Using the Support Functions to Embed the Families of Fuzzy Sets into Banach Spaces

Abstract

With the help of support functions of fuzzy sets, we are able to embed some interesting families of fuzzy sets into Banach spaces. It is better to make the embedded Banach spaces as small as possible, which can be achieved by considering the weak* topology. In this case, the weak* continuity of support functions should be studied. The concept of a maximal embedding behavior is also presented in this paper, which shows the maximal subset satisfying the completeness.

1. Introduction

Let $(U,\Vert ·\Vert )$ be a normed space. We denote by $\mathcal{F}\left(U\right)$ the family of all fuzzy sets in U. Given any $\tilde{A},\tilde{B}\in \mathcal{F}\left(U\right)$ and $\lambda \in \mathbb{R}$, the addition $\tilde{A}\oplus \tilde{B}$ and scalar multiplication $\lambda \tilde{A}$ are well known. Under these operations, the space $\mathcal{F}\left(U\right)$ cannot be a vector space. The main reason is that the additive inverse does not exist. For the purpose of applications, we expect to treat $\mathcal{F}\left(U\right)$ as a subset of a vector space. Since we have the difficulty of treating the whole family $\mathcal{F}\left(U\right)$ as a subset of a normed space, in this paper, we are going to present some specific subfamilies of $\mathcal{F}\left(U\right)$ which can be treated as the closed convex cone in some Banach spaces.

We denote by $\mathcal{P}\left(U\right)$ the family of all nonempty subsets of a normed space $(U,\Vert ·\Vert )$. Given any $A,B\in \mathcal{P}\left(U\right)$, the Hausdorff metric of A and B is defined by

$$d_H(A,B)=max\left\{\underset{a\in A}{sup}\underset{b\in B}{inf}\Vert a-b\Vert ,\underset{b\in B}{sup}\underset{a\in A}{inf}\Vert a-b\Vert \right\},$$

where $d_H(A,B)$ is an extended real number. By convention, we define

$$d_H(\varnothing ,\varnothing )=0 \mathrm{and} d_H(A,\varnothing )=\infty \mathrm{for} A\ne \varnothing .$$

We say that $\tilde{A}$ is a fuzzy set in a normed space $(U,\Vert ·{\Vert }_U)$ when a membership function ${\xi }_{\tilde{A}}:U\to [0,1]$ is associated with $\tilde{A}$. For $\alpha \in (0,1]$, the $\alpha$-level sets ${\tilde{A}}_{\alpha }$ of $\tilde{A}$ are defined by

$${\tilde{A}}_{\alpha }=\left\{x\in U:{\xi }_{\tilde{A}}\left(x\right)\ge \alpha \right\}.$$

For $\alpha \in [0,1)$, the strong $\alpha$-level sets ${\tilde{A}}_{\alpha +}$ of $\tilde{A}$ are defined by

$${\tilde{A}}_{\alpha +}=\left\{x\in U:{\xi }_{\tilde{A}}\left(x\right)>\alpha \right\}.$$

Then, we have

$${\tilde{A}}_{\alpha +}=\bigcup _{\beta \in (\alpha ,1]}{\tilde{A}}_{\beta } \mathrm{for} \alpha \in [0,1).$$

The support ${\tilde{A}}_{0+}$ of $\tilde{A}$ is given by

$${\tilde{A}}_{0+}=\left\{x\in U:{\xi }_{\tilde{A}}\left(x\right)>0\right\}.$$

In this case, we can define the 0-level set of $\tilde{A}$ by ${\tilde{A}}_0=cl\left({\tilde{A}}_{0+}\right)$, which means that the 0-level set of $\tilde{A}$ is the closure of the support of $\tilde{A}$.

We consider the following function

$${\eta }_{\tilde{A}}:[0,1]\to (\mathcal{P}\left(U\right),d_H) \mathrm{defined}\mathrm{by} \alpha \mapsto {\tilde{A}}_{\alpha },$$

where ${\tilde{A}}_{\alpha }$ denotes the $\alpha$-level set of $\tilde{A}$ for $\alpha \in [0,1]$. We say that ${\eta }_{\tilde{A}}$ is continuous with respect to the Hausdorff metric $d_H$ when, given any $ϵ>0$, there exists $\delta >0$ such that $|\alpha -\beta |<\delta$ implies $d_H({\tilde{A}}_{\alpha },{\tilde{A}}_{\beta })<ϵ$. Then, we adopt the following notations.

• We denote by ${\mathcal{F}}_{kc}\left(U\right)$ the family of all fuzzy sets in U with nonempty compact and convex $\alpha$-level sets for $\alpha \in [0,1]$.
• We denote by ${\mathcal{F}}_{kc}^C\left(U\right)$ the family of elements $\tilde{A}$ in ${\mathcal{F}}_{kc}\left(U\right)$ such that the function ${\eta }_{\tilde{A}}$ is continuous.

In this paper, we are going to use the support functions of fuzzy sets to embed the families ${\mathcal{F}}_{kc}\left(U\right)$ and ${\mathcal{F}}_{kc}^C\left(U\right)$ into some useful Banach spaces.

Example 1.

We say that $\tilde{A}$ is a fuzzy interval when its α-level sets are bounded closed intervals given by

$${\tilde{A}}_{\alpha }=\left[{\tilde{A}}_{\alpha }^L,{\tilde{A}}_{\alpha }^U\right] for all \alpha \in [0,1].$$

It is clear to see that the bounded closed interval ${\tilde{A}}_{\alpha }$ is a compact and convex set in $\mathbb{R}$, which means $\tilde{A}\in {\mathcal{F}}_{kc}\left(\mathbb{R}\right)$. We also have

$$\begin{array}{cc}\hfill d_H\left({\tilde{A}}_{\alpha },{\tilde{A}}_{\beta }\right)& =max\left\{\underset{a\in {\tilde{A}}_{\alpha }}{sup}\underset{b\in {\tilde{B}}_{\alpha }}{inf}|a-b|,\underset{b\in {\tilde{B}}_{\alpha }}{sup}\underset{a\in {\tilde{A}}_{\alpha }}{inf}|a-b|\right\}\hfill \\ \hfill & =max\left\{\left|{\tilde{A}}_{\alpha }^L-{\tilde{B}}_{\alpha }^L\right|,\left|{\tilde{A}}_{\alpha }^U-{\tilde{B}}_{\alpha }^U\right|\right\}.\hfill \end{array}$$

We can define two real-valued functions l and u on $[0,1]$ by

$$l\left(\alpha \right)={\tilde{A}}_{\alpha }^L and u\left(\alpha \right)={\tilde{A}}_{\alpha }^U.$$

It is clear to see that, if the functions l and u are continuous on $[0,1]$, then the function

$${\eta }_{\tilde{A}}:[0,1]\to (\mathcal{P}\left(U\right),d_H) \mathit{defined}\mathit{by} \alpha \mapsto {\tilde{A}}_{\alpha }=\left[{\tilde{A}}_{\alpha }^L,{\tilde{A}}_{\alpha }^U\right],$$

is also continuous on $[0,1]$, which means $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$.

In order to define the differentials of fuzzy functions, Puri and Ralescu [1] developed an embedding theorem to embed the space ${\mathcal{F}}_{kc}\left(U\right)$ into a normed space by following the approach of Rådström [2]. Also, Kaleva [3,4] and Román-Flores and Rojas-Medar [5] established the different kinds of embedding theorems to study the fuzzy differential equations. On the other hand, the embedding theorem is also applicable to study the fuzzy random variables by referring to Puri and Ralescu [6,7] and Klement et al. [8]. Recently, Wu [9] considered the equivalence classes of Cauchy sequences to study the embedding theorems of some interesting families of fuzzy sets.

To consider the families of fuzzy sets in a normed space as a subset of some Banach spaces is an important issue in applications. There are two ways to study the embedding theorems for some specific families of fuzzy sets in the literature. One is directly constructing some suitable Banach spaces such that the specific families of fuzzy sets can be isometrically and isomorphically embedded into the corresponding Banach space. Another approach is to use the support functions of fuzzy sets. In this case, the continuities of support functions should be studied under some suitable topologies. The weak* topology of the closed unit ball in the dual space of a normed space is considered in this paper for the purpose of constructing smaller Banach spaces. It is better to make the embedded Banach spaces as small as possible, which can be achieved by considering the weak* topology. The concept of a maximal embedding behavior is also presented in this paper. It means that the maximal subset satisfying the completeness in the sense that every Cauchy sequence is convergent.

The support functions of fuzzy sets are defined using the concept of supremum. Considering the supremum is equivalent to considering optimization problems. Solving the optimization problem is not an easy task. However, when the optimization problems have some elegant mathematical structures, the well-known techniques developed in the topic of optimization are useful for solving these well-structured optimization problems. Since the support functions of fuzzy sets own the elegant mathematical structures, studying the embedding theorems using the support functions of fuzzy sets is still worthy for real applications. Diamond and Kloeden [10,11], Ma [12], Puri and Ralescu [6,13], Román-Flores and Rojas-Medar [5], Wu and Ma [14,15,16], and Wu [17] presented the embedding theorems using the support functions of fuzzy sets. Those approaches were mostly based on the finite-dimensional Euclidean space. In this paper, we are going to study the embedding theorems based on the infinite-dimensional normed space. Although we expect to embed the families of fuzzy sets into some Banach spaces, it will be better to make these Banach spaces as small as possible. The advantage of this paper is to consider the weak* topology for the normed dual space of normed space, which can make the embedded Banach spaces smaller than that of the existing Banach spaces presented in the literature.

In Section 2, the dual space of normed space is considered. In order to make the embedded Banach space as small as possible, the weak* topology of closed unit ball in the dual space is discussed. Since the embedded Banach spaces studied in this paper consist of continuous functions, many spaces of continuous functions are studied in Section 3. The continuities of support functions of fuzzy sets with respect to the weak* topology are studied in Section 4. In Section 5, many embedding theorems of the families ${\mathcal{F}}_{kc}\left(U\right)$ and ${\mathcal{F}}_{kc}^C\left(U\right)$ are presented in which the embedded Banach spaces involve the weak* topology such that they can be smaller than that of the existing Banach space presented in the existing literature. Also, the concept of maximal embedding behavior is provided.

2. Topologies of the Closed Unit Ball in Norm Dual Space

Let $(U,\Vert ·{\Vert }_U)$ be a normed space. The closed unit ball and unit sphere are defined by

$$\overline{S}={\{x:\Vert x\Vert }_U\le 1\} \mathrm{and} \widehat{S}={\{x:\Vert x\Vert }_U=1\}$$

respectively. The topology ${\tau }_U^{\left(s\right)}$ induced by the norm of U is called the norm topology or strong topology. In this case, we have a topological space $(U,{\tau }_U^{\left(s\right)})$. Sometimes, we also write

$${(U,\Vert ·\Vert }_U)=(U,{\tau }_U^{\left(s\right)}).$$

The dual space $U^{*}$ is the family of all linear functionals defined on U such that they are continuous with respect to the norm topology. The weak topology ${\tau }_U^{\left(w\right)}$ for U is the weakest (coarsest) topology for U such that all linear functionals in $U^{*}$ are continuous.

Given any $x^{*}\in U^{*}$ with $x^{*}:U\to \mathbb{R}$, we write the functional value $x^{*}\left(x\right)=\langle x^{*},x\rangle \in \mathbb{R}$ for $x\in U$ and define

$$\Vert x^{*}{\Vert }_{U^{*}}=\underset{\Vert x\Vert =1}{sup}|\langle x^{*},x\rangle |=\underset{\Vert x\Vert \le 1}{sup}\left|\langle x^{*},x\rangle \right|.$$

Suppose that $\langle x^{*},x_0\rangle \le 0$ with $\Vert x_0\Vert \le 1$. Then, we take $x_1=-x_0$, which says $\Vert x_1\Vert \le 1$ and $\langle x^{*},x_1\rangle \ge 0$. Therefore, we obtain

$$\Vert x^{*}{\Vert }_{U^{*}}=\underset{\Vert x\Vert \le 1}{sup}|\langle x^{*},x\rangle |=\underset{\Vert x\Vert \le 1}{sup}\langle x^{*},x\rangle .$$

It is well known that $(U^{*},\Vert ·{\Vert }_{U^{*}})$ is also a normed space.

The norm topology ${\tau }_{U^{*}}^{\left(s\right)}$ for $U^{*}$ induced by the norm ${\Vert ·\Vert }_{U^{*}}$ is also called a strong topology. The weak topology ${\tau }_{U^{*}}^{\left(w\right)}$ for $U^{*}$ is the weakest (coarsest) topology for $U^{*}$ such that all linear functionals in $U^{**}$ are continuous. Let $\varphi$ be a natural embedding of U into $U^{**}$. Then, we have $\varphi \left(U\right)\subset U^{**}$. We say that the normed space U is reflexive when $\varphi \left(U\right)=U^{**}$. The weak* topology ${\tau }_{U^{*}}^{(w*)}$ for $U^{*}$ is the weakest topology for $U^{*}$ such that all linear functionals in $\varphi \left(U\right)$ are continuous. The weak* topology for $U^{*}$ is weaker than the weak topology for $U^{*}$. Therefore, we have that a weak* closed subset of $U^{*}$ is weakly closed, and that the weak convergence implies weak* convergence. When U is reflexive, the weak and weak* topologies for $U^{*}$ coincide.

The closed unit ball

$${\overline{S}}^{*}=\{x^{*}\in U^{*}:\Vert x^{*}{\Vert }_{U^{*}}\le 1\}$$

is not necessarily a compact subset of $U^{*}$ with respect to the norm topology for $U^{*}$. However, ${\overline{S}}^{*}$ is a weak* compact subset of $U^{*}$ by the Alaoglu’s theorem, which says that ${\overline{S}}^{*}$ is compact with respect to the weak* topology ${\tau }_{U^{*}}^{(w*)}$ for $U^{*}$. In this case, we can induce a topology ${\tau }_{{\overline{S}}^{*}}^{(w*)}$ for ${\overline{S}}^{*}$ such that $({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)})$ is a compact space, where, for each ${\overline{O}}^{*}\in {\tau }_{{\overline{S}}^{*}}^{(w*)}$, there exists $O^{*}\in {\tau }_{U^{*}}^{(w*)}$ satisfying ${\overline{O}}^{*}=O^{*}\cap {\overline{S}}^{*}$. However, the closed unit sphere

$${\widehat{S}}^{*}=\{x^{*}\in U^{*}:\Vert x^{*}{\Vert }_{U^{*}}=1\}$$

does not need to be weak* compact, since it is not even weak* closed (Aliprantis and Border [18], p. 250).

The closed unit sphere ${\widehat{S}}^{*}$ cannot be a vector subspace of $U^{*}$, since ${\widehat{S}}^{*}$ is not closed under the scalar multiplication. This says that we cannot define a norm on ${\widehat{S}}^{*}$. However, we can define a metric on ${\widehat{S}}^{*}$ by

$$d_{{\widehat{S}}^{*}}^{\left(s\right)}(x^{*},y^{*})={\Vert x^{*}-y^{*}\Vert }_{U^{*}}.$$

(1)

Therefore, we can form a metric space $({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)})$ such that it can also induce a metric topology ${\tau }_{{\widehat{S}}^{*}}^{\left(s\right)}$.

A topological space $(X,\tau )$ is metrizable when the topology $\tau$ can be generated by some metric. By considering the closed unit ball ${\overline{S}}^{*}$, we can also form a metric space $({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{\left(s\right)})$ such that it can induce a metric topology ${\tau }_{{\overline{S}}^{*}}^{\left(s\right)}$, where the metric $d_{{\overline{S}}^{*}}^{\left(s\right)}$ is also defined in the way of (1). Let ${\tau }_{U^{*}}^{\left(s\right)}$ be the norm topology for $U^{*}$. Then, we can also induce a topological subspace $({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}})$ of $U^{*}$, where ${\overline{O}}^{*}\in {\tau }_{{\overline{S}}^{*}}$ means ${\overline{O}}^{*}=O^{*}\cap {\overline{S}}^{*}$ for some $O^{*}\in {\tau }_{U^{*}}^{\left(s\right)}$. Using the routine argument, we can show that ${\tau }_{{\overline{S}}^{*}}^{\left(s\right)}={\tau }_{{\overline{S}}^{*}}$, which also means that ${\tau }_{{\overline{S}}^{*}}$ is metrizable with the metric $d_{{\overline{S}}^{*}}^{\left(s\right)}$ defined in the way of (1). The following lemma can be obtained by using the routine argument.

Lemma 1.

We have the following properties.

(i)

We have the equality ${\tau }_{{\overline{S}}^{*}}^{\left(s\right)}={\tau }_{{\overline{S}}^{*}}$ and the inclusion ${\tau }_{{\overline{S}}^{*}}^{(w*)}\subseteq {\tau }_{{\overline{S}}^{*}}^{\left(s\right)}$.

(ii)

Let Y be a topological space. Suppose that the function $f:{\overline{S}}^{*}\to Y$ is ${\tau }_{{\overline{S}}^{*}}^{(w*)}$-continuous on ${\overline{S}}^{*}$. Then, it is also ${\tau }_{{\overline{S}}^{*}}^{\left(s\right)}$-continuous on ${\overline{S}}^{*}$.

Lemma 2

(Aliprantis and Border [18], p. 254). Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. The closed unit ball ${\overline{S}}^{*}$ in $U^{*}$ is weak*-metrizable if and only if U is separable.

