {\mathcal {P}}$ for all $n\in {\mathbb {N}}$.

The following lemma is an special case of Lemma 5.5 of [6].

Lemma 1

Let ${\mathcal {P}}$ be a $\bigvee -$semilattice cone and $\{A_i\}_{i\in I}$ be a collection of non-empty subsets of ${\mathcal {P}}$. Then

$$\begin{aligned} \sup \left( \bigcup _{i \in I} A_i\right) =\sup \{\sup A_i \ | \ i \in I\}. \end{aligned}$$

Proof

Let $a \in \bigcup _{i\in I} A_i$ be arbitrary. Then there exists $i \in I$ such that $a \in A_i$. We have $a\le \sup A_i$ and so $a\le \sup \{\sup A_i \ | \ i \in I\}$. Then

$$\begin{aligned} \sup \left( \bigcup _{i \in I} A_i\right) \le \sup \{\sup A_i \ | \ i \in I\}. \end{aligned}$$

On the other hand, $\sup A_i\le \sup (\bigcup _{i \in I} A_i)$ for all $i \in I$. This conclude that

$$\begin{aligned} \{\sup A_i \ | \ i \in I\}\le \sup \left( \bigcup _{i \in I} A_i\right) . \end{aligned}$$

$\square$

Remark 3

We note that $A^2\in Conv^2({\mathcal {P}})$ but the elements of $A^2$ belong to $Conv^1({\mathcal {P}})= Conv({\mathcal {P}})$. This implies that the union of the elements of $A^2$ ($\bigcup _{A^1\in A^2} A^1$) belongs to the power set of ${\mathcal {P}}$. Also, $A^3\in Conv^3({\mathcal {P}})$ and the elements of $A^3$ belong to $Conv^2({\mathcal {P}})$. Then the union of the elements of $A^3$ ($\bigcup _{A^2\in A^3} A^2$) belongs to the power set of $Conv^2({\mathcal {P}})$ and the union of these sets ($\bigcup _{A^2\in A^3}\bigcup _{A^1\in A^2} A^1$) belongs again to the power set of ${\mathcal {P}}$. By continuing this process, we conclude that $A^n\in Conv^n({\mathcal {P}})$ and the elements of $A^n$ belong to $Conv^{n-1}({\mathcal {P}})$. Then

$$\begin{aligned} \bigcup _{A^{n-1}\in A^n}\cdots \bigcup _{A^2\in A^3}\bigcup _{A^1\in A^2} A^1 \end{aligned}$$

belongs to the power set of ${\mathcal {P}}$. By Lemma 1, we have

$$\begin{aligned} sup^s(A^n)=\sup \left( \bigcup _{A^{n-1}\in A^n}\cdots \bigcup _{A^2\in A^3}\bigcup _{A^1\in A^2} A^1\right) . \end{aligned}$$

Let ${\mathcal {P}}$ be a cone and $\mu :{\mathcal {P}}\rightarrow \overline{{\mathbb {R}}}$ be a functional. We define

$$\begin{aligned} {\overline{\mu }}(A):=\sup \{\mu (a)\ | \ a\in A \},\quad A\in Conv({\mathcal {P}}), \end{aligned}$$

moreover, if ${\mathcal {P}}$ is a $\bigvee$-semilattice cone, we define

$$\begin{aligned} \overline{{\overline{\mu }}}(A):=\mu (\sup A ), \quad A\in Conv({\mathcal {P}}). \end{aligned}$$

Lemma 2

Let $({\mathcal {P}}, {\mathcal {V}})$ be a locally convex cone and $\mu \in {\mathcal {P}}^*$. Then ${\overline{\mu }} \in Conv({\mathcal {P}})^*$. Moreover, if $({\mathcal {P}}, {\mathcal {V}})$ is $\bigvee$-semilattice locally convex cone, then $\overline{{\overline{\mu }}} \in Conv({\mathcal {P}})^*$.

Proof

We have

$$\begin{aligned} {\overline{\mu }}(\alpha A+B)&=\sup \{\mu (\alpha a+b)\ |\ a\in A, b\in B\}\\&=\sup \{\alpha \mu (a)+\mu (b)\ |\ a\in A, b\in B\}\\&=\alpha \sup \{\mu ( a)\ |\ a\in A\} +\sup \{\mu (b)\ |\ b\in B\}\\&=\alpha {\overline{\mu }}(A)+{\overline{\mu }}(B), \end{aligned}$$

for all $A,B \in Conv({\mathcal {P}})$ and all $\alpha \ge 0$. So ${\overline{\mu }}$ is linear.

