On fuzzy ellipsoid numbers and membership functions

Abstract

In this paper, the operations of fuzzy ellipsoid numbers and the direct relationship between the joint membership function and the edge membership functions are investigated. First, we prove that the general scalar multiplication (defined by Zadeh’s expansion) of fuzzy ellipsoid numbers preserves the closeness of the operation, but the general addition (defined by Zadeh’s expansion) does not preserve the closeness of the operation. Then, a new addition “⊕” of fuzzy ellipsoid numbers is defined, and it is shown that the new addition “⊕” preserves the closeness of the operation, and is the best approximation of the general addition operation in all addition operations which preserves the closeness of the operation for fuzzy ellipsoid numbers. Next, a bijection is established between fuzzy n-ellipsoid number space and n dimension fuzzy vector space. Furthermore a formula is obtained to express the edge membership functions with the joint membership function, and an iterative algorithm is proposed to calculate the joint membership function value as the edge membership functions are known, and the convergence of the iterative algorithm is demonstrated.

Keywords

Fuzzy numbers fuzzy cell numbers fuzzy ellipsoid numbers operations of fuzzy numbers membership functions

1 Introduction

The concept of fuzzy numbers was introduced by Chang and Zadeh [6] in 1972. Since then both the numbers and the problems in relation to them have been widely studied, see for example, [5 , 37] and the references therein. With the development of theories and applications of fuzzy numbers, this concept becomes more and more important [4 , 38]. Recently, there are still a lot of work in this area or related to this area. For example, in [2] Ban, etc. obtained a nearest trapezoidal approximation and the nearest symmetric trapezoidal approximation to a given fuzzy number; In [1], Adam and Pawel discussed a class of sequencing problems with uncertain parameters which is modeled by the usage of fuzzy intervals; In [16], Hosseini, etc. presented an automatic approach to learn and tune Gaussian interval type-2 membership functions; In [8, 9], Coroianu, etc. obtained a characterization of fuzzy number-valued Lipschitz functions, and studied the problem of the nearest approximation of fuzzy numbers by piecewise linear 1-knot fuzzy numbers; In [15], Hong considered the law of large numbers for T-related weighted fuzzy variables whose underlying spaces are R^p; In [21], Lodwick and Jamison constructed interval-valued probability measure, and used it to develop the extension of some measures; In [22], Moreno-Garcia, etc. proposed a method to construct trapezoidal fuzzy number approximations from raw discrete data; In [27], Tak $\overset{´}{a} \overset{˘}{c}$ proposed new subsethood measures for interval-valued fuzzy sets; In [11, 18], Eslamipoor, etc. set up some new methods for ranking two generalized fuzzy numbers; In [39], Zhou, etc. proposed an operational law for fuzzy arithmetic.

For general n dimensional fuzzy numbers, due to their structural complexity, they can not be used conveniently in some fields of applications and some researches of theory [30, 36]. In [3], Bandemer and Näther, from n 1-dimension fuzzy numbers u₁, ⋯ , u_n, defined a special kind n-dimensional fuzzy numbers whose membership function value u (x) at x = (x₁, ⋯ , x_n) is defined by u (x) = i =min_{1, 2 ⋯ , n}u_i (x_i), and Inuiguchi, Ramík and Tanino called it vector of non-interactive 1-dimension fuzzy numbers in [17]. In [30–34 , 36], we also carefully studied a special type of n-dimensional fuzzy numbers which are called fuzzy n-cell numbers, set up some methods of constructing thus fuzzy numbers to represent imprecise or uncertain multi-channel digital signals, and demonstrated that the fuzzy n-cell numbers are used much more conveniently than general n dimensional fuzzy numbers in theoretical investigations and in some fields of application. However, fuzzy n-cell numbers have some defects in structure, it will bring about some bad effects in applications [35]. In order to overcome the weakness, and make fuzzy numbers can better be used, a new special type of n dimension fuzzy numbers which are called fuzzy n-ellipsoid numbers in [35]. About crisp n-ellipsoid, ellipsoid numbers have been known in interval analysis for some time, for example, as early as in 1993, Neumaier has used interval ellipsoid arithmetic to give a method for reducing the wrapping effect in [24]. In addition, in 2007, Kreinovich, Beck and Nguyen also discussed ellipsoid-shaped fuzzy sets in [19], showed that ellipsoids are natural multi-variate generalization of intervals and ellipsoid-shaped fuzzy sets are a natural generalization of fuzzynumbers.