Lemma 2 says that, if the normed space $(U,\Vert ·\Vert )$ is separable, then there exists a metric $d_{{\overline{S}}^{*}}^{(w*)}$ defined on ${\overline{S}}^{*}$ such that the compact space $({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)})$ can be induced by the metric $d_{{\overline{S}}^{*}}^{(w*)}$, i.e., the metric topology induced by the metric $d_{{\overline{S}}^{*}}^{(w*)}$ coincides with the topology ${\tau }_{{\overline{S}}^{*}}^{(w*)}$. More precisely, the metric can be taken as

$$d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{2^n}}·\left|\langle x^{*},x_n\rangle -\langle y^{*},x_n\rangle \right|=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{2^n}}·\left|\langle x^{*}-y^{*},x_n\rangle \right|,$$

(2)

where $\{x_1,x_2,\cdots \}$ is a countable dense subset of the closed unit ball in U and $x^{*},y^{*}\in U^{*}$. Since $\Vert x_n\Vert \le 1$ for all n, we have

$$|\langle x^{*}-y^{*},x_n\rangle | \le \underset{\Vert x\Vert \le 1}{sup}|\langle x^{*}-y^{*},x\rangle |=\Vert x^{*}-y^{*}\Vert ,$$

which implies that

$$d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{2^n}}·\left|\langle x^{*}-y^{*},x_n\rangle \right|\le \sum _{n=1}^{\infty }{\displaystyle \frac{1}{2^n}}·\Vert x^{*}-y^{*}\Vert =\Vert x^{*}-y^{*}\Vert =d_{{\overline{S}}^{*}}^{\left(s\right)}(x^{*},y^{*}).$$

3. The Spaces of Continuous Functions

We denote by $C(X,Y)$ the family of all continuous functions from the topological space X to the topological space Y. When Y is taken to be a normed space, given any $f,g\in C(X,Y)$ and $\lambda \in \mathbb{R}$, the vector addition $f+g$ and scalar multiplication $\lambda f$ are defined by

$$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right) \mathrm{and} \left(\lambda f\right)\left(x\right)=\lambda f\left(x\right).$$

Then, $C(X,Y)$ is a vector space. For each $f\in C(X,Y)$, we define

$${\Vert f\Vert }_C=\underset{x\in X}{sup}{\Vert f\left(x\right)\Vert }_Y.$$

We also denote by $C_b(X,Y)$ the space of all continuous functions $f:X\to Y$ satisfying $\Vert f\Vert <+\infty$. The following results can be obtained by the routine argument.

Proposition 1.

Let $(Y,\Vert ·{\Vert }_Y)$ be a normed space. The spaces $C(X,Y)$ and $C_b(X,Y)$ with the norm

$${\Vert f\Vert }_C=\underset{x\in X}{sup}{\Vert f\left(x\right)\Vert }_Y$$

are normed spaces. We further assume that Y is a Banach space. Then, $C(X,Y)$ and $C_b(X,Y)$ are also Banach spaces.

Let X and Y be two topological spaces. We denote by $D\left(\right[0,1]\times X,Y)$ the space of all functions from $[0,1]\times X$ into Y such that the following conditions are satisfied.

• Given any fixed $\alpha \in [0,1]$, the function $f(\alpha ,·)$ is continuous on X.
• Given any fixed $x\in X$, the function $f(·,x)$ is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any $\alpha \in (0,1)$.

Let $(Y,\Vert ·{\Vert }_Y)$ be a normed space. Given any $f\in D\left(\right[0,1]\times X,Y)$, we define

$${\Vert f\Vert }_D=\underset{(\alpha ,x)\in [0,1]\times X}{sup}{\Vert f(\alpha ,x)\Vert }_Y.$$

The following results can be obtained by the routine argument.

Proposition 2.

Let $(Y,\Vert ·{\Vert }_Y)$ be a normed space. The space $D\left(\right[0,1]\times X,Y)$ with the norm

$${\Vert f\Vert }_D=\underset{(\alpha ,x)\in [0,1]\times X}{sup}{\Vert f(\alpha ,x)\Vert }_Y$$

is a normed space. We further assume that Y is a Banach space. Then, $D\left(\right[0,1]\times X,Y)$ is also a Banach space.

Let X and Y be two topological spaces.

• We denote by $\widehat{C}([0,1],C(X,Y))$ the space of all functions

  $$F:[0,1]\to \left({C(X,Y),\Vert ·\Vert }_C\right)$$

  such that F is left-continuous on $(0,1]$, right-continuous at 0 and has the right limit at any $\alpha \in (0,1)$.
• We denote by $C\left(\right[0,1],C(X,Y\left)\right)$ the space of all functions

  $$F:[0,1]\to \left({C(X,Y),\Vert ·\Vert }_C\right)$$

  such that F is continuous on $[0,1]$.

Let $(Y,\Vert ·{\Vert }_Y)$ be a normed space. Given any $F\in \widehat{C}([0,1],C(X,Y))$ or $F\in C\left(\right[0,1],C(X,Y\left)\right)$, we define

$$\Vert F\Vert =\underset{\alpha \in [0,1]}{sup}{\Vert F\left(t\right)\Vert }_C.$$

Then, we have

$$\Vert F\Vert =\underset{\alpha \in [0,1]}{sup}\underset{x\in X}{sup}{\Vert F\left(t\right)\left(x\right)\Vert }_Y.$$

The following results can be obtained by the routine argument.

Proposition 3.

Let $(Y,\Vert ·{\Vert }_Y)$ be a normed space. The spaces

$$\widehat{C}\left([0,1],C\left(X,Y\right)\right) and C\left([0,1],C\left(X,Y\right)\right)$$

with the norm

$$\Vert F\Vert =\underset{\alpha \in [0,1]}{sup}{\Vert F\left(t\right)\Vert }_C=\underset{\alpha \in [0,1]}{sup}\underset{x\in X}{sup}{\Vert F\left(t\right)\left(x\right)\Vert }_Y$$

are Banach spaces.

In Proposition 1, the space X is a general topological space. When X is a compact space, we have the following useful results.

Proposition 4

(Aliprantis and Border [18], pp. 119–120). Let X be a compact space, and let $(Y,d_Y)$ be a metric space. Given any $f,g\in C(X,Y)$, we define

$$d_C(f,g)=\underset{x\in X}{sup}{d}_Y(f\left(x\right),g\left(x\right)).$$

Then, we have the following properties.

(i)

The metric space $(C(X,Y),d_C)$ is complete if and only if the metric space $(Y,d)$ is complete.

(ii)

Suppose that X is metrizable and Y is separable. Then, the metric space $(C(X,Y),d_C)$ is separable.

If $Y=\mathbb{R}$, we simply write $C\left(X\right)=C(X,\mathbb{R})$. Then, $C_b\left(X\right)$ is the space of all continuous function $f:X\to \mathbb{R}$ satisfying

$$\Vert f\Vert =\underset{x\in X}{sup}\left|f\left(x\right)\right|<\infty .$$

We denote by $C_0\left(X\right)$ the space of all continuous function $f:X\to \mathbb{R}$ such that, for all $ϵ>0$, the set

$$\{x\in X:|f\left(x\right)|\ge ϵ\}$$

is compact. We denote by $C_c\left(X\right)$ the space of all continuous function $f:X\to \mathbb{R}$ such that the support

$$cl\left(\{x\in X:f(x)\ne 0\}\right)$$

is compact.

Proposition 5

(Conway [19] pp. 65–67). Let X be a locally compact space. Then, we have $C_c\left(X\right)$ is dense in $C_0\left(X\right)$, and that $C_0\left(X\right)$ is a closed subspace of $C_b\left(X\right)$. It also says that $C_0\left(X\right)$ is a Banach space. Moreover, when X is compact, we have

$$C_0\left(X\right)=C_b\left(X\right)=C\left(X\right).$$

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. The spaces of continuous functions defined on the closed unit ball ${\overline{S}}^{*}$ and the unit sphere ${\widehat{S}}^{*}$ in the dual space $U^{*}$ will be used to study the embedding theorems for fuzzy sets. Recall

$$C\left({\widehat{S}}^{*},{\tau }_{{\widehat{S}}^{*}}^{\left(s\right)}\right)=C\left({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)}\right),$$

which is the space of all continuous real-valued functions $f:({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)})\to \mathbb{R}$, where $d_{{\widehat{S}}^{*}}^{\left(s\right)}$ is given in (1) and ${\tau }_{{\widehat{S}}^{*}}^{\left(s\right)}$ is the metric topology induced by the metric $d_{{\widehat{S}}^{*}}^{\left(s\right)}$. Let $C_b({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)})$ denote the space of all ${\tau }_{{\widehat{S}}^{*}}^{\left(s\right)}$-continuous real-valued function

$$f:\left({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)}\right)\to \mathbb{R} \mathrm{satisfying} \Vert f\Vert =\underset{x^{*}\in {\widehat{S}}^{*}}{sup}\left|f\left(x^{*}\right)\right|<\infty .$$

Proposition 1 says that

$$\left(C\left({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right) \mathrm{and} \left(C_b\left({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right)$$

are Banach spaces.

The Alaoglu’s theorem says that $({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)})$ is a compact space. Assume that the normed space $(U,\Vert ·{\Vert }_U)$ is separable. Then, using Lemma 2, we have

$$C\left({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)}\right)=C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),$$

which is the space of all continuous real-valued functions $f:({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)})\to \mathbb{R}$, where $d_{{\overline{S}}^{*}}^{(w*)}$ is given in (2) such that the metric topology induced by the metric $d_{{\overline{S}}^{*}}^{(w*)}$ coincides with the topology ${\tau }_{{\overline{S}}^{*}}^{(w*)}$. Let $C_b({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)})$ denote the space of all ${\tau }_{{\overline{S}}^{*}}^{(w*)}$-continuous real-valued functions

$$f:\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\to \mathbb{R} \mathrm{satisfying} \Vert f\Vert =\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|f\left(x^{*}\right)\right|<\infty .$$

Proposition 6.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space. Regarding the closed unit ball ${\overline{S}}^{*}$ in $U^{*}$, we have that

$$\begin{array}{cc}\hfill \left(C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)& =\left(C\left({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)\hfill \\ & =\left(C_b\left({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)=\left(C_b\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)\hfill \end{array}$$

is a separable Banach space.

Proof. 

Proposition 1 says that

$$\left(C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right) \mathrm{and} \left(C_b\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

are Banach spaces. Since $({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)})$ is a compact metric space by Lemma 2, part (iii) of Proposition 4 says that $(C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}),\Vert ·\Vert )$ is separable. Part (ii) of Proposition 5 also says

$$\left(C\left({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)=\left(C_b\left({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right).$$

This completes the proof. □

4. Support Functions of Fuzzy Sets

Let $\mathcal{F}$ be a subfamily of $\mathcal{F}\left(U\right)$. In order to embed $\mathcal{F}$ into a normed space or Banach space. We need to guarantee that the space $\mathcal{F}$ is closed under the addition and scalar multiplication. In other words, given any $\tilde{A},\tilde{B}\in \mathcal{F}$ and any $\lambda \in \mathbb{R}$, we need to assure $\tilde{A}\oplus \tilde{B}\in \mathcal{F}$ and $\lambda \tilde{A}\in \mathcal{F}$. The following proposition provides some of these subfamilies.

Proposition 7.

We have the following properties:

(i)

Let U be a Hausdorff topological vector space. Then, the family ${\mathcal{F}}_{kc}\left(U\right)$ is closed under the addition and scalar multiplication.

(ii)

Let $(U,\Vert ·\Vert )$ be a normed space. Then, the family ${\mathcal{F}}_{kc}^C\left(U\right)$ is closed under the addition and scalar multiplication.

Moreover, given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_k\left(U\right)$, we have

$${\left(\tilde{A}\oplus \tilde{B}\right)}_{\alpha }={\tilde{A}}_{\alpha }+{\tilde{B}}_{\alpha } and {\left(\lambda \tilde{A}\right)}_{\alpha }=\lambda {\tilde{A}}_{\alpha }$$

(3)

for any $\lambda \in \mathbb{R}$ and for all $\alpha \in [0,1]$.

Let $(U,\Vert ·\Vert )$ be a normed space. Puri and Ralescu [1] considered the function ${\widehat{d}}_{\mathcal{F}}$ defined on $\mathcal{F}\left(U\right)\times \mathcal{F}\left(U\right)$ by

$${\widehat{d}}_{\mathcal{F}}\left(\tilde{A},\tilde{B}\right)=\underset{\alpha \in (0,1]}{sup}{d}_H\left({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha }\right).$$

In this paper, we define

$$d_{\mathcal{F}}\left(\tilde{A},\tilde{B}\right)d_{\mathcal{F}}=\underset{\alpha \in [0,1]}{sup}{d}_H\left({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha }\right),$$

where $\alpha =0$ is considered. Then, we have

$$\left({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}}\right)=\left({\mathcal{F}}_{kc}^C\left(U\right),{\widehat{d}}_{\mathcal{F}}\right).$$

Example 2.

Continued from Example 1, we assume that the functions l and u are continuous on $[0,1]$. This means that $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$. Therefore, we obtain

$$\left({\mathcal{F}}_{kc}^C\left(\mathbb{R}\right),d_{\mathcal{F}}\right)=\left({\mathcal{F}}_{kc}^C\left(\mathbb{R}\right),{\widehat{d}}_{\mathcal{F}}\right),$$

where

$$\begin{array}{cc}\hfill d_{\mathcal{F}}\left(\tilde{A},\tilde{B}\right)d_{\mathcal{F}}& =\underset{\alpha \in [0,1]}{sup}{d}_H\left({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha }\right)=\underset{\alpha \in (0,1]}{sup}{d}_H\left({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha }\right)={\widehat{d}}_{\mathcal{F}}\left(\tilde{A},\tilde{B}\right)\hfill \\ \hfill & =\underset{\alpha \in [0,1]}{sup}max\left\{\left|{\tilde{A}}_{\alpha }^L-{\tilde{B}}_{\alpha }^L\right|,\left|{\tilde{A}}_{\alpha }^U-{\tilde{B}}_{\alpha }^U\right|\right\}\hfill \\ \hfill & =\underset{\alpha \in (0,1]}{sup}max\left\{\left|{\tilde{A}}_{\alpha }^L-{\tilde{B}}_{\alpha }^L\right|,\left|{\tilde{A}}_{\alpha }^U-{\tilde{B}}_{\alpha }^U\right|\right\}.\hfill \end{array}$$

Proposition 8.

Let $(U,\Vert ·\Vert )$ be a Banach space. Then, the spaces $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ and $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ are complete metric spaces.

Let $(U,\tau )$ be a topological vector space, and let $U^{*}$ be the collection of all continuous linear functionals on U. Given a nonempty subset A of U, the function

$$s_A:U^{*}\to \mathbb{R} \mathrm{defined}\mathrm{by} s_A\left(x^{*}\right)=\underset{x\in A}{sup}{x}^{*}\left(x\right)=\underset{x\in A}{sup}\langle x^{*},x\rangle$$

(4)

is called the support function of A. Since each $x^{*}$ is also a bounded linear functional, given any bounded set A, we have $s_A\left(x^{*}\right)<\infty$ for any $x^{*}\in U^{*}$. Given any compact set A, the supremum in (4) is attained.

Example 3.

For $U=\mathbb{R}$, it follows that $U^{*}=\mathbb{R}$. In this case, we have ${\overline{S}}^{*}=[-1,1]$. Therefore, the support function $s_{\tilde{A}}$ is defined on $[0,1]\times [-1,1]$ and is given by

$$s_{\tilde{A}}(\alpha ,x)=\underset{y\in {\tilde{A}}_{\alpha }}{sup}\langle x,y\rangle =\underset{y\in {\tilde{A}}_{\alpha }}{sup}xy.$$

Let $\tilde{A}$ be a fuzzy interval. Then, the α-level set $\tilde{A}$ is a bounded closed interval given by

$${\tilde{A}}_{\alpha }=\left[{\tilde{A}}_{\alpha }^L,{\tilde{A}}_{\alpha }^U\right] \mathit{for} \mathit{all} \alpha \in [0,1].$$

In this case, we obtain

$$s_{\tilde{A}}(\alpha ,x)=\underset{y\in {\tilde{A}}_{\alpha }}{sup}xy=\underset{y\in [{\tilde{A}}_{\alpha }^L,{\tilde{A}}_{\alpha }^U]}{sup}xy=\left\{\begin{array}{cc}x·{\tilde{A}}_{\alpha }^U\hfill & ifx\in [0,1]\hfill \\ x·{\tilde{A}}_{\alpha }^L\hfill & ifx\in [-1,0].\hfill \end{array}\right.$$

(5)

The norm of $s_{\tilde{A}}$ is given by

$$\begin{array}{cc}\hfill \Vert s_{\tilde{A}}\Vert & =\underset{(\alpha ,x)\in [0,1]\times [-1,1]}{sup}\left|s_{\tilde{A}}(\alpha ,x)\right|\hfill \\ \hfill & =max\left\{\underset{(\alpha ,x)\in [0,1]\times [-1,0]}{sup}\left|x·{\tilde{A}}_{\alpha }^L\right|,\underset{(\alpha ,x)\in [0,1]\times [0,1]}{sup}\left|x·{\tilde{A}}_{\alpha }^U\right|\right\}.\hfill \end{array}$$

By referring to Hu and Papageorgiou [20] and Aliprantis and Border [18], the following basic results will be used.

Proposition 9.

We have the following properties:

(i)

Given any subsets A and B of U, we have

$$s_{A+B}\left(x^{*}\right)=s_A\left(x^{*}\right)+s_B\left(x^{*}\right) and s_{\lambda A}\left(x^{*}\right)=\lambda s_A\left(x^{*}\right)$$

for any $x^{*}\in U^{*}$ and $\lambda \ge 0$.

(ii)

Given any compact and convex subsets A and B in U, suppose that $s_A\left(x^{*}\right)=s_B\left(x^{*}\right)$ for all $x^{*}\in U^{*}$ with $\Vert x^{*}\Vert =1$. Then, we have $A=B$.

(iii)

Given any compact and convex subsets A and B in U, we have

$$d_H(A,B)=\underset{\Vert x^{*}\Vert \le 1}{sup}\left|s_A\left(x^{*}\right)-s_B\left(x^{*}\right)\right|=\underset{\Vert x^{*}\Vert =1}{sup}\left|s_A\left(x^{*}\right)-s_B\left(x^{*}\right)\right|.$$

Moreover, we have

$$d_H(hA,kA)\le |h-k|·\Vert A\Vert$$

for any $h,k\ge 0$.