Now, if $({\mathcal {P}}, {\mathcal {V}})$ is $\bigvee$-semilattice locally convex cone, then

$$\begin{aligned} \sup (A+B)&=\sup \left( \bigcup _{b\in B}(A+b)\right) \\&=\sup \{\sup (A+b)\ |\ b\in B\}\quad \text{(by } \text{ Lemma }\, 1\mathrm{)}\\&=\sup \{\sup A+b \ |\ b\in B\}\\&=\sup (A)+\sup (B). \end{aligned}$$

This yields that $\mu (\sup (A+B))=\mu (\sup (A))+\mu (\sup (B))$ and then $\overline{{\overline{\mu }}}(A+B)=\overline{{\overline{\mu }}}(A)+\overline{{\overline{\mu }}}(B)$ for all $A,B \in Conv({\mathcal {P}})$. Also,

$$\begin{aligned} \overline{{\overline{\mu }}}(\alpha A)=\mu (\sup (\alpha A))=\mu (\alpha \sup A )=\alpha \mu (\sup A )=\alpha \overline{{\overline{\mu }}}(A), \end{aligned}$$

for all $\alpha \ge 0$ and $A\in Conv({\mathcal {P}})$. Therefore $\overline{{\overline{\mu }}}$ is linear.

Now, we show that ${\overline{\mu }}$ and $\overline{{\overline{\mu }}}$ are u-continuous extensions of $\mu$ to $Conv({\mathcal {P}})$. Via of continuity of $\mu$, there is a $v\in {\mathcal {V}}$ such that $a\le b+v$ implies $\mu (a)\le \mu (b)+1$. Let $A\preceq B+\{v\}$. Then, for each $a\in A$ there exists $b\in B$ such that $a\le b+v$. We have

$$\begin{aligned} \mu (a)\le \mu (b)+1&\Rightarrow \mu (a)\le \sup \{\mu (b) \ | \ b\in B \}+1\\&\Rightarrow \sup \{\mu (a)\ | \ a\in A \}\le \sup \{\mu (b) \ | \ b\in B \}+1\\&\Rightarrow {\overline{\mu }}(A)\le {\overline{\mu }}(B)+1. \end{aligned}$$

This shows that ${\overline{\mu }}$ is u-continuous. Also if $({\mathcal {P}}, {\mathcal {V}})$ is $\bigvee$-semilattice locally convex cone, we have

$$\begin{aligned} a\le \sup (B)+v&\Rightarrow \sup (A)\le \sup (B) +v \quad ( \text{ by } \bigvee 2)\\&\Rightarrow \mu (\sup (A))\le \mu (\sup (B)) +1\\&\Rightarrow \overline{{\overline{\mu }}}(A)\le \overline{{\overline{\mu }}}(B)+1. \end{aligned}$$

This yields that $\overline{{\overline{\mu }}}$ is u-continuous.

$\square$

Proposition 1

Let ${\mathcal {P}}$ be a preordered cone, $\mu$ be a monotone functional on ${\mathcal {P}}$ and ${\widetilde{\mu }}$ be a monotone extension of $\mu$ on $Conv({\mathcal {P}})$. Then ${\overline{\mu }}\le {\widetilde{\mu }}$. Furthermore, if ${\mathcal {P}}$ is a $\bigvee$-semilattice cone, then

$$\begin{aligned} {\overline{\mu }}\le {\widetilde{\mu }}\le \overline{{\overline{\mu }}}. \end{aligned}$$

(2)

Proof

Let ${\overline{\mu }}\nleq {\widetilde{\mu }}$. Then there exists $A\in Conv({\mathcal {P}})$ such that ${\overline{\mu }}(A)\nleq {\widetilde{\mu }}(A)$ i.e. ${\widetilde{\mu }}(A)< {\overline{\mu }}(A)=\sup \{\mu (a)\ | \ a\in A \}$. Then there exists $a\in A$ such that ${\widetilde{\mu }}(A)<\mu (a)={\widetilde{\mu }}(\{a\})$ (by the supremum property). On the other hand, $\{a\}\preceq A$ and so ${\widetilde{\mu }}(\{a\})\le {\widetilde{\mu }}(A)$. This contradiction yields that ${\overline{\mu }}\le {\widetilde{\mu }}$.