In this paper, we investigate the operations of fuzzy ellipsoid numbers and the direct relationship between the joint membership function and the edge membership functions are investigated. In Section 2, we review some basic definitions and notations and results which will be used in this paper. In Section 3, we firstly obtain two Lemmas obtained in this paper. Then show that the general scalar multiplication preserves the closeness of the operation, but the general addition does not preserve the closeness of the operation. And then, we define a new addition “⊕”, and show the new addition “⊕” preserves the closeness of the operation, and is the best approximation of the general addition operation in the all addition operations which preserve the closeness of the operation for fuzzy ellipsoid numbers. In Section 4, we firstly establish a bijection between fuzzy n-ellipsoid number space and n dimension fuzzy vector space. Then we obtain a formula which is used to express the edge membership functions with the joint membership function. And then, we present a iterative algorithm to calculate the joint membership function value as the edge membership functions are known, and prove the convergence of the iterative algorithm. In Section 5, we make a summary of this paper.

2 Basic definitions and notations

Let n be a natural number, R be the real number set, and Rⁿ be the n-dimensional Euclidean space. And let K (Rⁿ) denote the collection of non-empty compact subsets of Rⁿ. The space K (Rⁿ) has a linear structure induced by the addition and scalar multiplication A + B = {a + b | a ∈ A, b ∈ B} and λA = {λa | a ∈ A} for any A, B ∈ K (Rⁿ) , λ ∈ R.

A fuzzy subset (in short, a fuzzy set) of Rⁿ is a function u : Rⁿ→ [0, 1]. For each such fuzzy set u, we denote by [u]^r= {x ∈ Rⁿ: u (x) ≥ r} for any r ∈ (0, 1], its r-level set. By suppu we denote the support of u, i.e., the {x ∈ Rⁿ: u (x) >0}. By [u]⁰ we denote the closure of the suppu, i.e., $[u]^{0} = \bar{{x \in R^{n} : u (x) > 0}}$ .

If u is a normal and fuzzy convex fuzzy set of Rⁿ, u (x) is upper semi-continuous, and [u]⁰ is compact, then we call u a n dimension fuzzy number, and denote the collection of all n dimension fuzzy numbers by Eⁿ.

It is known that if u ∈ Eⁿ, then for each r ∈ [0, 1], [u]^r is a compact set in Rⁿ.

Let u_i ∈ E (i . e . , E¹), i = 1, 2, …, n. We call the ordered one dimension fuzzy number class u₁, u₂, …, u_n (i.e., the Cartesian product of one-dimensional fuzzy numbers u₁, u₂, …, u_n) a n dimension fuzzy vector, denote it as (u₁, u₂, …, u_n), and call the collection of all n dimension fuzzy vectors (i.e., the Cartesian product ${\overset{︷}{E \times E \times \dots \times E}}^{n}$ ) n dimension fuzzy vector space, and denote it as (E)ⁿ.

For any a ∈ Rⁿ, define an n dimension fuzzy number $\hat{a}$ by

$\hat{a} (x) = {\begin{matrix} 1 & if & x = a \\ 0 & if & x \neq a \end{matrix}$ for any x ∈ Rⁿ.

The addition, scalar multiplication on Eⁿ are defined by

$(u + v) (x) = sup_{y + z = x} min (u (y), v (z))$ $(λ u) (x) = {\begin{matrix} u (λ^{- 1} x) & if & λ \neq 0 \\ \hat{0} (x) & if & λ = 0 \end{matrix}$

for u, v ∈ Eⁿ and λ ∈ R.

Let u, v ∈ Eⁿ and k ∈ R. Then u + v, ku ∈ Eⁿ and [u + v]^r= [u]^r+ [v]^r, [ku]^r= k [u]^r for any r ∈ [0, 1].