(iv)

Given any subset A of U, we have

$$s_{\mathrm{cl}\left(A\right)}\left(x^{*}\right)=s_A\left(x^{*}\right)$$

for all $x^{*}\in U^{*}$.

(v)

Given any bounded set A in U, the support function $s_A$ satisfies the uniform Lipschitz condition, i.e.,

$$|s_A\left(x^{*}\right)-s_A\left(y^{*}\right)|\le \Vert A\Vert ·\Vert x^{*}-y^{*}\Vert$$

(6)

for all $x^{*},y^{*}\in U^{*}$. This also says that $s_A$ is continuous on $U^{*}$.

Proposition 10.

Let $(U,\Vert ·\Vert )$ be a separable normed space. Given any compact set A in U, the support function $s_A$ is weak*-continuous on ${\overline{S}}^{*}$.

Proof. 

According to part (iii) of Proposition 9, we just need to show that $s_A$ is weak*-upper semi-continuous on ${\overline{S}}^{*}$, i.e., we want to show that the set

$$\Gamma =\{x^{*}\in {\overline{S}}^{*}:s_A\left(x^{*}\right)\ge r\}$$

is weak*-closed for each $r\in \mathbb{R}$. Suppose that $x^{*}$ is in the weak*-closure of $\Gamma$. Since ${\overline{S}}^{*}$ is weak*-metrizable by Lemma 2, there exists a sequence ${\left\{x_n^{*}\right\}}_{n=1}^{\infty }$ in $\Gamma$ satisfying $d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0$ as $n\to \infty$. Since A is a compact subset of U and $x_n^{*}$ is continuous on A, we have

$$r\le s_A\left(x_n^{*}\right)=\underset{x\in A}{sup}\langle x_n^{*},x\rangle =\underset{x\in A}{max}\langle x_n^{*},x\rangle =\langle x_n^{*},x_n\rangle$$

(7)

for some $x_n\in A$. Therefore, we can form a sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$. Since A is compact, i.e., sequentially compact, there exists a convergent subsequence ${\left\{x_{n_j}\right\}}_{j=1}^{\infty }$, i.e., there exists $x\in A$ satisfying $\Vert x_{n_j}-x\Vert \to 0$ as $j\to \infty$. We also have $d_{{\overline{S}}^{*}}^{(w*)}(x_{n_j}^{*},x^{*})\to 0$ as $j\to \infty$. Since x is weak*-continuous on $U^{*}$, it follows that $\langle x_{n_j}^{*},x\rangle \to \langle x^{*},x\rangle$. Now, we have

$$\begin{array}{cc}\hfill \left|\langle x^{*},x\rangle -\langle x_{n_j}^{*},x_{n_j}\rangle \right|& \le \left|\langle x^{*},x\rangle -\langle x_{n_j}^{*},x\rangle \right|+\left|\langle x_{n_j}^{*},x\rangle -\langle x_{n_j}^{*},x_{n_j}\rangle \right|\hfill \\ & =\left|\langle x^{*},x\rangle -\langle x_{n_j}^{*},x\rangle \right|+\left|\langle x_{n_j}^{*},x-x_{n_j}\rangle \right|\hfill \\ & \le \left|\langle x^{*},x\rangle -\langle x_{n_j}^{*},x\rangle \right|+\Vert x_{n_j}^{*}\Vert ·\Vert x-x_{n_j}\Vert \hfill \\ & \le \left|\langle x^{*},x\rangle -\langle x_{n_j}^{*},x\rangle \right|+\Vert x-x_{n_j}\Vert (\mathrm{since}{x}_{n_j}^{*}\in {\overline{S}}^{*}).\hfill \end{array}$$

Therefore, we obtain $\langle x_{n_j}^{*},x_{n_j}\rangle \to \langle x^{*},x\rangle$ as $j\to \infty$. Since $\langle x_{n_j}^{*},x_{n_j}\rangle \ge r$ for all $n_j$ by (7), it follows

$$s_A\left(x^{*}\right)\ge \langle x^{*},x\rangle \ge r,$$

which says $x^{*}\in \Gamma$. This shows that $\Gamma$ is weak*-closed, i.e., the support function $s_A$ is indeed weak*-upper semi-continuous on ${\overline{S}}^{*}$. This completes the proof. □

The concept of a support function of fuzzy sets was considered in Diamond and Kloeden [10] and Puri and Ralescu [6] based on the finite-dimensional Euclidean space ${\mathbb{R}}^n$. In this paper, we are going to consider the support functions based on the infinite-dimensional normed space.

Definition 1.

Let $\tilde{A}$ be a fuzzy set in a normed space $(U,\Vert ·{\Vert }_U)$, and let $(U^{*},\Vert ·{\Vert }_{U^{*}})$ be the normed dual space of U. The support function

$$s_{\tilde{A}}:[0,1]\times U^{*}\to \mathbb{R}$$

of $\tilde{A}$ is defined by

$$s_{\tilde{A}}(\alpha ,x^{*})=s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)=\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle .$$

(8)

The norm of support function $s_{\tilde{A}}$ is defined by

$$\Vert s_{\tilde{A}}\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times U^{*}}{sup}\left|s_{\tilde{A}}(\alpha ,x)\right|.$$

Sometimes, the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\overline{S}}^{*}$ or $[0,1]\times {\widehat{S}}^{*}$, where

$${\overline{S}}^{*}=\left\{x^{*}\in U^{*}:\Vert x^{*}\Vert \le 1\right\} \mathrm{and} {\widehat{S}}^{*}=\left\{x^{*}\in U^{*}:\Vert x^{*}\Vert =1\right\}$$

are the closed unit ball and unit sphere in $U^{*}$, respectively. In this case, the norms of $s_{\tilde{A}}$ are given by

$$\Vert s_{\tilde{A}}\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{sup}\left|s_{\tilde{A}}(\alpha ,x)\right| \mathrm{and} \Vert s_{\tilde{A}}\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\widehat{S}}^{*}}{sup}\left|s_{\tilde{A}}(\alpha ,x)\right|.$$

respectively.

Proposition 11.

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$, and let $\tilde{A}$ be a fuzzy set in U. We have the following properties.

(i)

Given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_k\left(U\right)$, for $\lambda \ge 0$, we have

$$s_{\tilde{A}\oplus \tilde{B}}=s_{\tilde{A}}+s_{\tilde{B}} and s_{\lambda \tilde{A}}=\lambda ·s_{\tilde{A}}.$$

(ii)

Given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_{kc}\left(U\right)$, suppose that $s_{\tilde{A}}(\alpha ,x^{*})=s_{\tilde{B}}(\alpha ,x^{*})$ for all $\alpha \in [0,1]$ and all $x^{*}\in {\overline{S}}^{*}$. Then, we have $\tilde{A}=\tilde{B}$.

(iii)

Given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_{kc}\left(U\right)$, we have

$$\begin{array}{cc}\hfill d_{\mathcal{F}}(\tilde{A},\tilde{B})& =\underset{\{(\alpha ,x^{*}):\alpha \in [0,1],\Vert x^{*}{\Vert }_{U^{*}}\le 1\}}{sup}\left|s_{\tilde{A}}(\alpha ,x^{*})-s_{\tilde{B}}(\alpha ,x^{*})\right|\hfill \\ \hfill & =\underset{\{(\alpha ,x^{*}):\alpha \in [0,1],\Vert x^{*}{\Vert }_{U^{*}}=1\}}{sup}\left|s_{\tilde{A}}(\alpha ,x^{*})-s_{\tilde{B}}(\alpha ,x^{*})\right|.\hfill \end{array}$$

Proof. 

To prove part (i), using (3), we have

$${(\tilde{A}\oplus \tilde{B})}_{\alpha }={\tilde{A}}_{\alpha }+{\tilde{B}}_{\alpha } \mathrm{and} {\left(\lambda \tilde{A}\right)}_{\alpha }=\lambda {\tilde{A}}_{\alpha }$$

for any $\alpha \in [0,1]$. Using part (i) of Proposition 9, we also have

$$s_{\tilde{A}\oplus \tilde{B}}(\alpha ,x^{*})=s_{{(\tilde{A}\oplus \tilde{B})}_{\alpha }}\left(x^{*}\right)=s_{{\tilde{A}}_{\alpha }+{\tilde{B}}_{\alpha }}\left(x^{*}\right)=s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)+s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)=s_{\tilde{A}}(\alpha ,x^{*})+s_{\tilde{B}}(\alpha ,x^{*})$$

and

$$s_{\lambda \tilde{A}}(\alpha ,x^{*})=s_{{\left(\lambda \tilde{A}\right)}_{\alpha }}\left(x^{*}\right)=s_{\lambda {\tilde{A}}_{\alpha }}\left(x^{*}\right)=\lambda ·s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)=\lambda ·s_{\tilde{A}}(\alpha ,x^{*})$$

for any $(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}$ and $\lambda \ge 0$.

To prove part (ii), suppose that $s_{\tilde{A}}(\alpha ,x^{*})=s_{\tilde{B}}(\alpha ,x^{*})$ for all $\alpha \in [0,1]$ and all $x^{*}\in {\overline{S}}^{*}$. Then, we have $s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)=s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)$. Since ${\tilde{A}}_{\alpha }$ and ${\tilde{B}}_{\alpha }$ are compact and convex sets in U for any $\alpha \in [0,1]$, part (ii) of Proposition 9 says ${\tilde{A}}_{\alpha }={\tilde{B}}_{\alpha }$ for all $\alpha \in [0,1]$. This shows $\tilde{A}=\tilde{B}$.

To prove part (iii), using part (iii) of Proposition 9, we have

$$\begin{array}{cc}\hfill d_{\mathcal{F}}(\tilde{A},\tilde{B})& =\underset{\alpha \in [0,1]}{sup}{d}_H({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha })=\underset{\alpha \in [0,1]}{sup}\underset{\Vert x^{*}{\Vert }_{U^{*}}\le 1}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)\right|\hfill \\ \hfill & =\underset{\alpha \in [0,1]}{sup}\underset{\Vert x^{*}{\Vert }_{U^{*}}=1}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)\right|.\hfill \end{array}$$

This completes the proof. □

4.1. Continuity Regarding the Weak* Topology

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space. Lemma 2 says that the closed unit ball ${\overline{S}}^{*}$ is weak*-compact and weak*-metrizable with the metric $d_{{\overline{S}}^{*}}^{(w*)}$ given in (2). This metric $d_{{\overline{S}}^{*}}^{(w*)}$ can induce the topology ${\tau }_{{\overline{S}}^{*}}^{(w*)}$ such that $({\overline{S}}^{*},{\tau }_{{\overline{S}}^{*}}^{(w*)})$ is a compact topological space. Since $[0,1]$ is a compact subset of $\mathbb{R}$ with respect to the usual topology ${\tau }_{\mathbb{R}}$, this means that we can endow a topology ${\tau }_{[0,1]}$ to $[0,1]$ such that $([0,1],{\tau }_{[0,1]})$ is a compact topological subspace of $(\mathbb{R},{\tau }_{\mathbb{R}})$. Using the topologies ${\tau }_{{\overline{S}}^{*}}^{(w*)}$ and ${\tau }_{[0,1]}$, we can induce a product topology ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ for the product space $[0,1]\times {\overline{S}}^{*}$. Tychonoff’s theorem (ref. Royden [21]) says that the product space

$$\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right).$$

is also a compact topological space.

Now, we consider the metric spaces

$$\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right) \mathrm{and} \left([0,1]\times {\overline{S}}^{*},{\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),$$

where the metrics are given by

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})$$

(9)

and

$${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=max\left\{|\alpha -\beta |,d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})\right\}.$$

(10)

For the metric $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ on $[0,1]\times {\overline{S}}^{*}$, we can induce a metric topology ${\tau }_d^{(w*)}$. For the metric ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ on $[0,1]\times {\overline{S}}^{*}$, we can also induce a metric topology ${\tau }_{\widehat{d}}^{(w*)}$. We want to show that the metric topologies ${\tau }_d^{(w*)}$ and ${\tau }_{\widehat{d}}^{(w*)}$ are equal to the product topology ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ for the product space $[0,1]\times {\overline{S}}^{*}$.

The open ball $B((\alpha ,x^{*});ϵ)$ in the metric space $([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$ is given by

$$B\left((\alpha ,x^{*});ϵ\right)=\left\{(\beta ,y^{*})\in [0,1]\times {\overline{S}}^{*}:d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)<ϵ\right\}.$$

Also, the open balls $B(\alpha ;ϵ)$ and $B(x^{*};ϵ)$ of $[0,1]$ and ${\overline{S}}^{*}$ are given by

$$B(\alpha ;ϵ)=\left\{\beta \in [0,1]:|\alpha -\beta |<ϵ\right\}$$

and

$$B(x^{*};ϵ)=\left\{y^{*}\in {\overline{S}}^{*}:d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})<ϵ\right\},$$

respectively.

Proposition 12.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We have the following properties:

(i)

Given any $ϵ>0$ and $(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}$, we have

$$B((\alpha ,x^{*});ϵ)\subseteq B(\alpha ;ϵ)\times B(x^{*};ϵ)$$

and

$$B(\alpha ;ϵ/2)\times B(x^{*};ϵ/2)\subseteq B((\alpha ,x^{*});ϵ).$$

(ii)

Let ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ be the product topology for $[0,1]\times {\overline{S}}^{*}$, and let ${\tau }_d^{(w*)}$ be the metric topology for $[0,1]\times {\overline{S}}^{*}$ induced by the metric $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ in (9). Then, we have ${\tau }_d^{(w*)}={\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$. This also says that the product topology ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ is metrizable with the metric $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$.

Proof. 

To prove part (i), given any $(\beta ,y^{*})\in B((\alpha ,x^{*});ϵ)$, we have

$$|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})<ϵ,$$

which says t

$$|\alpha -\beta |<ϵ \mathrm{and} d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})<ϵ;$$

that is,

$$\beta \in B(\alpha ;ϵ) \mathrm{and} y^{*}\in B(x^{*};ϵ).$$

Therefore, we obtain the inclusion

$$B((\alpha ,x^{*});ϵ)\subseteq B(\alpha ;ϵ)\times B(x^{*};ϵ).$$

On the other hand, given any

$$(\beta ,y^{*})\in B(\alpha ;ϵ/2)\times B(x^{*};ϵ/2),$$

we have

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})<{\displaystyle \frac{ϵ}{2}}+{\displaystyle \frac{ϵ}{2}}=ϵ.$$

Therefore, we obtain the inclusion

$$B(\alpha ;ϵ/2)\times B(x^{*};ϵ/2)\subseteq B((\alpha ,x^{*});ϵ).$$

To prove part (ii), since ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ is the product topology for $[0,1]\times {\overline{S}}^{*}$, according to the concept of product topology, it means that $O\in {\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ if and only if, given any $(\alpha ,x^{*})\in O$, there exist neighborhoods $N_1$ of $\alpha$ and $N_2$ of $x^{*}$ satisfying $N_1\times N_2\subseteq O$. There also exist ${ϵ}_1,{ϵ}_2>0$ satisfying

$$B(\alpha ;{ϵ}_1)\subseteq N_1 \mathrm{and} B (x^{*};{ϵ}_2)\subseteq N_2.$$

Let $ϵ=min\{{ϵ}_1,,{ϵ}_2\}$. Then, we have

$$B(\alpha ;ϵ)\subseteq B(\alpha ;{ϵ}_1)\subseteq N_1 \mathrm{and} B (x^{*};ϵ)\subseteq B(x^{*};{ϵ}_2)\subseteq N_2.$$

Using part (i), we have

$$B((\alpha ,x^{*});ϵ)\subseteq B(\alpha ;ϵ)\times B(x^{*};ϵ)\subseteq N_1\times N_2\subseteq O,$$

which says $O\in {\tau }_d^{(w*)}$. This shows the inclusion ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\subseteq {\tau }_d^{(w*)}$. On the other hand, given any $O\in {\tau }_d^{\left(s\right)}$ and $(\alpha ,x^{*})\in O$, there exists $ϵ>0$ satisfying $B((\alpha ,x^{*});ϵ)\subseteq O$. Using part (i), we also have

$$B(\alpha ;ϵ/2)\times B(x^{*};ϵ/2)\subseteq B((\alpha ,x^{*});ϵ)\subseteq O,$$

which says $O\in {\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$. Therefore, we obtain the inclusion ${\tau }_d^{(w*)}\subseteq {\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$, and the proof is complete. □

Suppose that we consider the metric ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ on $[0,1]\times {\overline{S}}^{*}$. The open ball $\widehat{B}((\alpha ,x^{*});ϵ)$ in the metric space

$$\left([0,1]\times {\overline{S}}^{*},{\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)$$

is given by

$$\widehat{B}\left((\alpha ,x^{*});ϵ\right)=\left\{(\beta ,y^{*})\in [0,1]\times {\overline{S}}^{*}:{\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)<ϵ\right\}.$$

According to the similar arguments of Proposition 12, we also have the following results.

Proposition 13.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Then, the following statements hold true.

(i)

Given any $ϵ>0$ and $(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}$, we have

$$\widehat{B}((\alpha ,x^{*});ϵ)=B(\alpha ;ϵ)\times B(x^{*};ϵ).$$

(ii)

Let ${\tau }_{\widehat{d}}^{(w*)}$ be the metric topology for $[0,1]\times {\overline{S}}^{*}$ induced by the metric ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ in (10). Then, we have ${\tau }_{\widehat{d}}^{(w*)}={\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$. This says that the product topology ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ is metrizable with the metric ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$.

The following lemmas are useful for proving the continuity of support functions.

Lemma 3.