Now, let ${\mathcal {P}}$ be a $\bigvee$-semilattice cone. Let $A\in Conv({\mathcal {P}})$ be arbitrary. We have $A\preceq \{\sup A\}$. Then ${\widetilde{\mu }}(A)\le {\widetilde{\mu }}(\{sup A\})=\mu (sup A)=\overline{{\overline{\mu }}}(A)$. $\square$

Let ${\mathcal {P}}$ be a $\bigvee$-semilattice cone. We denote

$$\begin{aligned} \Omega ({\mathcal {P}}):=\{\mu \in {\mathcal {L}}({\mathcal {P}}) \ | \ \mu \text{ is } \text{ monotone } \text{ and } {\overline{\mu }}(A)=\overline{{\overline{\mu }}}(A), \forall A\in Conv({\mathcal {P}})\}, \end{aligned}$$

where ${\mathcal {L}}({\mathcal {P}})$ is the cone of all linear functionals on ${\mathcal {P}}$.

Corollary 1

Let ${\mathcal {P}}$ be a $\bigvee -$semilattice cone. Then the elements of $\Omega ({\mathcal {P}})$ have unique extensions to ${Conv}({\mathcal {P}})$.

By the assumptions of the Corollary 1, we conclude that the elements of $\Omega ({\mathcal {P}})$ have unique extensions to ${Conv}^n({\mathcal {P}})$.

Proposition 2

Let ${\mathcal {P}}$ be a $\bigvee -$semilattice cone. Then

$$\begin{aligned} sup ^s(A^n)+sup ^s(B^n)=sup ^s(A^n+B^n), \end{aligned}$$

for all $n\in {\mathbb {N}}$ and $A^{n},B^{n}\in Conv^n({\mathcal {P}})$.

Proof

For $n=1$, let $A^1=A$ and $B^1=B$ be elements of $Conv^1({\mathcal {P}})=Conv({\mathcal {P}})$. We have

$$\begin{aligned} sup^s(A+B)&=\sup \left( \bigcup _{b\in B}(A+b)\right) \\&=\sup \{\sup (A+b)\ |\ b\in B\}\quad \text{(by } \text{ Lemma }\, 1\mathrm{)}\\&=\sup \{\sup A+b \ |\ b\in B\}\\&=sup^s(A)+sup^s(B). \end{aligned}$$

Now, let

$$\begin{aligned} sup ^s(A^{n-1})+sup ^s(B^{n-1})=sup ^s(A^{n-1}+B^{n-1}). \end{aligned}$$

Then

$$\begin{aligned} sup^s(A^{n}+B^{n})&=sup^s(\{A^{n-1}+B^{n-1}\ |\ A^{n-1}\in A^n ,\ B^{n-1}\in B^n \})\\&=\sup (\{sup^s(A^{n-1}+B^{n-1})\ |\ A^{n-1}\in A^n\ ,\ B^{n-1}\in B^n \})\\&=\sup (\{sup^s(A^{n-1})+sup^s(B^{n-1})\ |\ A^{n-1}\in A^n\ ,\ B^{n-1}\in B^n \})\\&=\sup (\{sup^s(A^{n-1})\ |\ A^{n-1}\in A^n \})\\&\quad +\sup (\{sup^s(B^{n-1})\ |\ B^{n-1}\in B^n \}) \\&=sup^s(A^{n})+sup^s(B^{n}). \end{aligned}$$

$\square$

Let coh(F) denote the convex hull of the set F, the smallest convex set containing F. We set

$$\begin{aligned} coh^s(A^n) :=coh(\{coh^s(A^{n-1}) \ | \ A^{n-1}\in A^n\}\cup \{sup^s(A^n)\}^n ) \end{aligned}$$

for $n=2,3, \ldots$ and $coh^s(A^1) =coh(A\cup \{sup(A)\})$.

Proposition 3

Let ${\mathcal {P}}$ be a $\bigvee -$semilattice cone. Then

$$\begin{aligned} \sup (coh^s(A^n))=\sup (A^n), \end{aligned}$$

all $A^{n}\in Conv^n({\mathcal {P}})$ and $n\in {\mathbb {N}}$.