If u ∈ Eⁿ, and for each r ∈ [0, 1], [u]^r is a n-ellipsoid, i.e., exist $\underline{u_{i}} (r), \bar{u_{i}} (r) \in R$ with $\underline{u_{i}} (r) \leq \bar{u_{i}} (r)$ for any r ∈ [0, 1], i = 1, 2, …, n, such that

$\begin{matrix} [u]^{r} \\ = {(x_{1}, \dots, x_{n}) \in R^{n} | \sum_{i = 1}^{n} \frac{(x_{i} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} \leq 1} \end{matrix}$

where, $\underline{u_{i}} (r) = \bar{u_{i}} (r)$ can be allowed for some i = 1, 2, ⋯ , n and r ∈ [0, 1], the meaning of $\sum_{i = 1}^{n} \frac{(x_{i} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} \leq 1$ can see Remark 1 in [35]), then we call u a fuzzy n-ellipsoid number [35]. And we denote the collection of all fuzzy n-ellipsoid numbers by E (Eⁿ).

Theorem 1. [35] If u ∈ E (Eⁿ), then

[u]^r is a non-empty n dimension closed ellipsoid, i.e.,

\begin{matrix} [u]^{r} \\ = {(x_{1}, \dots, x_{n}) \in R^{n} | \sum_{i = 1}^{n} \frac{(x_{i} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} \leq 1} \end{matrix}

for each r ∈ [0, 1];

[u]^{r
₂}⊂ [u]^{r
₁} for 0 ≤ r₁ ≤ r₂ ≤ 1;

$⋂_{m = 1}^{\infty} [u]^{r_{m}} = [u]^{r}$ for positive non-decreasing sequence {r_m} with $lim_{m \to \infty} r_{m} = r$ .

Conversely, if A_r, r ∈ [0, 1] satisfies the (1), (2) and (3) above, then there exists a unique fuzzy n-ellipsoid number u such that [u]^r= A_r for each r ∈ (0, 1] and $[u]^{0} = \bar{⋃_{0 < r \leq 1} A_{r}} \subset A_{0}$ .

Theorem 2. [35] If u ∈ E (Eⁿ), then for i = 1, 2, ⋯ , n, $\underline{u_{i}} (r)$ , $\bar{u_{i}} (r)$ are real-valued functions on [0, 1], and satisfy

$\underline{u_{i}} (r)$ are non-decreasing and left continuous;

$\bar{u_{i}} (r)$ are non-increasing and left continuous;

$\underline{u_{i}} (r) \leq \bar{u_{i}} (r)$ (it is equivalent to $\underline{u_{i}} (1) \leq \bar{u_{i}} (1)$ );

$\underline{u_{i}} (r)$ , $\bar{u_{i}} (r)$ are right continuous at r = 0.

Conversely if a_i (r), b_i (r), i = 1, 2, ⋯ , n are real-valued functions on [0, 1] which satisfy the (1)-(4) above, then there exists a unique u ∈ E (Eⁿ)such that $\begin{matrix} [u]^{r} \\ = {(x_{1}, \dots, x_{n}) \in R^{n} | \sum_{i = 1}^{n} \frac{(x_{i} - \frac{b_{i} (r) + a_{i} (r)}{2})^{2}}{(\frac{b_{i} (r) - a_{i} (r)}{2})^{2}} \leq 1} \end{matrix}$ for any r ∈ [0, 1].

3 Operations of fuzzy ellipsoid numbers

For a_i, b_i ∈ R with a_i ≤ b_i, i = 1, 2, ⋯ , n, we denote by $\prod_{i = 1}^{n} [a_{i}, b_{i}] = {(x_{1}, \dots, x_{n}) \in R^{n} | x_{i} \in [a_{i}, b_{i}], i = 1, 2, \dots, n}$ and also still use the notation in [35] $\begin{matrix} (\begin{matrix} a_{1} & b_{1} \\ a_{2} & b_{2} \\ ⋮ & ⋮ \\ a_{n} & b_{n} \end{matrix}) (in short, {(a_{i}, b_{i})}_{i = 1}^{n}) \\ = {(x_{1}, x_{2}, \dots, x_{n}) \in R^{n} | \sum_{i = 1}^{n} \frac{(x_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i}}{2})^{2}} \leq 1} \end{matrix}$ where a_i = b_i for some i = 1, 2, ⋯, n can beallowed, and we stipulate that if a_k = b_k for k ∈ K ⊂ {1, 2, ⋯, n}, then $\sum_{i = 1}^{n}$ $\frac{(x_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i}}{2})^{2}} \leq 1$ indicates ∑_{1≤i≤n,i∉K} $\frac{(x_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i}}{2})^{2}}$ ≤ 1 and x_k = a_k = b_k for any k ∈ K For example, as a₁ = b₁ and a₃ = b₃, we stipulate that the inequality $\sum_{i = 1}^{4}$ $\frac{(x_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i}}{2})^{2}}$ ≤ 1 indicates $\frac{(x_{2} - \frac{b_{2} + a_{2}}{2})^{2}}{(\frac{b_{2} - a_{2}}{2})^{2}}$ + $\frac{(x_{4} - \frac{b_{4} + a_{4}}{2})^{2}}{(\frac{b_{4} - a_{4}}{2})^{2}}$ ≤ 1 and x₁ = a₁ = b₁, x₃ = a₃ = b₃.