Let $\tilde{A}$ be a fuzzy subset of a universal set U If $\alpha \in [0,1]$ and $0\le {\alpha }_n\uparrow \alpha$, then

$${\tilde{A}}_{\alpha }=\bigcap _{n=1}^{\infty }{\tilde{A}}_{{\alpha }_n}.$$

Lemma 4

(Wu [22], pp. 5–7). Let ${\left\{A_n\right\}}_{n=1}^{\infty }$ be a sequence of subsets of universal set U satisfying $A_{n+1}\subseteq A_n$ for all n and ${\bigcap }_{n=1}^{\infty }{A}_n=A$, and let f be a real-valued function defined on U. Then, we have

$$\underset{n\to \infty }{lim}\underset{a\in A_n}{sup}f\left(a\right)=\underset{a\in A}{sup}f\left(a\right)and\underset{a\in A_n}{sup}f\left(a\right)\ge \underset{a\in A_{n+1}}{sup}f\left(a\right)$$

and

$$\underset{n\to \infty }{lim}\underset{a\in A_n}{inf}f\left(a\right)=\underset{a\in A}{inf}f\left(a\right)and\underset{a\in A_n}{inf}f\left(a\right)\le \underset{a\in A_{n+1}}{inf}f\left(a\right).$$

Lemma 5

(Wu [22], pp. 5–7). Let ${\left\{A_n\right\}}_{n=1}^{\infty }$ be a sequence of subsets of universal set U satisfying $A_n\subseteq A_{n+1}$ for all n and ${\bigcup }_{n=1}^{\infty }{A}_n=A$, and let f be a real-valued function defined on U. Then, we have

$$\underset{n\to \infty }{lim}\underset{a\in A_n}{sup}f\left(a\right)=\underset{a\in A}{sup}f\left(a\right) and \underset{a\in A_n}{sup}f\left(a\right)\le \underset{a\in A_{n+1}}{sup}f\left(a\right)$$

and

$$\underset{n\to \infty }{lim}\underset{a\in A_n}{inf}f\left(a\right)=\underset{a\in A}{inf}f\left(a\right)and\underset{a\in A_n}{inf}f\left(a\right)\ge \underset{a\in A_{n+1}}{inf}f\left(a\right).$$

Proposition 14.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space. For $\tilde{A}\in {\mathcal{F}}_k\left(U\right)$, we have the following properties:

(i)

The support function $s_{\tilde{A}}$ is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-upper semi-continuous on $[0,1]\times {\overline{S}}^{*}$, i.e., ${\tau }_d^{(w*)}$-upper semi-continuous on $[0,1]\times {\overline{S}}^{*}$.

(ii)

The support function $s_{\tilde{A}}$ is ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-upper semi-continuous on $[0,1]\times {\overline{S}}^{*}$, i.e., ${\tau }_{\widehat{d}}^{(w*)}$-upper semi-continuous on $[0,1]\times {\overline{S}}^{*}$.

(iii)

The support function $s_{\tilde{A}}$ is ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-upper semi-continuous on $[0,1]\times {\overline{S}}^{*}$.

Proof. 

To prove part (i), we want to show that the following set

$$\Gamma =\left\{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}:s_{\tilde{A}}(\alpha ,x^{*})\ge r\right\}$$

is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-closed for each $r\in \mathbb{R}$. Given any fixed $\widehat{\alpha }\in [0,1]$, suppose that there exists a sequence ${\left\{(\widehat{\alpha },x_n^{*})\right\}}_{n=1}^{\infty }$ in $\Gamma$ satisfying $d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0$ as $n\to \infty$. We want to show that $(\widehat{\alpha },x^{*})\in \Gamma$. Since ${\tilde{A}}_{\widehat{\alpha }}$ is a compact subset of U, we have

$$r\le s_{\tilde{A}}(\widehat{\alpha },x_n^{*})=\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{sup}\langle x_n^{*},x\rangle =\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{max}\langle x_n^{*},x\rangle =\langle x_n^{*},x_n\rangle$$

for some $x_n\in {\tilde{A}}_{\widehat{\alpha }}$. Since ${\tilde{A}}_{\widehat{\alpha }}$ is compact, it is also sequentially compact. Therefore, there exists a convergent subsequence ${\left\{x_{n_j}\right\}}_{j=1}^{\infty }$, i.e., there exists $\widehat{x}\in {\tilde{A}}_{\widehat{\alpha }}$ satisfying $\Vert x_{n_j}-\widehat{x}{\Vert }_U\to 0$ as $j\to \infty$. We also have $d_{{\overline{S}}^{*}}^{(w*)}(x_{n_j}^{*},x^{*})\to 0$ as $j\to \infty$. Since $\widehat{x}$ is weak*-continuous on $U^{*}$, it follows $|\langle x^{*}-x_{n_j}^{*},\widehat{x}\rangle |\to 0$ as $j\to \infty$. Now, we have

$$\begin{array}{cc}\hfill \left|\langle x^{*},\widehat{x}\rangle -\langle x_{n_j}^{*},x_{n_j}\rangle \right|& \le \left|\langle x^{*},\widehat{x}\rangle -\langle x_{n_j}^{*},\widehat{x}\rangle \right|+\left|\langle x_{n_j}^{*},\widehat{x}\rangle -\langle x_{n_j}^{*},x_{n_j}\rangle \right|\hfill \\ & =\left|\langle x^{*}-x_{n_j}^{*},\widehat{x}\rangle \right|+\left|\langle x_{n_j}^{*},\widehat{x}-x_{n_j}\rangle \right|\hfill \\ & \le \left|\langle x^{*}-x_{n_j}^{*},\widehat{x}\rangle \right|+\Vert x_{n_j}^{*}{\Vert }_{U^{*}}·{\Vert \widehat{x}-x_{n_j}\Vert }_U\hfill \\ & \le \left|\langle x^{*}-x_{n_j}^{*},\widehat{x}\rangle \right|+{\Vert \widehat{x}-x_{n_j}\Vert }_U (\mathrm{since} x_{n_j}^{*}\in {\overline{S}}^{*}).\hfill \end{array}$$

Therefore, we obtain $\langle x_{n_j}^{*},x_{n_j}\rangle \to \langle x^{*},\widehat{x}\rangle$ as $j\to \infty$. Since $\langle x_{n_j}^{*},x_{n_j}\rangle \ge r$ for all j, it follows that $\langle x^{*},\widehat{x}\rangle \ge r$, where $\widehat{x}\in {\tilde{A}}_{\widehat{\alpha }}$. Therefore, we conclude

$$s_{\tilde{A}}(\widehat{\alpha },x^{*})=\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{sup}\langle x^{*},x\rangle \ge \langle x^{*},\widehat{x}\rangle \ge r,$$

(11)

which says that $(\widehat{\alpha },x^{*})\in \Gamma$.

Suppose that $(\alpha ,x^{*})\in cl(\Gamma )$. Then, there exists a sequence ${\left\{({\alpha }_n,x_n^{*})\right\}}_{n=1}^{\infty }$ in $\Gamma$ satisfying

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left(({\alpha }_n,x_n^{*}),(\alpha ,x^{*})\right)\to 0 \mathrm{as} n\to \infty .$$

Using (9), we have $|{\alpha }_n-\alpha |\to 0$ and $d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0$ as $n\to \infty$, which also says that there exists a subsequence ${\left\{{\alpha }_{n_j}\right\}}_{j=1}^{\infty }$ satisfying ${\alpha }_{n_j}\uparrow \alpha$ or ${\alpha }_{n_j}\downarrow \alpha$ as $j\to \infty$.

• We assume that ${\alpha }_{n_j}\uparrow \alpha$ as $j\to \infty$ and write ${\beta }_j={\alpha }_{n_j}$ for all j. Then, we have ${\bigcap }_{j=1}^{\infty }{\tilde{A}}_{{\beta }_j}={\tilde{A}}_{\alpha }$ by Lemma 3. Using Lemma 4, we obtain

  $$\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle \to \underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle .$$

  (12)

  Now, using (11) and (12), we also obtain

  $$r\le s_{\tilde{A}}({\beta }_j,x^{*})=\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle \to \underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle =s_{\tilde{A}}(\alpha ,x^{*}),$$

  which shows $s_A(\alpha ,x^{*})\ge r$, i.e., $(\alpha ,x^{*})\in \Gamma$.
• We assume ${\alpha }_{n_j}\downarrow \alpha$ as $j\to \infty$. Then, we have ${\tilde{A}}_{{\alpha }_{n_j}}\subseteq {\tilde{A}}_{\alpha }$. We write $y_j^{*}=x_{n_j}^{*}$. Since ${\tilde{A}}_{\alpha }$ is a compact subset of U, we have

  $$r\le s_{\tilde{A}}({\alpha }_{n_j},x_{n_j}^{*})=s_{\tilde{A}}({\beta }_j,y_j^{*})=\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle y_j^{*},x\rangle \le \underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle y_j^{*},x\rangle =\underset{x\in {\tilde{A}}_{\alpha }}{max}\langle y_j^{*},x\rangle =\langle y_j^{*},y_j\rangle$$

  for some $y_j\in {\tilde{A}}_{\alpha }$. Since ${\tilde{A}}_{\alpha }$ is compact, there exists a convergent subsequence ${\left\{y_{j_k}\right\}}_{k=1}^{\infty }$, i.e., there exists $y\in {\tilde{A}}_{\alpha }$ satisfying $\Vert y_{j_k}{-y\Vert }_U\to 0$ as $k\to \infty$. Using the above same arguments, we can obtain $\langle x^{*},y\rangle \ge r$, where $y\in {\tilde{A}}_{\alpha }$. Therefore, we have

  $$s_{\tilde{A}}(\alpha ,x^{*})=\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle \ge \langle x^{*},y\rangle \ge r,$$

  which says $(\alpha ,x^{*})\in \Gamma$.

Therefore, we conclude that $\Gamma$ is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-closed.

To prove part (ii) using (10), we see that

$${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}(({\alpha }_n,x_n^{*}),(\alpha ,x^{*}))\to 0 \mathrm{as} n\to \infty$$

implies

$$|{\alpha }_n-\alpha |\to 0 \mathrm{and} d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0 \mathrm{as} n\to \infty .$$

Therefore, we can similarly show that $\Gamma$ is ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-closed. Finally, part (iii) follows immediately from Proposition 12. This completes the proof. □

Remark 1.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space. Given any $\tilde{A}\in {\mathcal{F}}_k\left(U\right)$, since $s_{\tilde{A}}$ is ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-upper semi-continuous on $[0,1]\times {\overline{S}}^{*}$ by Proposition 14 and the space $([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$ is compact, it follows that the supremum is attained. Therefore, the norm of the support function $s_{\tilde{A}}$ is given by

$$\Vert s_{\tilde{A}}\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{sup}\left|s_{\tilde{A}}(\alpha ,x^{*})\right|=\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{max}\left|s_{\tilde{A}}(\alpha ,x^{*})\right|=s_{\tilde{A}}({\alpha }_0,x_0^{*})=\underset{x\in {\tilde{A}}_{{\alpha }_0}}{sup}\langle x_0^{*},x\rangle$$

for some $({\alpha }_0,x_0^{*})\in [0,1]\times {\overline{S}}^{*}$. Since ${\tilde{A}}_{{\alpha }_0}$ is a compact set and $x_0^{*}$ is a continuous linear functional, we also have

$$\Vert s_{\tilde{A}}\Vert =\underset{x\in {\tilde{A}}_{{\alpha }_0}}{sup}\langle x_0^{*},x\rangle =\underset{x\in {\tilde{A}}_{{\alpha }_0}}{max}\langle x_0^{*},x\rangle =\langle x_0^{*},x_0\rangle <+\infty$$

for some $x_0\in {\tilde{A}}_{{\alpha }_0}$.

Next, we present the continuity of support functions.

Proposition 15.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space. Given any $\tilde{A}\in {\mathcal{F}}_k^C\left(U\right)$, we have the following properties:

(i)

The support function $s_{\tilde{A}}$ is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-continuous on $[0,1]\times {\overline{S}}^{*}$, i.e., ${\tau }_d^{(w*)}$-continuous on $[0,1]\times {\overline{S}}^{*}$.

(ii)

The support function $s_{\tilde{A}}$ is ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-continuous on $[0,1]\times {\overline{S}}^{*}$, i.e., ${\tau }_{\widehat{d}}^{(w*)}$-continuous on $[0,1]\times {\overline{S}}^{*}$.

(iii)

The support function $s_{\tilde{A}}$ is ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-continuous on $[0,1]\times {\overline{S}}^{*}$.

Proof. 

To prove part (i), according to Proposition 14, we just need to show that $s_A$ is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-lower semi-continuous, i.e., we want to show that the following set

$$\Gamma =\left\{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}:s_{\tilde{A}}(\alpha ,x^{*})\le r\right\}$$

is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-closed for each $r\in \mathbb{R}$. Given any fixed $\widehat{\alpha }\in [0,1]$, suppose that there exists a sequence $\left\{(\widehat{\alpha },x_n^{*})\right\}$ in $\Gamma$ satisfying $d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0$ as $n\to \infty$. We want to claim that $(\widehat{\alpha },x^{*})\in \Gamma$. Since ${\tilde{A}}_{\widehat{\alpha }}$ is a compact subset of U, we have

$$r\ge s_{\tilde{A}}(\widehat{\alpha },x_n^{*})=\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{sup}\langle x_n^{*},x\rangle =\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{max}\langle x_n^{*},x\rangle =\langle x_n^{*},x_n\rangle$$

(13)

for some $x_n\in {\tilde{A}}_{\widehat{\alpha }}$. Since ${\tilde{A}}_{\widehat{\alpha }}$ is compact, there exists a convergent subsequence ${\left\{x_{n_j}\right\}}_{j=1}^{\infty }$, i.e., there exists $\widehat{x}\in {\tilde{A}}_{\widehat{\alpha }}$ satisfying $\Vert x_{n_j}-\widehat{x}{\Vert }_U\to 0$ as $j\to \infty$. Using the same arguments of Proposition 14, we can obtain

$$\langle x^{*},\widehat{x}\rangle =\underset{j\to \infty }{lim}\langle x_{n_j}^{*},x_{n_j}\rangle .$$

(14)

Since $\langle x_{n_j}^{*},x_{n_j}\rangle \le r$ for all j, it follows that $\langle x^{*},\widehat{x}\rangle \le r$, where $\widehat{x}\in {\tilde{A}}_{\widehat{\alpha }}$. Since

$$s_{\tilde{A}}(\widehat{\alpha },x^{*})=\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{sup}\langle x^{*},x\rangle =\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{max}\langle x^{*},x\rangle =\langle x^{*},x_0\rangle$$

(15)

for some $x_0\in {\tilde{A}}_{\widehat{\alpha }}$, we have

$$\langle x^{*},x_0\rangle \ge \langle x^{*},\widehat{x}\rangle .$$

(16)

On the other hand, since $x_0$ is weak*-continuous on $U^{*}$, we have

$$\underset{j\to \infty }{lim}\langle x_{n_j}^{*},x_0\rangle =\langle x^{*},x_0\rangle .$$

(17)

Therefore, using (14), we obtain

$$\begin{array}{cc}\hfill \langle x^{*},\widehat{x}\rangle & =\underset{j\to \infty }{lim}\langle x_{n_j}^{*},x_{n_j}\rangle =\underset{j\to \infty }{lim}\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{sup}\langle x_{n_j}^{*},x\rangle (\mathrm{using} (13\left)\right)\hfill \\ \hfill & \ge \underset{j\to \infty }{lim}\langle x_{n_j}^{*},x_0\rangle =\langle x^{*},x_0\rangle , (\mathrm{using} (17\left)\right),\hfill \end{array}$$

which implies that $\langle x^{*},x_0\rangle =\langle x^{*},\widehat{x}\rangle$ by (16). Using (15), we also obtain

$$s_{\tilde{A}}(\widehat{\alpha },x^{*})=\underset{x\in {\tilde{A}}_{\widehat{\alpha }}}{sup}\langle x^{*},x\rangle =\langle x^{*},x_0\rangle =\langle x^{*},\widehat{x}\rangle \le r,$$

(18)

which shows that $(\widehat{\alpha },x^{*})\in \Gamma$.

Suppose that $(\alpha ,x^{*})\in \mathrm{cl}(\Gamma )$. There exists a sequence ${\left\{({\alpha }_n,x_n^{*})\right\}}_{n=1}^{\infty }$ in $\Gamma$ satisfying

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left(({\alpha }_n,x_n^{*}),(\alpha ,x^{*})\right)\to 0 \mathrm{as} n\to \infty .$$

Using (9), we have $|{\alpha }_n-\alpha |\to 0$ and $d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0$ as $n\to \infty$, which also says that there exists a subsequence ${\left\{{\alpha }_{n_j}\right\}}_{j=1}^{\infty }$ satisfying ${\alpha }_{n_j}\uparrow \alpha$ or ${\alpha }_{n_j}\downarrow \alpha$ as $j\to \infty$.