Proof

First we show that $\sup (coh^s(A^1))=\sup (A^1)$. Let $x\in coh^s(A^1)$ be arbitrary. Then there are $\lambda _1,\lambda _2,\ldots ,\lambda _k\ge 0$ and $a_1,a_2,\ldots ,a_k\in A^1\cup \{\sup (A^1)\}$ such that $\sum _{i=1}^k \lambda _i=1$ and $x=\sum _{i=1}^k \lambda _i a_i$. On the other hand, $\lambda _i a_i\le \lambda _i \sup (A^1)$ for all $i=1,2,\ldots ,k$. We have

$$\begin{aligned} x=\sum _{i=1}^k \lambda _i a_i \le \sum _{i=1}^k\lambda _i \sup A^1=\sup (A^1). \end{aligned}$$

This yields that

$$\begin{aligned} \sup (coh^s(A^1))=\sup (A^1), \end{aligned}$$

since $\sup (A^1)\in coh^s(A^1)$.

Now, let $\sup (coh^s(A^{n-1}))=\sup (A^{n-1})$ for all $A^{n-1}\in Conv^{n-1}({\mathcal {P}})$. Consider $A^{n}\in Conv^{n}({\mathcal {P}})$ and ${\mathcal {X}}\in coh^s(A^n)$. Then there are $\lambda _1,\lambda _2,\ldots ,\lambda _k\ge 0$ and $A^{n-1}_1,A^{n-1}_2,\ldots ,A^{n-1}_k\in A^n\cup \{\sup (A^n)\}^n$ such that $\sum _{i=1}^k \lambda _i=1$ and ${\mathcal {X}}=\sum _{i=1}^k \lambda _i coh^s(A^{n-1}_i).$ On the other hand,

$$\begin{aligned} \lambda _i coh^s(A^{n-1}_i)\preceq \lambda _i \{sup^s (coh^s( A^{n-1}_i))\}^n=\{sup^s ( A^{n-1}_i)\}^n\preceq \lambda _i \{sup^s ( A^n)\}^n, \end{aligned}$$

for all $i=1,2,\ldots ,k$. So

$$\begin{aligned} {\mathcal {X}}=\sum _{i=1}^k \lambda _i coh^s(A^{n-1}_i) \preceq \sum _{i=1}^k\lambda _i \{sup^s ( A^n)\}^n=\{sup^s ( A^n)\}^n, \end{aligned}$$

and so

$$\begin{aligned} sup^s({\mathcal {X}})\le sup^s ( A^n). \end{aligned}$$

Since $\{sup^s ( A^n)\}^n\in coh^s(A^{n})$, we have

$$\begin{aligned} sup^s(coh^s(A^{n}))=sup^s(A^{n}). \end{aligned}$$

$\square$

Remark 4

By Proposition 3 and by considering the construction of $coh^s(A^{n})$, we have

$$\begin{aligned} \{sup^s(coh^s(A^{n}))\}^n\in coh^s(A^{n}) \end{aligned}$$

for all $n\in {\mathbb {N}}$.

Example 3

For the cone $\overline{{\mathbb {R}}}$, we have $\{0\},\{0,+\infty \} \in {Conv}(\overline{{\mathbb {R}}})$ and $A^2=\{ \{0\},\{0,+\infty \} \}$ is an element of $Conv^2(\overline{{\mathbb {R}}})$. We have $sup ^s(A^2)=\sup \{0,+\infty \}=+\infty$ and $coh^s(A^2)=\{\{0\},\{0,+\infty \} ,\{+\infty \} \}$.

For every positive integer n we introduce

$$\begin{aligned} {Conv_s}^n({\mathcal {P}}):= \{coh^s(A^n) \ | \ A^n\in Conv^n({\mathcal {P}})\}. \end{aligned}$$

Theorem 1

Let ${\mathcal {P}}$ be a $\bigvee -$semilattice cone. Then ${Conv_s}^n({\mathcal {P}})$ is a subcone of ${Conv}^n({\mathcal {P}})$ for all $n\in {\mathbb {N}}$.