Lemma 1. Let $a_{i}, b_{i}, a_{i}^{'}, b_{i}^{'} \in R$ with a_i ≤ b_i and $a_{i}^{'} \leq b_{i}^{'}$ , i = 1, 2, ⋯ , n. Then $\begin{matrix} {(a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'})}_{i = 1}^{n} \subset {(a_{i}, b_{i})}_{i = 1}^{n} + {(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n} \\ \underset{\neq}{\subset} \prod_{i = 1}^{n} [a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'}] \end{matrix}$ (1)

Proof. For any $a_{i}, a_{i}^{'}, \in R$ with $a_{i}, a_{i}^{'} > 0$ , i = 1, 2, ⋯ , n, we first show that

$\begin{matrix} {(- (a_{i} + a_{i}^{'}, a_{i} + a_{i}^{'})}_{i = 1}^{n} \subset {(a_{i}, b_{i})}_{i = 1}^{n} + {(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n} \\ \underset{\neq}{\subset} \prod_{i = 1}^{n} [a_{i} + a_{i}^{'}, a_{i} + a_{i}^{'}] \end{matrix}$ (2)

Let x⁰ = $(x_{1}^{0}$ , $x_{2}^{0}$ , ⋯, $x_{n}^{0})$ ∈ $(- (a_{i} + a_{i}^{'})$ , a_i + ${a_{i}^{'})}_{i = 1}^{n}$ . Then we have $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0})^{2}}{(a_{i} + a_{i}^{'})^{2}}$ ≤ 1. It is obvious that $x_{i}^{0}$ ∈ $[- (a_{i} + a_{i}^{'})$ , $a_{i} + a_{i}^{'}]$ for any i = 1, 2, ⋯, n, so there exists λ_i ∈ [0, 1] such that $x_{i}^{0}$ = $λ_{i} [- (a_{i} + a_{i}^{'})]$ + (1 - λ_i) $(a_{i} + a_{i}^{'})$ = (1 -2λ_i) $(a_{i} + a_{i}^{'})$ . Therefore, from $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0})^{2}}{(a_{i} + a_{i}^{'})^{2}}$ ≤ 1, we know $\sum_{i = 1}^{n}$ $\frac{[(1 - 2 λ_{i}) (a_{i} + a_{i}^{'})]^{2}}{(a_{i} + a_{i}^{'})^{2}}$ ≤ 1, i.e., $\sum_{i = 1}^{n} (1 - 2 λ_{i})^{2} \leq 1$ (3) Let $y_{i}^{0}$ = λ_i (- a_i) + (1 - λ_i) a_i = (1 -2λ_i) a_i and $z_{i}^{0}$ = $λ_{i} (- a_{i}^{'})$ + $(1 - λ_{i}) a_{i}^{'}$ = $(1 - 2 λ_{i}) a_{i}^{'}$ , i = 1, 2, ⋯ , n. Then $x_{i}^{0}$ = $y_{i}^{0}$ + $z_{i}^{0}$ , i = 1, 2, ⋯ , n. Denote y⁰ = $(y_{1}^{0}$ , $y_{2}^{0}$ , ⋯, $y_{n}^{0})$ , z⁰ = $(z_{1}^{0}$ , $z_{2}^{0}$ , ⋯, $z_{n}^{0})$ , then we have x⁰ = y⁰ + z⁰. On the other hand, from Equation (3), we know $\sum_{i = 1}^{n}$ $\frac{(y_{i}^{0})^{2}}{a_{i}^{2}}$ = $\sum_{i = 1}^{n}$ $\frac{[(1 - 2 λ_{i}) a_{i}]^{2}}{a_{i}^{2}}$ = $\sum_{i = 1}^{n}$ (1 -2λ_i)² ≤ 1 and $\sum_{i = 1}^{n}$ $\frac{(z_{i}^{0})^{2}}{(a_{i}^{'})^{2}}$ = $\sum_{i = 1}^{n}$ $\frac{[(1 - 2 λ_{i}) a_{i}^{'}]^{2}}{(a_{i}^{'})^{2}}$ $= \sum_{i = 1}^{n}$ (1 -2λ_i)² ≤ 1, so y⁰ ∈ (- a_i, ${a_{i})}_{i = 1}^{n}$ , z⁰ ∈ $(- a_{i}^{'}$ , ${a_{i}^{'})}_{i = 1}^{n}$ . Thus, we have x⁰ = y⁰ + z⁰ ∈ (- a₁, ${a_{1})}_{i = 1}^{n}$ + $(- a_{i}^{'}$ , ${a_{i}^{'})}_{i = 1}^{n}$ . Therefore, (- (a_i + $a_{i}^{'})$ , a_i + ${a_{i}^{'})}_{i = 1}^{n}$ ⊂ (- a_i, ${a_{i})}_{i = 1}^{n}$ + $(- a_{i}^{'}$ , ${a_{i}^{'})}_{i = 1}^{n}$ holds. From (- a_i, ${a_{i})}_{i = 1}^{n}$ $\underset{\neq}{\subset}$ $\prod_{i = 1}^{n}$ [- a_i, a_i] and $(- a_{i}^{'}$ , ${a_{i}^{'})}_{i = 1}^{n}$ $\underset{\neq}{\subset}$ $\prod_{i = 1}^{n}$ $[- a_{i}^{'}$ , $a_{i}^{'}]$ , we have (- a_i, ${a_{i})}_{i = 1}^{n}$ + $(- a_{i}^{'}$ , ${a_{i}^{'})}_{i = 1}^{n}$ $\underset{\neq}{\subset}$ $\prod_{i = 1}^{n}$ [- a_i, a_i] + $\prod_{i = 1}^{n}$ $[- a_{i}^{'}$ , $a_{i}^{'}]$ = $\prod_{i = 1}^{n}$ [- (a_i + $a_{i}^{'})$ , a_i + $a_{i}^{'}]$ . Thus we have completed the proof of Equation (2). We can easily see that ${(\frac{y_{1} + x_{1}}{2}$ , $\frac{y_{2} + x_{2}}{2}$ , ⋯, $\frac{y_{n} + x_{n}}{2})}$ + $(- \frac{y_{i} - x_{i}}{2}$ , ${\frac{y_{i} - x_{i}}{2})}_{i = 1}^{n}$ = (x_i, ${y_{i})}_{i = 1}^{n}$ for any x_i, y_i ∈ R with x_i ≤ y_i, i = 1, 2, ⋯, n, so by Equation (2), we have $\begin{matrix} {(a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'})}_{i = 1}^{n} \\ = & {(\frac{b_{1} + a_{1} + b_{1}^{'} + a_{1}^{'}}{2}, \frac{b_{2} + a_{2} + b_{2}^{'} + a_{2}^{'}}{2}, \dots, \frac{b_{n} + a_{n} + b_{n}^{'} + a_{n}^{'}}{2})} \\ + {(- \frac{(b_{i} + b_{i}^{'}) - (a_{i} + a_{i}^{'})}{2}, \frac{(b_{i} + b_{i}^{'}) - (a_{i} + a_{i}^{'})}{2})}_{i = 1}^{n} \\ = & {(\frac{b_{1} + a_{1} + b_{1}^{'} + a_{1}^{'}}{2}, \frac{b_{2} + a_{2} + b_{2}^{'} + a_{2}^{'}}{2}, \dots, \frac{b_{n} + a_{n} + b_{n}^{'} + a_{n}^{'}}{2})} \\ + {(- \frac{(b_{i} - a_{i}) + (b_{i}^{'} - a_{i}^{'})}{2}, \frac{(b_{i} - a_{i}) + (b_{i}^{'} - a_{i}^{'})}{2})}_{i = 1}^{n} \\ \subset & {(\frac{b_{1} + a_{1}}{2}, \dots, \frac{b_{n} + a_{n}}{2})} + {(\frac{b_{1}^{'} + a_{1}^{'}}{2}, \dots, \frac{b_{n}^{'} + a_{n}^{'}}{2})} \\ + {(- \frac{b_{i} - a_{i}}{2}, \frac{b_{i} - a_{i}}{2})}_{i = 1}^{n} + {(- \frac{b_{i}^{'} - a_{i}^{'}}{2}, \frac{b_{i}^{'} - a_{i}^{'}}{2})}_{i = 1}^{n} \\ = & {(\frac{b_{1} + a_{1}}{2}, \frac{b_{2} + a_{2}}{2}, \dots, \frac{b_{n} + a_{n}}{2})} + {(- \frac{b_{i} - a_{i}}{2}, \frac{b_{i} - a_{i}}{2})}_{i = 1}^{n} \\ + {(\frac{b_{1}^{'} + a_{1}^{'}}{2}, \frac{b_{2}^{'} + a_{2}^{'}}{2}, \dots, \frac{b_{n}^{'} + a_{n}^{'}}{2})} + {(- \frac{b_{i}^{'} - a_{i}^{'}}{2}, \frac{b_{i}^{'} - a_{i}^{'}}{2})}_{i = 1}^{n} \\ = & {(a_{i}, b_{i})}_{i = 1}^{n} + {(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n} \end{matrix}$ and $\begin{matrix} {(a_{i}, b_{i})}_{i = 1}^{n} + {(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n} \\ = & {(\frac{b_{1} + a_{1}}{2}, \dots, \frac{b_{n} + a_{n}}{2})} + {(\frac{b_{1}^{'} + a_{1}^{'}}{2}, \dots, \frac{b_{n}^{'} + a_{n}^{'}}{2})} \\ + {(- \frac{b_{i} - a_{i}}{2}, \frac{b_{i} - a_{i}}{2})}_{i = 1}^{n} + {(- \frac{b_{i}^{'} - a_{i}^{'}}{2}, \frac{b_{i}^{'} - a_{i}^{'}}{2})}_{i = 1}^{n} \\ \underset{\neq}{\subset} & {(\frac{b_{1} + a_{1} + b_{1}^{'} + a_{1}^{'}}{2}, \frac{b_{2} + a_{2} + b_{2}^{'} + a_{2}^{'}}{2}, \dots, \frac{b_{n} + a_{n} + b_{n}^{'} + a_{n}^{'}}{2})} \\ + \prod_{i = 1}^{n} [- \frac{(b_{i} - a_{i}) + (b_{i}^{'} - a_{i}^{'})}{2}, \frac{(b_{i} - a_{i}) + (b_{i}^{'} - a_{i}^{'})}{2}] \\ = & \prod_{i = 1}^{n} [a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'}] \end{matrix}$ The proof of the lemma is completed.