• We assume that ${\alpha }_{n_j}\downarrow \alpha$ as $j\to \infty$. We write ${\beta }_j={\alpha }_{n_j}$ and let $A={\bigcup }_{j=1}^{\infty }{\tilde{A}}_{{\beta }_j}$. Since ${\tilde{A}}_{{\beta }_j}$ is increasing in the sense of set inclusion, using Lemma 5, we have

  $$\underset{j\to \infty }{lim}\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle =\underset{x\in A}{sup}\langle x^{*},x\rangle =\underset{x\in {\bigcup }_{j=1}^{\infty }{\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle .$$

  (19)

  Using part (iv) of Proposition 9, we also have

  $$\underset{x\in {\bigcup }_{j=1}^{\infty }{\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle =\underset{x\in cl\left({\bigcup }_{j=1}^{\infty }{\tilde{A}}_{{\beta }_j}\right)}{sup}\langle x^{*},x\rangle \mathrm{and} \underset{x\in cl\left({\tilde{A}}_{\alpha }\right)}{sup}\langle x^{*},x\rangle =\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle .$$

  (20)

  The definition of $\overline{F}\left(U\right)$ says

  $$\mathrm{cl}\left(\bigcup _{j=1}^{\infty }{\tilde{A}}_{{\beta }_j}\right)=\mathrm{cl}\left({\tilde{A}}_{\alpha }\right).$$

  Therefore, using (19) and (20), we obtain

  $$\underset{j\to \infty }{lim}\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle =\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle .$$

  (21)

  From (18) and (21), we also obtain

  $$r\ge \underset{j\to \infty }{lim}{s}_{\tilde{A}}({\beta }_j,x^{*})=\underset{j\to \infty }{lim}\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle x^{*},x\rangle =\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle =s_{\tilde{A}}(\alpha ,x^{*}),$$

  which show $(\alpha ,x^{*})\in \Gamma$.
• We assume ${\alpha }_{n_j}\uparrow \alpha$ as $j\to \infty$. Then, we have ${\tilde{A}}_{\alpha }\subseteq {\tilde{A}}_{{\alpha }_{n_j}}$. We write $y_j^{*}=x_{n_j}^{*}$. Since ${\tilde{A}}_{\alpha }$ is a compact subset of U, we have

  $$r\ge s_{\tilde{A}}({\alpha }_{n_j},x_{n_j}^{*})=s_{\tilde{A}}({\beta }_j,y_j^{*})=\underset{x\in {\tilde{A}}_{{\beta }_j}}{sup}\langle y_j^{*},x\rangle \ge \underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle y_j^{*},x\rangle =\underset{x\in {\tilde{A}}_{\alpha }}{max}\langle y_j^{*},x\rangle =\langle y_j^{*},y_j\rangle$$

  for some $y_j\in {\tilde{A}}_{\alpha }$. Since ${\tilde{A}}_{\alpha }$ is compact, there exists a convergent subsequence ${\left\{y_{j_k}\right\}}_{k=1}^{\infty }$, i.e., there exists $y\in {\tilde{A}}_{\alpha }$ satisfying $\Vert y_{j_k}{-y\Vert }_U\to 0$ as $k\to \infty$. Using the above same arguments, we can obtain $\langle x^{*},y\rangle \le r$ and

  $$s_{\tilde{A}}(\alpha ,x^{*})=\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle =\langle x^{*},y\rangle \le r,$$

  which says $(\alpha ,x^{*})\in \Gamma$.

Therefore, we conclude that $\Gamma$ is $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-closed.

To prove part (ii), from (10), we have

$${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left(({\alpha }_n,x_n^{*}),(\alpha ,x^{*})\right)\to 0 \mathrm{as} n\to \infty$$

which implies

$$|{\alpha }_n-\alpha |\to 0 \mathrm{and} d_{{\overline{S}}^{*}}^{(w*)}(x_n^{*},x^{*})\to 0 \mathrm{as} n\to \infty .$$

Therefore, we can similarly show that $\Gamma$ is ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$-closed. Finally, part (iii) follows from Proposition 13. This completes the proof. □

Proposition 16.

Let $(U,\Vert ·{\Vert }_U)$ be a normed space. We have the following properties:

(i)

For $\tilde{A}\in \mathcal{F}\left(U\right)$ with the support function $s_{\tilde{A}}$, given any fixed $\alpha \in [0,1]$, the function $s_{\tilde{A}}(\alpha ,·)$ is lower semi-continuous and weak*-lower semi-continuous on $U^{*}$. We also have the following properties.

(a)

For $\tilde{A}\in {\mathcal{F}}_b\left(U\right)$, the function $s_{\tilde{A}}(\alpha ,·)$ is continuous on $U^{*}$.

(b)

Suppose that the normed space U is separable. For $\tilde{A}\in {\mathcal{F}}_k\left(U\right)$, the function $s_{\tilde{A}}(\alpha ,·)$ is weak*-continuous on ${\overline{S}}^{*}$.

(ii)

For $\tilde{A}\in \mathcal{F}\left(U\right)$ with the support function $s_{\tilde{A}}$, given any fixed $x^{*}\in U^{*}$, the function $s_{\tilde{A}}(·,x^{*})$ is left-continuous on $(0,1]$ and right-continuous at 0, and has the right limit

$$\underset{x\in {\tilde{A}}_{\alpha +}}{sup}\langle x^{*},x\rangle$$

at any $\alpha \in (0,1)$, where

$${\tilde{A}}_{\alpha +}=\bigcup _{\beta >\alpha }{\tilde{A}}_{\beta }.$$

Proof. 

To prove part (i), given any fixed $\alpha \in [0,1]$, part (iii) of Proposition 9 says that the function $s_{\tilde{A}}(\alpha ,·)$ is lower semi-continuous and weak*-lower semi-continuous on $U^{*}$. For $\tilde{A}\in {\mathcal{F}}_b\left(U\right)$, part (v) of Proposition 9 says that the function $s_{\tilde{A}}(\alpha ,·)$ is continuous on $U^{*}$. Suppose that U is separable and $\tilde{A}\in {\mathcal{F}}_k\left(U\right)$. Proposition 10 says that the function $s_{\tilde{A}}(\alpha ,·)$ is weak*-continuous on ${\overline{S}}^{*}$.

To prove part (ii), given any fixed $x^{*}\in U^{*}$, let

$$f\left(\alpha \right)=s_{\tilde{A}}(\alpha ,x^{*})=\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle .$$

Given any $\alpha \in (0,1]$, Lemma 3 says ${\tilde{A}}_{\alpha }={\bigcap }_{n=1}^{\infty }{\tilde{A}}_{{\alpha }_n}$ for ${\alpha }_n\uparrow \alpha$. Using Lemma 4, we have

$$\left|f\left({\alpha }_n\right)-f\left(\alpha \right)\right|=\left|\underset{x\in {\tilde{A}}_{{\alpha }_n}}{sup}\langle x^{*},x\rangle -\underset{x\in {\tilde{A}}_{\alpha }}{sup}\langle x^{*},x\rangle \right|\to 0 \mathrm{as} {\alpha }_n\uparrow \alpha ,$$

which shows that f is left-continuous at $\alpha$. Since

$${\tilde{A}}_{0+}=\bigcup _{\alpha \in (0,1]}{\tilde{A}}_{\alpha },$$

it follows that

$${\tilde{A}}_{0+}=\bigcup _{n=1}^{\infty }{\tilde{A}}_{{\alpha }_n}$$

for ${\alpha }_n\downarrow 0$. Using Lemma 5, we have

$$\left|\underset{x\in {\tilde{A}}_{{\alpha }_n}}{sup}\langle x^{*},x\rangle -\underset{x\in {\tilde{A}}_{0+}}{sup}\langle x^{*},x\rangle \right|\to 0 \mathrm{as} {\alpha }_n\downarrow 0.$$

Since ${\tilde{A}}_0=\mathrm{cl}\left({\tilde{A}}_{0+}\right)$, part (iv) of Proposition 9 says

$$\begin{array}{cc}\hfill \left|f\left({\alpha }_n\right)-f\left(0\right)\right|& =\left|\underset{x\in {\tilde{A}}_{{\alpha }_n}}{sup}\langle x^{*},x\rangle -\underset{x\in {\tilde{A}}_0}{sup}\langle x^{*},x\rangle \right|=\left|\underset{x\in {\tilde{A}}_{{\alpha }_n}}{sup}\langle x^{*},x\rangle -\underset{x\in cl\left({\tilde{A}}_{0+}\right)}{sup}\langle x^{*},x\rangle \right|\hfill \\ \hfill & =\left|\underset{x\in {\tilde{A}}_{{\alpha }_n}}{sup}\langle x^{*},x\rangle -\underset{x\in {\tilde{A}}_{0+}}{sup}\langle x^{*},x\rangle \right|\to 0,\hfill \end{array}$$

which shows that f is right-continuous at 0. Now, given any $\alpha \in [0,1)$, since

$${\tilde{A}}_{\alpha +}=\bigcup _{\beta >\alpha }{\tilde{A}}_{\beta },$$

we have

$${\tilde{A}}_{\alpha +}=\bigcup _{n=1}^{\infty }{\tilde{A}}_{\alpha } \mathrm{as} {\alpha }_n\downarrow \alpha .$$

Using Lemma 5, we have

$$\left|\underset{x\in {\tilde{A}}_{{\alpha }_n}}{sup}\langle x^{*},x\rangle -\underset{x\in {\tilde{A}}_{\alpha +}}{sup}\langle x^{*},x\rangle \right|\to 0 \mathrm{as} {\alpha }_n\downarrow \alpha ,$$

(22)

which shows that f has the right limit

$$\underset{x\in {\tilde{A}}_{\alpha +}}{sup}\langle x^{*},x\rangle$$

at $\alpha$. This completes the proof. □

4.2. Continuity Regarding the Norm Topology

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Suppose that $[0,1]$ is endowed with the usual topology ${\tau }_{[0,1]}$. Considering the unit sphere ${\widehat{S}}^{*}$ in $U^{*}$, we can form a metric space $({\widehat{S}}^{*},d_{{\widehat{S}}^{*}}^{\left(s\right)})$ where the metric $d_{{\widehat{S}}^{*}}^{\left(s\right)}$ is defined by

$$d_{{\widehat{S}}^{*}}^{\left(s\right)}(x^{*},y^{*})={\Vert x^{*}-y^{*}\Vert }_{U^{*}}.$$

This metric $d_{{\widehat{S}}^{*}}^{\left(s\right)}$ can induce a metric topology ${\tau }_{{\widehat{S}}^{*}}^{\left(s\right)}$. Using the topologies ${\tau }_{[0,1]}$ and ${\tau }_{{\widehat{S}}^{*}}^{\left(s\right)}$, we can form the product topology ${\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ for the product space $[0,1]\times {\widehat{S}}^{*}$. We define a metric on $[0,1]\times {\widehat{S}}^{*}$ by

$$d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=\left|\alpha -\beta \right|+d_{{\widehat{S}}^{*}}^{\left(s\right)}\left(x^{*},y^{*}\right)=\left|\alpha -\beta \right|+{\Vert x^{*}-y^{*}\Vert }_{U^{*}},$$

(23)

which can induce a metric topology ${\tau }_d^{\left(s\right)}$. We want to claim ${\tau }_d^{\left(s\right)}={\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$.

The open ball $B((\alpha ,x^{*});ϵ)$ in the metric space $([0,1]\times {\widehat{S}}^{*},d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)})$ is given by

$$B\left((\alpha ,x^{*});ϵ\right)=\left\{(\beta ,y^{*})\in [0,1]\times {\widehat{S}}^{*}:d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)<ϵ\right\}.$$

Also, the open balls $B(\alpha ;ϵ)$ and $B(x^{*};ϵ)$ of $[0,1]$ and $U^{*}$ are given by

$$B(\alpha ;ϵ)=\left\{\beta \in [0,1]:|\alpha -\beta |<ϵ\right\} \mathrm{and} B(x^{*};ϵ)=\left\{y^{*}\in {\widehat{S}}^{*}:d_{{\widehat{S}}^{*}}^{\left(s\right)}(x^{*},y^{*})<ϵ\right\},$$

respectively. According to the similar argument of Proposition 12.

Proposition 17.

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We have the following properties.

(i)

Given any $ϵ>0$ and $(\alpha ,x^{*})\in [0,1]\times {\widehat{S}}^{*}$, we have

$$B((\alpha ,x^{*});ϵ)\subseteq B(\alpha ;ϵ)\times B(x^{*};ϵ)$$

and

$$B(\alpha ;ϵ/2)\times B(x^{*};ϵ/2)\subseteq B((\alpha ,x^{*});ϵ).$$

(ii)

Let ${\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ be the product topology for the product space $[0,1]\times {\widehat{S}}^{*}$, and let ${\tau }_d^{\left(s\right)}$ be the metric topology for $[0,1]\times {\widehat{S}}^{*}$ with metric $d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ defined in (23). We have ${\tau }_d^{\left(s\right)}={\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$. This also says that the product topology ${\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ is metrizable with the metric $d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$.

Suppose that we define another metric ${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ on $[0,1]\times {\widehat{S}}^{*}$ by

$${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=max\left\{|\alpha -\beta |,d_{U^{*}}^{\left(s\right)}(x^{*},y^{*})\right\}=max\left\{|\alpha -\beta |,\Vert x^{*}-y^{*}{\Vert }_{U^{*}}\right\}.$$

(24)

The open ball $\widehat{B}((\alpha ,x^{*});ϵ)$ in the metric space $([0,1]\times {\widehat{S}}^{*},{\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)})$ is given by

$$\widehat{B}((\alpha ,x^{*});ϵ)=\left\{(\beta ,y^{*})\in [0,1]\times {\widehat{S}}^{*}:{\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)<ϵ\right\}.$$

According to the similar arguments of Proposition 12, we also have the following results.

Proposition 18.

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We have the following properties:

(i)

Given any $ϵ>0$ and $(\alpha ,x^{*})\in [0,1]\times {\widehat{S}}^{*}$, we have

$$\widehat{B}((\alpha ,x^{*});ϵ)=B(\alpha ;ϵ)\times B(x^{*};ϵ).$$

(ii)

Let ${\tau }_{\widehat{d}}^{\left(s\right)}$ be the metric topology for $[0,1]\times {\widehat{S}}^{*}$ with metric ${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ defined in (24). We have ${\tau }_{\widehat{d}}^{\left(s\right)}={\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$. This says that the product topology ${\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ is metrizable with the metric ${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$.

Proposition 19.

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Given any $\tilde{A}\in {\mathcal{F}}_k^C\left(U\right)$, we have the following properties:

(i)

The support function $s_{\tilde{A}}$ is $d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$-continuous on $[0,1]\times {\widehat{S}}^{*}$, i.e., ${\tau }_d^{\left(s\right)}$-continuous on $[0,1]\times {\widehat{S}}^{*}$.

(ii)

The support function $s_{\tilde{A}}$ is ${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$-continuous on $[0,1]\times {\widehat{S}}^{*}$, i.e., ${\tau }_{\widehat{d}}^{\left(s\right)}$-continuous on $[0,1]\times {\widehat{S}}^{*}$.

(iii)

The support function $s_{\tilde{A}}$ is ${\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$-continuous on $[0,1]\times {\widehat{S}}^{*}$.

Proof. 

The results follow from the similar proof of Proposition 15. □

5. Embedding Theorems

We denote by

$$C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)$$

the space of all continuous real-valued functions

$$f:[0,1]\times {\overline{S}}^{*}\to \mathbb{R} \mathrm{defined}\mathrm{on} \left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)$$

that is a compact space. We also denote by

$$C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right)$$

the space of all continuous real-valued functions

$$f:[0,1]\times {\widehat{S}}^{*}\to \mathbb{R} \mathrm{defined}\mathrm{on} ([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)})$$

that is not necessarily a compact space.

For the spaces

$$C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right) \mathrm{and} C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),$$

we consider the norms

$$\Vert f\Vert =\underset{(\alpha ,x)\in [0,1]\times {\overline{S}}^{*}}{sup}\left|f(\alpha ,x)\right| \mathrm{and} \Vert f\Vert =\underset{(\alpha ,x)\in [0,1]\times {\widehat{S}}^{*}}{sup}\left|f(\alpha ,x)\right|,$$

respectively. For the spaces

$$C_b\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right) \mathrm{and} C_b\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),$$

we mean that $\Vert f\Vert <+\infty$. The following results are needed.

Proposition 20.

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We have the following properties.

(i)

The space

$$\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right)$$

is metrizable with the metrics $d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ and ${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}$ defined in (23) and (24), respectively.

(ii)

Suppose that the normed space $(U,\Vert ·{\Vert }_U)$ is separable. Then, the compact space

$$\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)$$

is metrizable with the metrics $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ and ${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ defined in (9) and (10), respectively.

(iii)

We have that

$$\left(C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right) and \left(C_b\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right)$$

constitute a Banach space.

(iv)

Suppose that the normed space $(U,\Vert ·{\Vert }_U)$ is separable. We have that

$$\left(C_b\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)=\left(C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right).$$

is a separable Banach space.

Proof. 

Part (i) follows from Propositions 17 and 18, and part (ii) follows from Propositions 12 and 13. Part (iii) follows from Proposition 1. To prove part (iv), since $([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$ is a compact space and $\mathbb{R}$ is separable, Propositions 4 and 5 say that

$$\left(C_b\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)=\left(C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

is a separable Banach space. This completes the proof. □

5.1. Embedding Theorems for the Family ${\mathcal{F}}_{kc}^C\left(U\right)$

Now, we consider the metric spaces

$$\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right) \mathrm{and} \left([0,1]\times {\overline{S}}^{*},{\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),$$

which can induce the metric topologies ${\tau }_d^{(w*)}$ and ${\tau }_{\widehat{d}}^{(w*)}$, respectively. Propositions 12 and 13 say

$$\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)=\left([0,1]\times {\overline{S}}^{*},{\tau }_d^{(w*)}\right)=\left([0,1]\times {\overline{S}}^{*},{\tau }_{\widehat{d}}^{(w*)}\right).$$

Proposition 15 says that if U is separable, then, given any $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$, the support function $s_{\tilde{A}}$ is continuous on $([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$. Since the space ${\mathcal{F}}_{kc}^C\left(U\right)$ is closed under the addition and scalar multiplication by Proposition 7, we can consider the following embedding theorem

Theorem 1 (The Embedding Theorem). 

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We consider the metric spaces

$$\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right) \mathit{and} \left([0,1]\times {\overline{S}}^{*},{\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),$$

where

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})$$

and

$${\widehat{d}}_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=max\left\{|\alpha -\beta |,d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})\right\},$$

respectively. We define the function

$$\pi :\left({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}}\right)\to \left(C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right) by \tilde{A}\mapsto s_{\tilde{A}},$$

(25)

where the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\overline{S}}^{*}$. Then, we have the following properties.