Proof

Let ${\mathcal {A}},{\mathcal {B}}\in {Conv_s}^1({\mathcal {P}})$. Then there exist $A^1,B^1\in Canv({\mathcal {P}})$ such that

$$\begin{aligned} {\mathcal {A}}=coh^s(A^1) =coh(A^1\cup \{sup(A^1)\}) \end{aligned}$$

and

$$\begin{aligned} {\mathcal {B}}=coh^s(B^1) =coh(B^1\cup \{sup(B^1)\}). \end{aligned}$$

We conclude that ${\mathcal {A}},{\mathcal {B}} \in Canv({\mathcal {P}})$. Put ${\mathcal {A}}+{\mathcal {B}}={\mathcal {C}}$. We have

$$\begin{aligned} \sup ({\mathcal {A}})+\sup ({\mathcal {B}})= \sup ({\mathcal {A}}+{\mathcal {B}})=\sup ({\mathcal {C}}), \end{aligned}$$

by Proposition 2 (for case $n=1$). Since ${\mathcal {A}},{\mathcal {B}}$ contain their suprema, then ${\mathcal {C}}$ contains its supremum. Hence

$$\begin{aligned} {\mathcal {C}} =coh({\mathcal {C}}\cup \{\sup ({\mathcal {C}})\})=coh^s({\mathcal {C}}), \end{aligned}$$

which conclude that ${\mathcal {C}}\in {Conv_s}^1({\mathcal {P}})$. On the other hand, for each $\alpha \ge 0$,

$$\begin{aligned} \alpha {\mathcal {A}}=\alpha coh^s(A^1)= coh^s(\alpha A^1) =coh( \alpha A^1\cup \{sup( \alpha A^1)\}), \end{aligned}$$

and so $\alpha {\mathcal {A}}\in {Conv_s}^1({\mathcal {P}})$. Hence ${Conv_s}^1({\mathcal {P}})$ is a subcone of ${Conv}({\mathcal {P}})$. For completion of induction, first we show that ${Conv_s}^{n+1}({\mathcal {P}})\subseteq {Conv}^{n+1}({\mathcal {P}})$. For this, let ${\mathcal {A}}\in {Conv_s}^{n+1}({\mathcal {P}})$. There is $A^{n+1}\in Canv^{n+1}({\mathcal {P}})$ such that ${\mathcal {A}}=coh^s(A^{n+1})$ and so

$$\begin{aligned} {\mathcal {A}}=coh^s(A^{n+1}) =coh(\{coh^s(A^{n}) \ | \ A^{n}\in A^{n+1}\}\cup \{sup^s(A^{n+1})\}^{n} ). \end{aligned}$$

Let ${\mathcal {X}}\in {\mathcal {A}}$ be arbitrary. There exist $\lambda _1,\lambda _2,\ldots ,\lambda _k\ge 0$ and $A^{n}_1,A^{n}_2,\ldots ,A^{n}_k\in A^{n+1}\cup \{\sup (A^{n+1})\}^{n}$ such that $\sum _{i=1}^k \lambda _i=1$ and ${\mathcal {X}}=\sum _{i=1}^k \lambda _i coh^s(A^{n}_i).$ On the other hand, $\lambda _i coh^s(A^{n}_i)\in {Conv_s}^{n}({\mathcal {P}})$, for all $i=1,2,\ldots ,k$. Hence

$$\begin{aligned} {\mathcal {X}}=\sum _{i=1}^k \lambda _i coh^s(A^{n}_i) \in {Conv_s}^{n}({\mathcal {P}}), \end{aligned}$$

and then ${\mathcal {A}}\in Canv^{n+1}({\mathcal {P}})$.

Now, let ${\mathcal {A}},{\mathcal {B}}\in {Conv_s}^{n+1}({\mathcal {P}})$. Then for all ${\mathcal {X}}\in {\mathcal {A}}\subseteq Canv^{n}_s({\mathcal {P}})$ and ${\mathcal {Y}}\in {\mathcal {B}}\subseteq Canv^{n}_s({\mathcal {P}})$, we have ${\mathcal {X}}+{\mathcal {Y}}\in Canv^{n}_s({\mathcal {P}})$. By Proposition 3 and Remark 4, we have $\{\sup ^s({\mathcal {A}})\}^n\in {\mathcal {A}}$ and $\{\sup ^s({\mathcal {B}})\}^n\in {\mathcal {B}}$. Also, by Proposition 2, we have

$$\begin{aligned} \{sup^s({\mathcal {A}}+{\mathcal {B}})\}^{n+1}\in {\mathcal {A}}+{\mathcal {B}}, \end{aligned}$$

and then

$$\begin{aligned} {\mathcal {A}}+{\mathcal {B}} =coh(\{coh^s({\mathcal {Z}}) \ | \ {\mathcal {Z}}\in {\mathcal {A}}+{\mathcal {B}} \}\cup \{sup^s({\mathcal {A}}+{\mathcal {B}})\}^{n+1} ). \end{aligned}$$

Now, by considering the properties of sup and coh (convex hull of a set), we have $\alpha {\mathcal {A}}\in {Conv_s}^{n+1}({\mathcal {P}})$ for all $\alpha \ge 0$ and ${\mathcal {A}}\in {Conv_s}^{n+1}({\mathcal {P}})$. $\square$

Now, we characterize the elements of ${Conv_s}^n({\mathcal {P}})^*$. First we recall a theorem.

Theorem 2

([4], II, 2.9) Let ${\mathcal {Q}}$ be subcone of the locally convex cone $({\mathcal {P}}, {\mathcal {V}})$. Then every u-continuous linear functional on ${\mathcal {Q}}$ can be extended to a u-continuous linear functional on ${\mathcal {P}}$.

Theorem 3

If $({\mathcal {P}},{\mathcal {V}})$ is a $\bigvee$-semilattice locally convex cone, then for all $n\in {\mathbb {N}}$, $(Conv^n({\mathcal {P}}))^*$ and ${\mathcal {P}}^*$ coincide, in the sense that any vector of ${\mathcal {P}}^*$ has a unique extension to a vector of $(Conv^n({\mathcal {P}}))^*$ and conversely any vector $(Conv^n({\mathcal {P}}))^*$ can be restricted to a vector of ${\mathcal {P}}^*$.

Proof

By considering (1) we can embed ${\mathcal {P}}$ into ${Conv_s}^n({\mathcal {P}})$. It is easy to see that the restriction of each element of ${Conv_s}^n({\mathcal {P}})^*$ on ${\mathcal {P}}$ belongs to ${\mathcal {P}}^*$ and by Theorem 2, the extension of each element of ${\mathcal {P}}^*$ to ${Conv_s}^n({\mathcal {P}})$ is an element of ${Conv_s}^n({\mathcal {P}})^*$. So it is sufficient to show that each element of ${\mathcal {P}}^*$ has a unique extension in ${Conv_s}^n({\mathcal {P}})^*$. Let $\mu \in {\mathcal {P}}^*$. Define $({\bar{\mu }})^n$ as follows:

$$\begin{aligned} ({\bar{\mu }})^1(A):={\bar{\mu }}(A)=\sup \{\mu (a)\ |\ a\in A \} \quad ( A\in Conv_s({\mathcal {P}})), \end{aligned}$$

(3)

and

$$\begin{aligned} \begin{aligned} ({\bar{\mu }})^n(A^n):=\sup \{({\bar{\mu }})^{n-1} (A^{n-1})\ |\ A^{n-1}\in A^n\} \quad (A^n\in {Conv_s}^n({\mathcal {P}})), \end{aligned} \end{aligned}$$

(4)

for $n=2,3,\ldots$. By Lemma 2, the functional $({\bar{\mu }})^1$ is u-continuous and by repeating this process $({\bar{\mu }})^n$ is u-continuous too. We have $({\bar{\mu }})^1(A)=\mu (sup^s(A))$ and $({\bar{\mu }})^n(A^n)=({\bar{\mu }})^{n-1}(\{sup^s({A^n})\}^{n-1})$, since A contains $\sup A$. By Remark 4 and Proposition 2 the mapping $({\bar{\mu }})^n$ is an extension of $\mu$ to ${Conv_s}^n({\mathcal {P}})$. Let $\vartheta _n$ be another u-continuous extension of $\mu$ to ${Conv_s}^n({\mathcal {P}})$ (which exists by Theorem 2). We show that $\vartheta _n={\bar{\mu }}^n$.