(i)

The function π is one-to-one and isometric in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}).$$

For $\lambda \ge 0$, we also have

$$\pi (\tilde{A}\oplus \tilde{B})=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right) and \pi \left(\lambda \tilde{A}\right)=\lambda \pi \left(\tilde{A}\right).$$

(ii)

Given any $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)\setminus {\mathcal{F}}_{kc}^C\left(U\right)$, we have

$$\pi \left(\tilde{A}\right)\notin C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right).$$

In other words, the space ${\mathcal{F}}_{kc}^C\left(U\right)$ is the maximal subset of ${\mathcal{F}}_{kc}\left(U\right)$ satisfying

$$\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)\subseteq C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right).$$

Suppose that $(U,\Vert ,·\Vert )$ is a separable Banach space. Then, the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the separable Banach space

$$\left(C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

isometrically and isomorphically such that the space ${\mathcal{F}}_{kc}^C\left(U\right)$ is the maximal subset of ${\mathcal{F}}_{kc}\left(U\right)$ satisfying

$$\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)\subseteq C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),$$

and that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is complete, i.e., every Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ converges to an element in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$.

Proof. 

We first note that Proposition 20 says that

$$\left(C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

is a separable Banach space. Since ${\mathcal{F}}_{kc}^C\left(U\right)\subseteq {\mathcal{F}}_{kc}\left(U\right)$, part (ii) of Proposition 11 says that the function $\pi$ is one-to-one. Using part (iii) of Proposition 11, we also obtain

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =\Vert s_{\tilde{A}}-s_{\tilde{B}}\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{sup}\left|s_{\tilde{A}}(\alpha ,x^{*})-s_{\tilde{B}}(\alpha ,x^{*})\right|=d_{\mathcal{F}}\left(\tilde{A},\tilde{B}\right),$$

which shows that $\pi$ is isometric. Using part (i) of Proposition 11, we have

$$\pi (\tilde{A}\oplus \tilde{B})=s_{\tilde{A}\oplus \tilde{B}}=s_{\tilde{A}}+s_{\tilde{B}}=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right)$$

and

$$\pi \left(\lambda \tilde{A}\right)=s_{\lambda \tilde{A}}=\lambda s_{\tilde{A}}=\lambda \pi \left(\tilde{A}\right)for\lambda \ge 0.$$

Given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_{kc}^C\left(U\right)$ and $\lambda \in (0,1)$, Proposition 7 says

$$\lambda \tilde{A}\oplus (1-\lambda )\tilde{B}\in {\mathcal{F}}_{kc}^C\left(U\right).$$

Since

$$\lambda \pi \left(\tilde{A}\right)+(1-\lambda )\pi \left(\tilde{B}\right)=\pi (\lambda \tilde{A}+(1-\lambda )\tilde{B})\in \pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right),$$

it follows that $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is a convex cone in $C([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$.

Now, we assume that $(U,\Vert ·{\Vert }_U)$ is a separable Banach space. Proposition 8 says that the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ is complete. Let ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ be a Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$. Given any $ϵ>0$, there exists an integer $n_0$ satisfying $\Vert s_{{\tilde{A}}^{\left(n\right)}}-s_{{\tilde{A}}^{\left(m\right)}}\Vert <ϵ$ for $n,m>n_0$. Then, we have

$$d_{\mathcal{F}}\left({\tilde{A}}^{\left(n\right)},{\tilde{A}}^{\left(m\right)}\right)=\Vert \pi \left({\tilde{A}}^{\left(n\right)}\right)-\pi \left({\tilde{A}}^{\left(m\right)}\right)\Vert =\Vert s_{{\tilde{A}}^{\left(n\right)}}-s_{{\tilde{A}}^{\left(m\right)}}\Vert <ϵ,$$

which says that $\left\{{\tilde{A}}^{\left(n\right)}\right\}$ is a Cauchy sequence in $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$. The completeness of ${\mathcal{F}}_{kc}^C\left(U\right)$ says that there exists $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$ satisfying $d_{\mathcal{F}}({\tilde{A}}^{\left(n\right)},\tilde{A})\to 0$ as $n\to \infty$. Therefore, we obtain

$$\Vert s_{{\tilde{A}}^{\left(n\right)}}-s_{\tilde{A}}\Vert =\Vert \pi \left({\tilde{A}}^{\left(n\right)}\right)-\pi \left(\tilde{A}\right)\Vert =d_{\mathcal{F}}\left({\tilde{A}}^{\left(n\right)},\tilde{A}\right)\to 0 \mathrm{as} n\to \infty ,$$

which says that the sequence ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to

$$s_{\tilde{A}}=\pi \left(\tilde{A}\right)\in \pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right).$$

This shows the completeness of $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$. Next, we want to show that $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is a closed subset of the separable Banach space

$$\left(C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right).$$

Given any $s_{\tilde{A}}\in \mathrm{cl}\left(\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)\right)$, there exists a sequence ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ satisfying that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to $s_{\tilde{A}}$, which also says that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$. The completeness of $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ says that there exists $s_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ such that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to $s_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$. The uniqueness of that limit says that $s_{\tilde{A}}=s_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$. This shows that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is a closed set.

Assume $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)\setminus {\mathcal{F}}_{kc}^C\left(U\right)$. We want to claim

$$\pi \left(\tilde{A}\right)\notin C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right).$$

Suppose that we have

$$\pi \left(\tilde{A}\right)\in C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right).$$

Since

$$\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)=\left([0,1]\times {\overline{S}}^{*},{\tau }_d^{(w*)}\right),$$

where ${\tau }_d^{(w*)}$ is the metric topology induced by the metric $d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$. The continuity of the support function $s_{\tilde{A}}$ says that, given any $ϵ>0$, there exists $\delta >0$ such that

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})<\delta$$

implies

$$|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{A}}_{\beta }}\left(y^{*}\right)|=|s_{\tilde{A}}(\alpha ,x^{*})-s_{\tilde{A}}(\beta ,y^{*})|<ϵ.$$

(26)

Since $\tilde{A}\notin {\mathcal{F}}_{kc}^C\left(U\right)$, it means that the function $\alpha \mapsto {\tilde{A}}_{\alpha }$ is not continuous. Therefore, there exists ${ϵ}_0>0$ such that, given any $\delta >0$, there exist $\alpha ,\beta \in [0,1]$ such that $|\alpha -\beta |<\delta$ implies $d_H({\tilde{A}}_{\alpha },{\tilde{A}}_{\beta })\ge {ϵ}_0$. Since ${\tilde{A}}_{\alpha }$ and ${\tilde{A}}_{\beta }$ are compact and convex subsets of U, part (iii) of Proposition 9 says

$$\underset{x^{*}\in {\overline{S}}^{*}}{sup}|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{A}}_{\beta }}\left(x^{*}\right)|=d_H({\tilde{A}}_{\alpha },{\tilde{A}}_{\beta })\ge {ϵ}_0.$$

Proposition 10 says that the functions $s_{{\tilde{A}}_{\alpha }}$ and $s_{{\tilde{A}}_{\beta }}$ are weak*-continuous on ${\overline{S}}^{*}$, i.e., $s_{{\tilde{A}}_{\alpha }}-s_{{\tilde{A}}_{\beta }}$ is weak*-continuous on ${\overline{S}}^{*}$. Since ${\overline{S}}^{*}$ is a weak*-compact subset of $U^{*}$, it follows that the above supremum is attained. Therefore, there exists $x_0^{*}\in {\overline{S}}^{*}$ satisfying

$$\left|s_{\tilde{A}}(\alpha ,x_0^{*})-s_{\tilde{A}}(\beta ,x_0^{*})\right|=\left|s_{{\tilde{A}}_{\alpha }}\left(x_0^{*}\right)-s_{{\tilde{A}}_{\beta }}\left(x_0^{*}\right)\right|=\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{A}}_{\beta }}\left(x^{*}\right)\right|\ge {ϵ}_0,$$

which contradicts (26), since

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x_0^{*}),(\beta ,x_0^{*})\right)=|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x_0^{*},x_0^{*})=|\alpha -\beta |<\delta .$$

This completes the proof. □

Example 4.

Continued from Examples 2 and 3, we can consider the family ${\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$. Given any $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$, the support function $s_{\tilde{A}}$ is given in (5). Since the weak* topology and norm topology in $\mathbb{R}$ are identical, we have

$$d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})=d_{{\overline{S}}^{*}}^{\left(s\right)}(x^{*},y^{*})=\left|x^{*}-y^{*}\right| \mathrm{for} x^{*},y^{*}\in {\overline{S}}^{*}=[-1,1].$$

Using (9), we have

$$d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=|\alpha -\beta |+d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})=|\alpha -\beta |+\left|x^{*}-y^{*}\right|.$$

By referring to (25), the embedding function $\pi$ is given by

$$\pi :\left({\mathcal{F}}_{kc}^C\left(\mathbb{R}\right),d_{\mathcal{F}}\right)\to \left(C\left([0,1]\times [-1,1],d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right) \mathrm{with} \pi \left(\tilde{A}\right)=s_{\tilde{A}}.$$

Although the family ${\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$ is not a vector space, Theorem 1 says that the family $({\mathcal{F}}_{kc}^C\left(\mathbb{R}\right),d_{\mathcal{F}})$ can be treated as a closed convex cone of the separable Banach space $(C([0,1]\times [-1,1],d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}),\Vert ·\Vert )$ via the embedding function $\pi$ such that the distance between $\tilde{A}$ and $\tilde{B}$ in ${\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$ is equal to the distance between $\pi \left(\tilde{A}\right)$ and $\pi \left(\tilde{B}\right)$ in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}).$$

When the normed space $(U,\Vert ·{\Vert }_U)$ is not separable, Proposition 19 says that, if $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$, then the support function $s_{\tilde{A}}$ is continuous on $([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)})$. Therefore, we can also have the following embedding theorem by considering the unit sphere ${\widehat{S}}^{*}$ instead of the closed unit ball ${\overline{S}}^{*}$.

Theorem 2 (The Embedding Theorem). 

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We consider the metric spaces

$$\left([0,1]\times {\widehat{S}}^{*},d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right) and \left([0,1]\times {\widehat{S}}^{*},{\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),$$

where

$$d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=|\alpha -\beta |+\Vert x^{*}-y^{*}{\Vert }_{U^{*}}$$

and

$${\widehat{d}}_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\left((\alpha ,x^{*}),(\beta ,y^{*})\right)=max\left\{|\alpha -\beta |,\Vert x^{*}-y^{*}{\Vert }_{U^{*}}\right\},$$

respectively. We define the function

$$\pi :\left({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}}\right)\to \left(C\left([0,1]\times {\widehat{S}}^{*},d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right) by \tilde{A}\mapsto s_{\tilde{A}},$$

(27)

where the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\widehat{S}}^{*}$. Then, the function π is one-to-one and isometric in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}).$$

For $\lambda \ge 0$, we also have

$$\pi (\tilde{A}\oplus \tilde{B})=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right) and \pi \left(\lambda \tilde{A}\right)=\lambda \pi \left(\tilde{A}\right).$$

Suppose that $(U,\Vert ,·\Vert )$ is a Banach space. Then, the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

$$\left(C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right)$$

isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is complete.

Proof. 

The results follow from the similar proof of Theorem 1. □

Remark 2.

We have some observations regarding Theorems 1 and 2.

• When the unit sphere ${\widehat{S}}^{*}$ in Theorem 2 is replaced by the closed unit ball ${\overline{S}}^{*}$, we have the same embedding results.
• In Theorem 2, since the closed unit ball ${\widehat{S}}^{*}$ is not a compact subset of $U^{*}$, the involved supremum in the proof of Theorem 1 is not attained. In other words, for $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)\setminus {\mathcal{F}}_{kc}^C\left(U\right)$, we cannot have

  $$\pi \left(\tilde{A}\right)\notin C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right).$$
  Therefore, the space ${\mathcal{F}}_{kc}^C\left(U\right)$ is not the maximal subset of ${\mathcal{F}}_{kc}\left(U\right)$ satisfying

  $$\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)\subseteq C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right).$$

We can also consider the embedding Theorem 2 by using the closed unit ball ${\overline{S}}^{*}$. In other words, we can similarly obtain the embedding theorem by considering the following embedding function

$$\pi :\left({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}}\right)\to \left(C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right)$$

(28)

in Theorem 2. Since ${\widehat{S}}^{*}\subset {\overline{S}}^{*}$, we have the following inclusion

$$C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{\left(s\right)}\right)\subseteq C\left([0,1]\times {\widehat{S}}^{*},d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right).$$

(29)

Lemma 1 also says that we have the inclusion

$$C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)\subseteq C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{\left(s\right)}\right).$$

(30)

Remark 3.

The inclusions shown in (29) and (30) suggest the usage of embedding theorems depending on the separability of normed space U.

• Suppose that the normed space U is separable. Then, Theorem 1 is more preferable by considering the unit closed ball  ${\overline{S}}^{*}$ and the weak* topology for $U^{*}$, since it can be embedded into a smaller space $C([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$ by referring to the inclusion (30).
• Suppose that the normed space U is not necessarily separable. Then, we must consider the strong (norm) topology for  $U^{*}$. In this case, we can consider the embedding theorems using the unit sphere ${\widehat{S}}^{*}$ given in Theorem 2 or using the closed unit ball ${\overline{S}}^{*}$ by similarly applying the embedding functions (28) to the embedding Theorem 2, respectively. By referring to the inclusion (29), we may prefer to use the closed unit ball ${\overline{S}}^{*}$ by taking the smaller space $C([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{\left(s\right)})$. However, there is a dilemma in this situation, since checking the continuity on the smaller set $[0,1]\times {\widehat{S}}^{*}$ is easier than that of checking on the larger set $[0,1]\times {\overline{S}}^{*}$. In this case, the usage of the unit sphere ${\widehat{S}}^{*}$ or the closed unit ball ${\overline{S}}^{*}$ may depend on the problems that planned to be studied.

5.2. Embedding Theorems for the Family ${\mathcal{F}}_{kc}\left(U\right)$

We consider the embedding theorem for family ${\mathcal{F}}_{kc}\left(U\right)$. We denote by $D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*})$ the space of all real-valued functions

$$f:[0,1]\times {\widehat{S}}^{*}\to \mathbb{R}$$

defined on $[0,1]\times {\widehat{S}}^{*}$ such that the following conditions are satisfied.

• Given any fixed $\alpha \in [0,1]$, the function $f(\alpha ,·)$ is continuous on ${\widehat{S}}^{*}$ (with respect to the strong topology).
• Given any fixed $x^{*}\in {\widehat{S}}^{*}$, the function $f(·,x^{*})$ is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any $\alpha \in (0,1)$.

We also denote by $D^{(w*)}([0,1]\times {\overline{S}}^{*})$ the space of all real-valued functions f defined on $[0,1]\times {\overline{S}}^{*}$ such that the following conditions are satisfied.

• Given any fixed $\alpha \in [0,1]$, the function $f(\alpha ,·)$ is weak*-continuous on ${\overline{S}}^{*}$.
• Given any fixed $x^{*}\in {\overline{S}}^{*}$, the function $f(·,x^{*})$ is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any $\alpha \in (0,1)$.

Proposition 2 says that the spaces

$$D^{(w*)}([0,1]\times {\overline{S}}^{*}) \mathrm{and} D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*})$$

(31)

with the norms

$$\Vert f\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{sup}\left|f(\alpha ,x^{*})\right| \mathrm{and} \Vert f\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\widehat{S}}^{*}}{sup}\left|f(\alpha ,x^{*})\right|,$$

(32)

respectively, are Banach spaces.

Each compact set in a normed space is also closed and bounded. According to Proposition 16, we can consider the following function

$$\pi :{\mathcal{F}}_{kc}\left(U\right)\to D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*}) \mathrm{by} \tilde{A}\mapsto s_{\tilde{A}},$$

where the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\widehat{S}}^{*}$. Assume that the normed space $(U,\Vert ·{\Vert }_U)$ is separable. Proposition 16 also says that we can consider the following function

$$\pi :{\mathcal{F}}_{kc}\left(U\right)\to D^{(w*)}([0,1]\times {\overline{S}}^{*}) \mathrm{by} \tilde{A}\mapsto s_{\tilde{A}},$$

where the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\overline{S}}^{*}$.

Theorem 3 (The Embedding Theorem). 

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We define a function

$$\pi :\left({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}}\right)\to \left(D^{(w*)}([0,1]\times {\overline{S}}^{*}),\Vert ·\Vert \right) by \tilde{A}\mapsto s_{\tilde{A}},$$

where the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\overline{S}}^{*}$. Then, the function π is one-to-one and isometric in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}).$$

For $\lambda \ge 0$, we also have

$$\pi (\tilde{A}\oplus \tilde{B})=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right) and \pi \left(\lambda \tilde{A}\right)=\lambda \pi \left(\tilde{A}\right).$$

(i)

The space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a convex cone into the Banach space

$$\left(D^{(w*)}([0,1]\times {\overline{S}}^{*}),\Vert ·\Vert \right)$$

isometrically and isomorphically.

(ii)

Suppose that $(U,\Vert ,·\Vert )$ is a separable Banach space. Then, the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

$$\left(D^{(w*)}([0,1]\times {\overline{S}}^{*}),\Vert ·\Vert \right)$$

isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is complete.

Proof. 

Assume that $s_{\tilde{A}}=s_{\tilde{B}}$. This means that, given any fixed $\alpha \in [0,1]$, we have $s_{\tilde{A}}(\alpha ,x^{*})=s_{\tilde{B}}(\alpha ,x^{*})$ for all $x^{*}\in {\overline{S}}^{*}$, which also says $s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)=s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)$ for all $x^{*}\in {\overline{S}}^{*}$. Part (ii) of Proposition 9 says ${\tilde{A}}_{\alpha }={\tilde{B}}_{\alpha }$ for any $\alpha \in [0,1]$, which implies $\tilde{A}=\tilde{B}$. This shows that the function $\pi$ is one-to-one.