Let $A^n \in Conv^n_s ({\mathcal {P}})$. Since $A^n\preceq \{ {sup^s(A^n)}\}^n$ and $\{ {sup^s(A^n)}\}^n\preceq A^n$, then $\vartheta _n(A^n)\le \vartheta _n(\{ {sup^s(A^n)}\}^n)$ and $\vartheta _n(\{ {sup^s(A^n)}\}^n)\le \vartheta _n(A_n)$ and so

$$\begin{aligned} \vartheta _n(A^n)=\vartheta _n(\{ {sup^s(A^n)}\}^n)=\mu (\{ {sup^s(A^n)}\}^n)=({\bar{\mu }})^n(\{ {sup^s(A^n)}\}^n)=({\bar{\mu }})^n(A^n) . \end{aligned}$$

This completes the proof. $\square$

In the following example we consider the locally convex cone $\overline{{\mathbb {R}}}$ and we characterize all elements of the dual of the locally convex cone $(Conv^n(\overline{{\mathbb {R}}}),{\overline{{\mathcal {V}}}}^n)$, where ${\mathcal {V}}=\{\epsilon >0\ |\ \epsilon \in {\mathbb {R}} \}$.

Example 4

We know that $\overline{{\mathbb {R}}}$ is a $\bigvee -$semilattice locally convex cone. It is easy to see that

$$\begin{aligned} Conv_s(\overline{{\mathbb {R}}})&=\{[a,b],(c,d], (-\infty ,d], \{e\}, A\cup \{+\infty \} \ | \ A\in Conv({\mathbb {R}}),\\ {}&\ \ \ \ \ \ a,b,c,d,e \in \overline{{\mathbb {R}}} \text{ with } a<b \text{ and } c<d \}\\&=Conv(\overline{{\mathbb {R}}}){\setminus }\{(a,b), (-\infty ,b), [c,d) \ | \ a,b,c,d\in \overline{{\mathbb {R}}}\}. \end{aligned}$$

According to Theorem 3, $(Conv^n({\mathbb {R}}))^*$ and ${\mathbb {R}}^*$ coincide, in the sense that any vector of ${\mathbb {R}}^*$ has a unique extension to a vector of $(Conv^n({\mathbb {R}}))^*$ and conversely any vector $(Conv^n({\mathbb {R}}))^*$ can be restricted to a vector of ${\mathbb {R}}^*$ for all $n\in {\mathbb {N}}$.

Since $\Omega (\overline{{\mathbb {R}}})=\overline{{\mathbb {R}}}^* {\setminus } \{0_\infty \}={\mathbb {R}}^*$, every element of ${\mathbb {R}}^*$ has a unique extension in $(Conv^n(\overline{{\mathbb {R}}}))^*$ by Corollary 1. The element $0_\infty$ violates the $\Omega$ condition at just one point $+\infty$. So two different extensions $\overline{ 0_\infty }(A)$ and $\overline{\overline{0_\infty }}$ can be written for it in $Conv (\overline{{\mathbb {R}}})^*$ as the following:

$$\begin{aligned} \overline{ 0_\infty }(A)&=\sup \{0_\infty (a)|a\in A\}=0,\\ \overline{\overline{0_\infty }}(A)&= 0_\infty (\sup A)=0, \end{aligned}$$

for all $A \in Conv(\overline{{\mathbb {R}}})$ which $\sup (A)\ne +\infty$,

$$\begin{aligned} \overline{ 0_\infty }(A)&=\sup \{0_\infty (a)|a\in A\}=+\infty ,\\ \overline{\overline{0_\infty }}(A)&= 0_\infty (\sup A)=+\infty , \end{aligned}$$

for $A\in Conv(\overline{{\mathbb {R}}})$ with $+\infty \in A$ and

$$\begin{aligned}&\overline{0_\infty }(A)=\sup \{0_\infty (a)|a\in A\}=0,\\&\overline{\overline{0_\infty }}(A)=0_\infty (\sup A)=0_\infty (\infty )=+\infty , \end{aligned}$$

for all $A\in {\mathcal {Q}}$, where ${\mathcal {Q}}:=\{ A\in Conv(\overline{{\mathbb {R}}}) \ | \ \sup (A)= +\infty \text{ and } +\infty \notin A\}$. Let $\gamma$ be another extension of $0_\infty$ to $Conv(\overline{{\mathbb {R}}})$. Then $\gamma (A)=\overline{ 0_\infty }(A)=\overline{\overline{0_\infty }}(A)=0$ for all $A \in Conv(\overline{{\mathbb {R}}})$ which $\sup (A)\ne +\infty$ and $\gamma (A)=\overline{ 0_\infty }(A)=\overline{\overline{0_\infty }}(A)=+\infty$ for $A\in Conv(\overline{{\mathbb {R}}})$ with $+\infty \in A$, by Theorem 3. Now, let $A,B\in {\mathcal {Q}}$. It is easy to see that $A\preceq B$ and $B\preceq A$ and then $\gamma (A)=\gamma (B)$. In particular, $\gamma (A)=\gamma (\alpha A)=\alpha \gamma (A)$ since $\alpha A\in {\mathcal {Q}}$ for all positive reals $\alpha$. By the above consideration $\gamma =\overline{ 0_\infty }=0$ or $\gamma =\overline{\overline{0_\infty }} =+\infty$ on ${\mathcal {Q}}$. Therefore $\overline{ 0_\infty }$ and $\overline{\overline{0_\infty }}$ are only extensions of $0_\infty$ on $Conv (\overline{{\mathbb {R}}})$. This yields that $(Conv (\overline{{\mathbb {R}}}))^*{\setminus } \{\overline{ 0_\infty },\overline{\overline{0_\infty }}\}$ and $\overline{{\mathbb {R}}}^*$ coincide.

Now, we show that the extensions of the mappings $\overline{ 0_\infty }$ and $\overline{\overline{0_\infty }}$ to the cone $Conv^n(\overline{{\mathbb {R}}})$ are unique: Let $\overline{ 0_\infty }^n$ and $\overline{\overline{0_\infty }}^n$ be the extensions of $\overline{ 0_\infty }$ and $\overline{\overline{0_\infty }}$ on $Conv^n(\overline{{\mathbb {R}}})$, respectively. Let $A\in Conv^n(\overline{{\mathbb {R}}}){\setminus } Conv^n({{\mathbb {R}}})$. Then $\{+\infty \}^n \preceq A$ and $A \preceq \{+\infty \}^n$. These yield that

$$\begin{aligned} \overline{ 0_\infty }^n(A)= & {} \overline{ 0_\infty }^n(\{\infty \}^n)= 0_\infty (+\infty )=+\infty ,\\ \overline{\overline{0_\infty }}^n(A)= & {} \overline{\overline{0_\infty }}^n(\{+\infty \}^n)= 0_\infty (+\infty )=+\infty . \end{aligned}$$

On the other hand, if $A\in Conv^n({{\mathbb {R}}})$, then $A\preceq \{(0,+\infty )\}^{n-1}$ and so $\overline{ 0_\infty }^n(A)\le 0$. Also there exists $a\in \mathbb {R}$ such that $\{a\}^n\preceq A$. Then $0=\overline{ 0_\infty }^n(\{a\}^n)\le \overline{ 0_\infty }^n(A)$. We conclude that $\overline{ 0_\infty }^n(A)=0$ for all $A\in Conv^n({{\mathbb {R}}})$.

If there is $b\in \mathbb {R}$ such that $A\preceq \{b\}^n$, then $\overline{\overline{0_\infty }}(A)\le 0$ and so $\overline{\overline{0_\infty }}(A)= 0$ by the similar way which applied for $\overline{ 0_\infty }^n(A)$. Otherwise $\{b\}^n\preceq A$ for all $b\in {\mathbb {R}}$. Then $\{(0,+\infty )\}^{n-1}\preceq A$ and so $+\infty =\overline{\overline{0_\infty }}(\{(0,+\infty )\}^{n-1})\le \overline{\overline{0_\infty }}(A)$. This yields that $\overline{\overline{0_\infty }}(A)= + \infty$. We conclude that the elements of $(Conv^n(\overline{{\mathbb {R}}}))^*$ are all non-negative reals, $\overline{ 0_\infty }^n$ and $\overline{\overline{0_\infty }}^n$ for all $n\in {\mathbb {N}}$. Also we have showed that the cones $(Conv(\overline{{\mathbb {R}}}))^*$ and $(Conv^n(\overline{{\mathbb {R}}}))^*$ coincide.

We conclude that $(Conv^n (\overline{{\mathbb {R}}}))^*{\setminus } \{\overline{ 0_\infty }^n,\overline{\overline{0_\infty }}^n\}$ and $\overline{{\mathbb {R}}}^*$ coincide.