Using part (i) of Proposition 11, we have

$$\pi (\tilde{A}\oplus \tilde{B})=s_{\tilde{A}\oplus \tilde{B}}=s_{\tilde{A}}+s_{\tilde{B}}=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right)$$

and

$$\pi \left(\lambda \tilde{A}\right)=s_{\lambda \tilde{A}}=\lambda s_{\tilde{A}}=\lambda \pi \left(\tilde{A}\right) \mathrm{for} \lambda \ge 0.$$

Given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_{kc}\left(U\right)$ and $\lambda \in (0,1)$, using Proposition 7, we have

$$\lambda \tilde{A}\oplus (1-\lambda )\tilde{B}\in {\mathcal{F}}_{kc}\left(U\right).$$

Since

$$\lambda \pi \left(\tilde{A}\right)+(1-\lambda )\pi \left(\tilde{B}\right)=\pi (\lambda \tilde{A}+(1-\lambda )\tilde{B})\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right),$$

it shows that $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is a convex cone in $D^{(w*)}([0,1]\times {\overline{S}}^{*})$. Using part (iii) of Proposition 11, we have

$$\begin{array}{cc}\hfill \Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert & =\Vert s_{\tilde{A}}-s_{\tilde{B}}\Vert =\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{sup}\left|s_{\tilde{A}}(\alpha ,x^{*})-s_{\tilde{B}}(\alpha ,x^{*})\right|\hfill \\ \hfill & =\underset{(\alpha ,x^{*})\in [0,1]\times {\overline{S}}^{*}}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)\right|=\underset{\alpha \in [0,1]}{sup}\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)\right|\hfill \\ \hfill & =\underset{\alpha \in [0,1]}{sup}{d}_H({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha })=d_{\mathcal{F}}(\tilde{A},\tilde{B}).\hfill \end{array}$$

This shows that $\pi$ is isometric.

Now, we assume that $(U,\Vert ,·\Vert )$ is a separable Banach space. Part (iii) of Proposition 8 says that the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ is complete. Let ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ be a Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. Given any $ϵ>0$, there exists an integer $n_0$ satisfying $\Vert s_{{\tilde{A}}^{\left(n\right)}}-s_{{\tilde{A}}^{\left(m\right)}}\Vert <ϵ$ for $n,m>n_0$. Then, we have

$$d_{\mathcal{F}}\left({\tilde{A}}^{\left(n\right)},{\tilde{A}}^{\left(m\right)}\right)=\Vert \pi \left({\tilde{A}}^{\left(n\right)}\right)-\pi \left({\tilde{A}}^{\left(m\right)}\right)\Vert =\Vert s_{{\tilde{A}}^{\left(n\right)}}-s_{{\tilde{A}}^{\left(m\right)}}\Vert <ϵ,$$

which says that ${\left\{{\tilde{A}}^{\left(n\right)}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence in $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$. The completeness of ${\mathcal{F}}_{kc}\left(U\right)$ says that there exists $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)$ satisfying $d_{\mathcal{F}}({\tilde{A}}^{\left(n\right)},\tilde{A})\to 0$ as $n\to \infty$. Therefore, we obtain

$$\Vert s_{{\tilde{A}}^{\left(n\right)}}-s_{\tilde{A}}\Vert =\Vert \pi \left({\tilde{A}}^{\left(n\right)}\right)-\pi \left(\tilde{A}\right)\Vert =d_{\mathcal{F}}({\tilde{A}}^{\left(n\right)},\tilde{A})\to 0 \mathrm{as} n\to \infty ,$$

which says that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to

$$s_{\tilde{A}}=\pi \left(\tilde{A}\right)\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right).$$

This proves the completeness.

Next, we want to show that $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is a closed subset of $(D^{(w*)}([0,1]\times {\overline{S}}^{*}),\Vert ·\Vert )$. Given any $s_{\tilde{A}}\in \mathrm{cl}\left(\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)\right)$, there exists a sequence ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ in $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ such that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to $s_{\tilde{A}}$. This also says that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. The completeness says that there exists $s_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ such that ${\left\{s_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to $s_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. The uniqueness of the limit says that $s_{\tilde{A}}=s_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. This shows that $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is a closed subset of $(D^{(w*)}([0,1]\times {\overline{S}}^{*}),\Vert ·\Vert )$, and the proof is complete. □

When the normed space $(U,\Vert ·{\Vert }_U)$ is not separable, the similar arguments of Theorem 3 can also apply to the space $(D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*}),\Vert ·\Vert )$, where the unit sphere ${\widehat{S}}^{*}$ is considered instead of the closed unit ball ${\overline{S}}^{*}$. Therefore, we can obtain the following embedding theorem.

Theorem 4. (The Embedding Theorem). 

Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We define a function

$$\pi :\left({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}}\right)\to \left(D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*}),\Vert ·\Vert \right) by \tilde{A}\mapsto s_{\tilde{A}},$$

where the support function $s_{\tilde{A}}$ is restricted on $[0,1]\times {\widehat{S}}^{*}$. Then, the function π is one-to-one and isometric in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}),$$

For $\lambda \ge 0$, we also have

$$\pi (\tilde{A}\oplus \tilde{B})=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right) and \pi \left(\lambda \tilde{A}\right)=\lambda \pi \left(\tilde{A}\right).$$

(i)

The space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a convex cone into the Banach space

$$\left(D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*}),\Vert ·\Vert \right)$$

isometrically and isomorphically.

(ii)

Suppose that $(U,\Vert ,·\Vert )$ is a Banach space. Then, the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

$$\left(D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*}),\Vert ·\Vert \right)$$

isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is complete.

Theorem 3 is based on the weak* topology for $U^{*}$ and the closed unit ball ${\overline{S}}^{*}$, and Theorem 4 is based on the norm (strong) topology for $U^{*}$ and the unit sphere ${\widehat{S}}^{*}$. The usage of Theorems 3 and 4 can also refer to Remark 3.

5.3. Embedding Theorems into Different Banach Spaces

Next, we consider the embedding theorems for the different kinds of Banach spaces. Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Proposition 6 says that $C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)})$ is a separable Banach space with the norm given by

$$\Vert f\Vert =\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|f\left(x^{*}\right)\right|.$$

(33)

• We denote by

  $$\widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

  the space of all functions

  $$F:[0,1]\to \left(C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

  such that F is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any $\alpha \in (0,1)$.
• We denote by

  $$C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

  the space of all functions

  $$F:[0,1]\to \left(C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

  such that F is continuous on $[0,1]$.

Then, we have the following inclusion

$$C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)\subset \widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

Proposition 1 says that the spaces

$$\widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right) \mathrm{and} C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

with the norm

$$\Vert F\Vert =\underset{\alpha \in [0,1]}{sup}\Vert F\left(\alpha \right)\Vert =\underset{\alpha \in [0,1]}{sup}\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|F\left(\alpha \right)\left(x^{*}\right)\right|$$

are Banach spaces.

Proposition 21.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Given any $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)$, the function

$${\zeta }_{\tilde{A}}:[0,1]\to \left(C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}),\Vert ·\Vert \right) \mathit{defined}\mathit{by} \alpha \mapsto s_{{\tilde{A}}_{\alpha }}$$

is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any ${\alpha }_0\in (0,1)$, where the support function $s_{{\tilde{A}}_{\alpha }}$ is restricted on ${\overline{S}}^{*}$.

Proof. 

Let A be a nonempty subset of U. Given any $x\in U$, we define

$$\mathrm{dis}(x,A)=\underset{a\in A}{inf}{\Vert x-a\Vert }_U.$$

Let A and B be any two nonempty subsets of U. We define

$$\rho (A,B)=\underset{a\in A}{sup}\mathrm{dis}(a,B)=\underset{a\in A}{sup}\underset{b\in B}{inf}{\Vert a-b\Vert }_U$$

and

$$d_H(A,B)=max\{\rho (A,B),\rho (B,A)\}.$$

Then, we have

$$d_H(A,B)=d_H(A,\mathrm{cl}\left(B\right))=d_H(\mathrm{cl}\left(A\right),B)=d_H(\mathrm{cl}\left(A\right),\mathrm{cl}\left(B\right)).$$

(34)

Let $\mathcal{P}\left(U\right)$ denote the family of all nonempty subsets of U. Given any $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)$, we define the function

$${\eta }_{\tilde{A}}:[0,1]\to \left(\mathcal{P}\left(U\right),d_H\right) \mathrm{by} \alpha \mapsto {\tilde{A}}_{\alpha }.$$

We are going to show that the function ${\eta }_{\tilde{A}}$ is left-continuous on $(0,1]$ and right-continuous at 0, and has the right limit at any ${\alpha }_0\in (0,1)$.

Given any fixed ${\alpha }_0\in (0,1]$, for $\alpha <{\alpha }_0$, since ${\tilde{A}}_{{\alpha }_0}\subseteq {\tilde{A}}_{\alpha }$, we have $\rho ({\tilde{A}}_{{\alpha }_0},{\tilde{A}}_{\alpha })=0$. Since ${\tilde{A}}_{\alpha }$ decreases to ${\tilde{A}}_{{\alpha }_0}$ as $\alpha \uparrow {\alpha }_0$, using Lemmas 4 and 3, we have

$$\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{{\alpha }_0})=\underset{a\in {\tilde{A}}_{\alpha }}{sup}\mathrm{dis}(a,{\tilde{A}}_{{\alpha }_0})\downarrow \underset{a\in {\tilde{A}}_{{\alpha }_0}}{sup}\mathrm{dis}(a,{\tilde{A}}_{{\alpha }_0})=\rho ({\tilde{A}}_{{\alpha }_0},{\tilde{A}}_{{\alpha }_0})=0 \mathrm{as} \alpha \uparrow {\alpha }_0.$$

Therefore, we obtain

$$d_H({\tilde{A}}_{{\alpha }_0},{\tilde{A}}_{\alpha })=max\left\{\rho ({\tilde{A}}_{{\alpha }_0},{\tilde{A}}_{\alpha }),\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{{\alpha }_0})\right\}=\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{{\alpha }_0})\downarrow 0 \mathrm{as} \alpha \uparrow {\alpha }_0.$$

(35)

This shows that ${\eta }_{\tilde{A}}$ is left-continuous at ${\alpha }_0$.

For $\alpha >0$, since

$${\tilde{A}}_{\alpha }\subseteq {\tilde{A}}_{0+}=\bigcup _{\alpha \in (0,1]}{\tilde{A}}_{\alpha },$$

we have $\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{0+})=0$. Since ${\tilde{A}}_{\alpha }$ increases to ${\tilde{A}}_{0+}$ as $\alpha \downarrow 0$, using Lemmas 5 and 3, we have

$$\begin{array}{cc}\hfill \rho ({\tilde{A}}_{0+},{\tilde{A}}_{\alpha })& =\underset{a\in {\tilde{A}}_{0+}}{sup}\mathrm{dis}(a,{\tilde{A}}_{\alpha })=\underset{a\in {\tilde{A}}_{0+}}{sup}\underset{x\in {\tilde{A}}_{\alpha }}{inf}d(a,x)\downarrow \underset{a\in {\tilde{A}}_{0+}}{sup}\underset{x\in {\tilde{A}}_{0+}}{inf}d(a,x)\hfill \\ \hfill & =\underset{a\in {\tilde{A}}_{0+}}{sup}\mathrm{dis}(a,{\tilde{A}}_{0+})=\rho ({\tilde{A}}_{0+},{\tilde{A}}_{0+})=0 \mathrm{as} \alpha \downarrow 0.\hfill \end{array}$$

Now, we have

$$d_H({\tilde{A}}_{0+},{\tilde{A}}_{\alpha })=max\left\{\rho ({\tilde{A}}_{0+},{\tilde{A}}_{\alpha }),\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{0+})\right\}=\rho ({\tilde{A}}_{0+},{\tilde{A}}_{\alpha })\downarrow 0 \mathrm{as} \alpha \downarrow 0.$$

Using (34), we have

$$d_H({\tilde{A}}_0,{\tilde{A}}_{\alpha })=d_H(\mathrm{cl}\left({\tilde{A}}_{0+}\right),{\tilde{A}}_{\alpha })=d_H({\tilde{A}}_{0+},{\tilde{A}}_{\alpha })\downarrow 0 \mathrm{as} \alpha \downarrow 0.$$

This shows that ${\eta }_{\tilde{A}}$ is right-continuous at 0.

In general, given any ${\alpha }_0\in (0,1)$, for $\alpha >{\alpha }_0$, since

$${\tilde{A}}_{\alpha }\subseteq {\tilde{A}}_{{\alpha }_0+}=\bigcup _{\alpha >{\alpha }_0}{\tilde{A}}_{\alpha },$$

we have $\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{{\alpha }_0+})=0$. Since ${\tilde{A}}_{\alpha }$ increases to ${\tilde{A}}_{{\alpha }_0+}$ as $\alpha \downarrow {\alpha }_0$, using Lemmas 5 and 3, we have

$$\begin{array}{cc}\hfill \rho ({\tilde{A}}_{{\alpha }_0+},{\tilde{A}}_{\alpha })& =\underset{a\in {\tilde{A}}_{{\alpha }_0+}}{sup}\mathrm{dis}(a,{\tilde{A}}_{\alpha })=\underset{a\in {\tilde{A}}_{{\alpha }_0+}}{sup}\underset{x\in {\tilde{A}}_{\alpha }}{inf}d(a,x)\downarrow \underset{a\in {\tilde{A}}_{{\alpha }_0+}}{sup}\underset{x\in {\tilde{A}}_{{\alpha }_0+}}{inf}d(a,x)\hfill \\ \hfill & =\underset{a\in {\tilde{A}}_{{\alpha }_0+}}{sup}\mathrm{dis}(a,{\tilde{A}}_{{\alpha }_0+})=\rho ({\tilde{A}}_{{\alpha }_0+},{\tilde{A}}_{{\alpha }_0+})=0 \mathrm{as} \alpha \downarrow {\alpha }_0.\hfill \end{array}$$

(36)

Now, we have

$$d_H({\tilde{A}}_{{\alpha }_0+},{\tilde{A}}_{\alpha })=max\left\{\rho ({\tilde{A}}_{{\alpha }_0+},{\tilde{A}}_{\alpha }),\rho ({\tilde{A}}_{\alpha },{\tilde{A}}_{{\alpha }_0+})\right\}=\rho ({\tilde{A}}_{{\alpha }_0+},{\tilde{A}}_{\alpha })\downarrow 0 \mathrm{as} \alpha \downarrow {\alpha }_0.$$

This shows that ${\eta }_{\tilde{A}}$ has the right limit ${\tilde{A}}_{{\alpha }_0+}$ at ${\alpha }_0$.

Part (iii) of Proposition 9 says

$$d_H({\tilde{A}}_{\alpha },{\tilde{A}}_{\beta })=\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{A}}_{\beta }}\left(x^{*}\right)\right|=\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|(s_{{\tilde{A}}_{\alpha }}-s_{{\tilde{A}}_{\beta }})\left(x^{*}\right)\right|=\Vert s_{{\tilde{A}}_{\alpha }}-s_{{\tilde{A}}_{\beta }}\Vert .$$

It follows that the function ${\zeta }_{\tilde{A}}$ is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any ${\alpha }_0\in (0,1)$. This completes the proof. □

Since the normed space $(U,\Vert ·{\Vert }_U)$ is separable, given any $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)$, Proposition 21 says that the function

$${\zeta }_{\tilde{A}}:[0,1]\to C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right) \mathrm{defined}\mathrm{by} \alpha \mapsto s_{{\tilde{A}}_{\alpha }}$$

is left-continuous on $(0,1]$, right-continuous at 0, and has the right limit at any $\alpha \in (0,1)$, where the support function $s_{{\tilde{A}}_{\alpha }}$ is restricted on ${\overline{S}}^{*}$. In this case, we can define a function

$$\pi :{\mathcal{F}}_{kc}\left(U\right)\to \widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right) \mathrm{by} \tilde{A}\mapsto {\zeta }_{\tilde{A}}.$$

Then, we can have the following embedding theorem.

Theorem 5 (The Embedding Theorem). 

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We define a function

$$\pi :{\mathcal{F}}_{kc}\left(U\right)\to \widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right) \mathit{by} \tilde{A}\mapsto {\zeta }_{\tilde{A}}.$$

Then, the function π is one-to-one and isometric in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}).$$

For $\lambda \ge 0$, we also have

$$\pi (\tilde{A}\oplus \tilde{B})=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right) \mathit{and} \pi \left(\lambda \tilde{A}\right)=\lambda \pi \left(\tilde{A}\right).$$

(i)

The space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a convex cone into the Banach space

$$\widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

isometrically and isomorphically.

(ii)

Suppose that $(U,\Vert ,·\Vert )$ is a separable Banach space. Then, the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

$$\widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is complete.

Proof. 

Assume that ${\zeta }_{\tilde{A}}={\zeta }_{\tilde{B}}$. This means that $s_{{\tilde{A}}_{\alpha }}=s_{{\tilde{B}}_{\alpha }}$ for all $\alpha \in [0,1]$, i.e.,

$$s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)=s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right) \mathrm{for} \mathrm{all} s^{*}\in {\overline{S}}^{*} \mathrm{and} \mathrm{all} \alpha \in [0,1].$$

Using part (ii) of Proposition 9, we have ${\tilde{A}}_{\alpha }={\tilde{B}}_{\alpha }$ for all $\alpha \in [0,1]$, which implies that $\tilde{A}=\tilde{B}$. This says that the function $\pi$ is one-to-one.

Using part (i) of Proposition 9, for $\lambda \ge 0$, we have

$${\zeta }_{\tilde{A}\oplus \tilde{B}}\left(\alpha \right)=s_{{(\tilde{A}\oplus \tilde{B})}_{\alpha }}=s_{{\tilde{A}}_{\alpha }+{\tilde{B}}_{\alpha }}=s_{{\tilde{A}}_{\alpha }}+s_{{\tilde{B}}_{\alpha }}={\zeta }_{\tilde{A}}\left(\alpha \right)+{\zeta }_{\tilde{B}}\left(\alpha \right)$$

and

$${\zeta }_{\lambda \tilde{A}}\left(\alpha \right)=s_{\lambda {\tilde{A}}_{\alpha }}=\lambda s_{{\tilde{A}}_{\alpha }}=\lambda {\zeta }_{\tilde{A}}\left(\alpha \right).$$

Therefore, we obtain

$$\pi (\tilde{A}\oplus \tilde{B})={\zeta }_{\tilde{A}\oplus \tilde{B}}={\zeta }_{\tilde{A}}+{\zeta }_{\tilde{B}}=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right)$$

and

$$\pi \left(\lambda \tilde{A}\right)={\zeta }_{\lambda \tilde{A}}=\lambda {\zeta }_{\tilde{A}}=\lambda \pi \left(\tilde{A}\right) \mathrm{for} \lambda \ge 0.$$

Given any $\tilde{A},\tilde{B}\in {\mathcal{F}}_{kc}\left(U\right)$ and $\lambda \in (0,1)$, Proposition 7 says

$$\lambda \tilde{A}\oplus (1-\lambda )\tilde{B}\in {\mathcal{F}}_{kc}\left(U\right).$$

Since

$$\lambda \pi \left(\tilde{A}\right)+(1-\lambda )\pi \left(\tilde{B}\right)=\pi (\lambda \tilde{A}+(1-\lambda )\tilde{B})\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right),$$

it follows that $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is a convex cone in $\widehat{C}([0,1],C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}))$. From part (iii) of Proposition 9, we obtain

$$\begin{array}{cc}\hfill \Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert & =\Vert {\zeta }_{\tilde{A}}-{\zeta }_{\tilde{B}}\Vert =\underset{\alpha \in [0,1]}{sup}\Vert {\zeta }_{\tilde{A}}\left(\alpha \right)-{\zeta }_{\tilde{B}}\left(\alpha \right)\Vert =\underset{\alpha \in [0,1]}{sup}\Vert s_{{\tilde{A}}_{\alpha }}-s_{{\tilde{B}}_{\alpha }}\Vert \hfill \\ & =\underset{\alpha \in [0,1]}{sup}\underset{x^{*}\in {\overline{S}}^{*}}{sup}\left|s_{{\tilde{A}}_{\alpha }}\left(x^{*}\right)-s_{{\tilde{B}}_{\alpha }}\left(x^{*}\right)\right|=\underset{\alpha \in [0,1]}{sup}{d}_H({\tilde{A}}_{\alpha },{\tilde{B}}_{\alpha })=d_{\mathcal{F}}(\tilde{A},\tilde{B}).\hfill \end{array}$$

This shows that $\pi$ is isometric.

Now, we assume that $(U,\Vert ,·\Vert )$ is a separable Banach space. Part (iii) of Proposition 8 says that the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ is complete. Let ${\left\{{\zeta }_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ be a Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. Given any $ϵ>0$, there exists an integer $n_0$ satisfying $\Vert {\zeta }_{{\tilde{A}}^{\left(n\right)}}-{\zeta }_{{\tilde{A}}^{\left(m\right)}}\Vert <ϵ$ for $n,m>n_0$. Then, we have

$$d_{\mathcal{F}}\left({\tilde{A}}^{\left(n\right)},{\tilde{A}}^{\left(m\right)}\right)=\Vert \pi \left({\tilde{A}}^{\left(n\right)}\right)-\pi \left({\tilde{A}}^{\left(m\right)}\right)\Vert =\Vert {\zeta }_{{\tilde{A}}^{\left(n\right)}}-{\zeta }_{{\tilde{A}}^{\left(m\right)}}\Vert <ϵ,$$

which says that ${\left\{{\tilde{A}}^{\left(n\right)}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence in $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$. The completeness of ${\mathcal{F}}_{kc}\left(U\right)$ says that there exists $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)$ satisfying $d_{\mathcal{F}}({\tilde{A}}^{\left(n\right)},\tilde{A})\to 0$ as $n\to \infty$. Therefore, we obtain

$$\Vert {\zeta }_{{\tilde{A}}^{\left(n\right)}}-{\zeta }_{\tilde{A}}\Vert =\Vert \pi \left({\tilde{A}}^{\left(n\right)}\right)-\pi \left(\tilde{A}\right)\Vert =d_{\mathcal{F}}\left({\tilde{A}}^{\left(n\right)},\tilde{A}\right)\to 0 \mathrm{as} n\to \infty ,$$

which says that the sequence ${\left\{{\zeta }_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to

$${\zeta }_{\tilde{A}}=\pi \left(\tilde{A}\right)\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right).$$

This proves the completeness.

Next, we want to show that $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is a closed subset of $(\widehat{C}([0,1],C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)})),\Vert ·\Vert )$. Given any ${\zeta }_{\tilde{A}}\in \mathrm{cl}\left(\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)\right)$, there exists a sequence ${\left\{{\zeta }_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ in $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ such that ${\left\{{\zeta }_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to ${\zeta }_{\tilde{A}}$. This also says that ${\left\{{\zeta }_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ is a Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. Therefore, there exists ${\zeta }_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ such that the sequence ${\left\{{\zeta }_{{\tilde{A}}^{\left(n\right)}}\right\}}_{n=1}^{\infty }$ converges to ${\zeta }_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. The uniqueness of the limit says ${\zeta }_{\tilde{A}}={\zeta }_{\tilde{B}}\in \pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$. This shows that $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is a closed subset of $\widehat{C}([0,1],C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}))$. This completes the proof. □

Next, we are going to consider the embedding theorem of the family ${\mathcal{F}}_{kc}^C\left(U\right)$ into the Banach space

$$C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right).$$

Proposition 22.

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Given any $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$, the function

$${\zeta }_{\tilde{A}}:[0,1]\to \left(C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}),\Vert ·\Vert \right) \mathrm{defined} \mathrm{by} \alpha \mapsto s_{{\tilde{A}}_{\alpha }}$$

is continuous on $[0,1]$.

Proof. 

Since $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$, we have

$$cl\left({\tilde{A}}_{\alpha +}\right)=cl\left({\tilde{A}}_{\alpha }\right) \mathrm{for} \mathrm{all} \alpha \in [0,1).$$

Given any ${\alpha }_0\in [0,1)$, using (34) and (36), we have

$$\begin{array}{cc}\hfill d_H({\tilde{A}}_{{\alpha }_0},{\tilde{A}}_{\alpha })& =d_H(\mathrm{cl}\left({\tilde{A}}_{{\alpha }_0}\right),{\tilde{A}}_{\alpha })=d_H(\mathrm{cl}\left({\tilde{A}}_{{\alpha }_0+}\right),{\tilde{A}}_{\alpha })(\mathrm{since} \tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right))\hfill \\ \hfill & =d_H({\tilde{A}}_{{\alpha }_0+},{\tilde{A}}_{\alpha })\downarrow 0 \mathrm{as} \alpha \downarrow {\alpha }_0.\hfill \end{array}$$

This shows that ${\eta }_{\tilde{A}}$ is right-continuous at any ${\alpha }_0\in [0,1)$. Using Proposition 22, we conclude that ${\eta }_{\tilde{A}}$ is continuous at ${\alpha }_0$. This completes the proof. □

Given any $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$, Proposition 22 says that the function

$${\zeta }_{\tilde{A}}:[0,1]\to C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right) \mathrm{defined}\mathrm{by} \alpha \mapsto s_{{\tilde{A}}_{\alpha }}$$

is continuous on $[0,1]$, where the support function $s_{{\tilde{A}}_{\alpha }}$ is restricted on ${\overline{S}}^{*}$. In this case, we can define a function

$$\pi :{\mathcal{F}}_{kc}^C\left(U\right)\to \left(C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right) \mathrm{by} \tilde{A}\mapsto {\zeta }_{\tilde{A}}.$$

Suppose that $(U,\Vert ·{\Vert }_U)$ is a Banach space. Since ${\mathcal{F}}_{kc}^C\left(U\right)$ is complete by Proposition 8, from the same arguments of Theorem 5 by considering the space $C([0,1],C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}))$ that is smaller than the space $\widehat{C}([0,1],C({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}))$, we can also have the following embedding theorem.

Theorem 6 (The Embedding Theorem). 

Let $(U,\Vert ·{\Vert }_U)$ be a separable normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. We define a function

$$\pi :{\mathcal{F}}_{kc}^C\left(U\right)\to \left(C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right).$$

Then, the function π is one-to-one and isometric in the sense of

$$\Vert \pi \left(\tilde{A}\right)-\pi \left(\tilde{B}\right)\Vert =d_{\mathcal{F}}(\tilde{A},\tilde{B}).$$

For $\lambda \ge 0$, we also have

$$\pi (\tilde{A}\oplus \tilde{B})=\pi \left(\tilde{A}\right)+\pi \left(\tilde{B}\right) \mathit{and} \pi \left(\lambda \tilde{A}\right)=\lambda \pi \left(\tilde{A}\right).$$

(i)

The space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a convex cone into the Banach space

$$\left(C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right)$$

isometrically and isomorphically.

(ii)

Suppose that $(U,\Vert ,·\Vert )$ is a Banach space. Then, the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

$$\left(C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right)$$

isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is complete.

Example 5.

Continued from Examples 4, Theorem 6 says that the family ${\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$ can be embedded into the Banach space

$$\left(C\left([0,1],C\left([-1,1],d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right) with d_{{\overline{S}}^{*}}^{(w*)}(x^{*},y^{*})=\left|x^{*}-y^{*}\right| for x^{*},y^{*}\in {\overline{S}}^{*}=[-1,1]$$

via the embedding function π given by

$$\pi \left(\tilde{A}\right)={\zeta }_{\tilde{A}} with {\zeta }_{\tilde{A}}\left(\alpha \right)=s_{{\tilde{A}}_{\alpha }} for all \alpha \in [0,1].$$

Given an open interval $(a,b)$, we consider the following fuzzy function

$$\tilde{f}:(a,b)\to \left({\mathcal{F}}_{kc}^C\left(\mathbb{R}\right),d_{\mathcal{F}}\right),$$

which says that each function value $\tilde{f}\left(t\right)\in {\mathcal{F}}_{kc}^C\left(\mathbb{R}\right)$ is a fuzzy interval for any $t\in (a,b)$. In this case, we say that the fuzzy function $\tilde{f}$ is differentiable at $t_0\in (a,b)$ when the function $\widehat{f}=\pi \circ \tilde{f}$ is Fréchet differentiable at $t_0$, which means that the function

$$\widehat{f}:(a,b)\to \left(C\left([0,1],C\left([-1,1],d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right) defined by \widehat{f}\left(t\right)=\pi \left(\tilde{f}\left(t\right)\right)$$

is differentiable at $t_0$. This means that there exists a linear bounded function

$${\widehat{f}}^{\prime }\left(t_0\right):(a,b)\to \left(C\left([0,1],C\left([-1,1],d_{{\overline{S}}^{*}}^{(w*)}\right)\right),\Vert ·\Vert \right)$$

(37)

satisfying

$$\underset{t\to t_0}{lim}{\displaystyle \frac{\Vert \widehat{f}\left(t\right)-\widehat{f}\left(t_0\right)-{\widehat{f}}^{\prime }\left(t_0\right)(t-t_0)\Vert }{|t-t_0|}}=0.$$

Therefore, we conclude that the fuzzy function $\tilde{f}$ is differentiable at $t_0$ when there exists a linear bounded function ${\widehat{f}}^{\prime }\left(t_0\right)$ given in (37) satisfying

$$\underset{t\to t_0}{lim}{\displaystyle \frac{\Vert \pi \left(\tilde{f}\left(t\right)\right)-\pi \left(\tilde{f}\left(t_0\right)\right)-{\widehat{f}}^{\prime }\left(t_0\right)(t-t_0)\Vert }{|t-t_0|}}=0,$$

where

$$\pi \left(\tilde{f}\left(t\right)\right)={\zeta }_{\tilde{f}\left(t\right)} with {\zeta }_{\tilde{f}\left(t\right)}\left(\alpha \right)=s_{{\left(\tilde{f}\left(t\right)\right)}_{\alpha }} for all \alpha \in [0,1].$$

6. Conclusions

Let $(U,\Vert ·\Vert )$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Many embedding theorems have been presented in this paper to embed the families ${\mathcal{F}}_{kc}\left(U\right)$ and ${\mathcal{F}}_{kc}^C\left(U\right)$ into the different Banach space of continuous functions. Let $(U,\Vert ·{\Vert }_U)$ be a normed space with the dual space $(U^{*},\Vert ·{\Vert }_{U^{*}})$. Recall that the closed unit ball ${\overline{S}}^{*}$ is a weak* compact subset of $U^{*}$; that is, the closed unit ball ${\overline{S}}^{*}$ is a compact subset of $U^{*}$ with respect to the weak* topology ${\tau }_{U^{*}}^{w*}$, which can induce a topology ${\tau }_{{\overline{S}}^{*}}^{w*}$ for $S^{*}$ such that $(S^{*},{\tau }_{{\overline{S}}^{*}}^{w*})$ is a compact topological subspace of $(U^{*},{\tau }_{U^{*}}^{w*})$. Since $[0,1]$ is a compact subset of $\mathbb{R}$ with respect to the usual topology ${\tau }_{\mathbb{R}}$ for $\mathbb{R}$, the Tychonoff’s theorem says that the topology ${\tau }_{{\overline{S}}^{*}}^{w*}$ for ${\overline{S}}^{*}$ and the topology ${\tau }_{[0,1]}$ for $[0,1]$ can induce a product topology ${\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}$ for the product space $[0,1]\times {\overline{S}}^{*}$ such that $([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$ is a compact topological space. When the normed space $(U,\Vert ·{\Vert }_U)$ is separable, Lemmas 12 and 13 say that the compact space $([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)})$ is metrizable, which is also a Hausdorff and locally compact space. The embedding properties are summarized below.

Consider the embedding theorems of the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$. Given any $\tilde{A}\in {\mathcal{F}}_{kc}^C\left(U\right)$, this means that the $\alpha$-level sets ${\tilde{A}}_{\alpha }$ are compact and convex sets in U for all $\alpha \in [0,1]$, and that the function ${\eta }_{\tilde{A}}$ is continuous with respect to the Hausdorff metric $d_H$. The separability of the normed space is an important issue.

• Suppose that $(U,\Vert ,·\Vert )$ is a Banach space. Theorem 6 says that the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

  $$C\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

  isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is complete.
• Suppose that $(U,\Vert ,·\Vert )$ is a Banach space. Theorem 2 says that the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

  $$\left(C\left([0,1]\times {\widehat{S}}^{*},{\tau }_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right),\Vert ·\Vert \right)$$

  isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is complete. We can have the same embedding results when the unit sphere ${\widehat{S}}^{*}$ is replaced by the closed unit ball ${\overline{S}}^{*}$.
• Suppose that $(U,\Vert ,·\Vert )$ is a separable Banach space. Theorem 1 says that the space $({\mathcal{F}}_{kc}^C\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the separable Banach space

  $$\left(C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),\Vert ·\Vert \right)$$

  isometrically and isomorphically such that the space ${\mathcal{F}}_{kc}^C\left(U\right)$ is the maximal subset of ${\mathcal{F}}_{kc}\left(U\right)$ satisfying

  $$\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)\subseteq C\left([0,1]\times {\overline{S}}^{*},{\tau }_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right),$$

  and that the image $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ is complete, i.e., every Cauchy sequence in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$ converges to an element in $\pi \left({\mathcal{F}}_{kc}^C\left(U\right)\right)$.

Since ${\widehat{S}}^{*}\subset {\overline{S}}^{*}$, we have the following inclusion

$$C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{\left(s\right)}\right)\subseteq C\left([0,1]\times {\widehat{S}}^{*},d_{[0,1]\times {\widehat{S}}^{*}}^{\left(s\right)}\right).$$

Lemma 1 also says that we have the inclusion

$$C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{(w*)}\right)\subseteq C\left([0,1]\times {\overline{S}}^{*},d_{[0,1]\times {\overline{S}}^{*}}^{\left(s\right)}\right).$$

The above inclusions suggest the usage of embedding Theorems 1 and 2 depending on the separability of normed space U, which is presented in Remark 3.

Consider the embedding theorems of the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$. Given any $\tilde{A}\in {\mathcal{F}}_{kc}\left(U\right)$, it means that the $\alpha$-level sets ${\tilde{A}}_{\alpha }$ are compact and convex sets in U for all $\alpha \in [0,1]$. The separability of the normed space is an important issue.

• Suppose that $(U,\Vert ,·\Vert )$ is a Banach space. Theorem 4 says that the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

  $$\left(D^{\left(s\right)}([0,1]\times {\widehat{S}}^{*}),\Vert ·\Vert \right)$$

  isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is complete. We can have the same embedding results when the unit sphere ${\widehat{S}}^{*}$ is replaced by the closed unit ball ${\overline{S}}^{*}$.
• Suppose that $(U,\Vert ,·\Vert )$ is a separable Banach space. Theorem 3 says that the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

  $$\left(D^{(w*)}([0,1]\times {\overline{S}}^{*}),\Vert ·\Vert \right)$$

  isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is complete.
• Suppose that $(U,\Vert ,·\Vert )$ is a separable Banach space. Theorem 5 says that the space $({\mathcal{F}}_{kc}\left(U\right),d_{\mathcal{F}})$ can be embedded as a closed convex cone into the Banach space

  $$\widehat{C}\left([0,1],C\left({\overline{S}}^{*},d_{{\overline{S}}^{*}}^{(w*)}\right)\right)$$

  isometrically and isomorphically such that the image $\pi \left({\mathcal{F}}_{kc}\left(U\right)\right)$ is complete.

Theorem 3 is based on the weak* topology for $U^{*}$ and the closed unit ball ${\overline{S}}^{*}$, and Theorem 4 is based on the norm (strong) topology for $U^{*}$ and the unit sphere ${\widehat{S}}^{*}$. The usage of Theorems 3 and 4 can also refer to Remark 3. In the future research, we try to use these embedding theorems to study the practical problems that involve fuzzy uncertainty.