Lemma 2.

The addition of two closed n-ellipsoids is not always a closed n-ellipsoid;

The scalar multiplication of a number and a closed n-ellipsoid must be still a closed n-ellipsoid, and

λ {(a_{i}, b_{i})}_{i = 1}^{n} = {\begin{matrix} {(λ a_{i}, λ b_{i})}_{i = 1}^{n} & if & λ \geq 0 \\ {(λ b_{i}, λ a_{i})}_{i = 1}^{n} & if & λ < 0 \end{matrix}

(4)

Proof. It is obvious that $(0, 1) \in (\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix}), (9, 0) \in (\begin{matrix} - 9 & 9 \\ - 1 & 1 \end{matrix})$ we know $(9, 1) = (0, 1) + (9, 0) \in (\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix}) + (\begin{matrix} - 9 & 9 \\ - 1 & 1 \end{matrix})$ But from $\frac{9^{2}}{10^{2}} + \frac{1^{2}}{2^{2}} = \frac{81}{100} + \frac{25}{100} = \frac{106}{100} > 1$ , we see $(9, 1) \notin (\begin{matrix} - 10 & 10 \\ - 2 & 2 \end{matrix}) = (\begin{matrix} (- 1) + (- 9) & 1 + 9 \\ (- 1) + (- 1) & 1 + 1 \end{matrix})$ so $(\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix}) + (\begin{matrix} - 9 & 9 \\ - 1 & 1 \end{matrix}) \neq (\begin{matrix} (- 1) + (- 9) & 1 + 9 \\ (- 1) + (- 1) & 1 + 1 \end{matrix})$ In the following, we use counterevidence to prove Conclusion (1): Assume Conclusion (1) is not valid, then for any two closed n-ellipsoids ${(a_{i}, b_{i})}_{i = 1}^{n}$ and $(a_{i}^{'}, b_{i}^{'}) i = 1 n, (a_{i}, b_{i}) i = 1 n + (a_{i}^{'}, b_{i}^{'}) i = 1 n$ is a closed n-ellipsoid, so from $(a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'}) i = 1 n \subset (a_{i}, b_{i}) i = 1 n$ + ${(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n}$ ⊂ $\prod_{i = 1}^{n} [a_{i}$ + $a_{i}^{'}$ , b_i + $b_{i}^{'}]$ (Lemma 1), we know that ${(a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'})}_{i = 1}^{n}$ and ${(a_{i}, b_{i})}_{i = 1}^{n} + {(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n}$ are the (same) closed n-ellipsoid with the center $(\frac{a_{1} + a_{1}^{'} + b_{1} + b_{1}^{'}}{2}, \dots, \frac{a_{n} + a_{n}^{'} + b_{n} + b_{n}^{'}}{2})$ and the coordinates of the vertexes on coordinate axis x₁, x₂, ⋯, x_n are in turn $a_{1} + a_{1}^{'}, b_{1} + b_{1}^{'}$ , $a_{2} + a_{2}^{'}, b_{2} + b_{2}^{'}$ , ⋯, $a_{n} + a_{n}^{'}, b_{n} + b_{n}^{'}$ . Therefore, we obtain that ${(a_{i}, b_{i})}_{i = 1}^{n} + {(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n} = {(a_{i} + a_{i}^{'}, b_{i} + b_{i}^{'})}_{i = 1}^{n}$ for any two closed n-ellipsoids ${(a_{i}, b_{i})}_{i = 1}^{n}$ and ${(a_{i}^{'}, b_{i}^{'})}_{i = 1}^{n}$ . This is a contradiction with $(\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix}) + (\begin{matrix} - 9 & 9 \\ - 1 & 1 \end{matrix}) \neq (\begin{matrix} (- 1) + (- 9) & 1 + 9 \\ (- 1) + (- 1) & 1 + 1 \end{matrix})$ so the conclusion (1) of the theorem holds.

Let λ > 0. If $x = (x_{1}, x_{2}, \dots, x_{n}) \in λ {(a_{i}, b_{i})}_{i = 1}^{n}$ , then there exist $y = (y_{1}, y_{2}, \dots, y_{n}) \in {(a_{i}, b_{i})}_{i = 1}^{n}$ such that x = λy, i.e., x_i = λy_i (i = 1, 2, ⋯ , n). From $\sum_{i = 1}^{n} \frac{(y_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i})}{2})^{2}} \leq 1$ , we see $\sum_{i = 1}^{n} \frac{(x_{i} - \frac{λ b_{i} + λ a_{i}}{2})^{2}}{(\frac{λ b_{i} - λ a_{i})}{2})^{2}}$ $= \sum_{i = 1}^{n} \frac{(λ y_{i} - \frac{λ b_{i} + λ a_{i}}{2})^{2}}{(\frac{λ b_{i} - λ a_{i})}{2})^{2}} = \sum_{i = 1}^{n} \frac{(y_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i})}{2})^{2}} \leq 1$ , so $x = (x_{1}, x_{2}, \dots, x_{n}) \in {(λ a_{i}, λ b_{i})}_{i = 1}^{n}$ . Conversely, if $x = (x_{1}, x_{2}, \dots, x_{n}) \in {(λ a_{i}, λ b_{i})}_{i = 1}^{n}$ , then $\sum_{i = 1}^{n} \frac{(x_{i} - \frac{λ b_{i} + λ a_{i}}{2})^{2}}{(\frac{λ b_{i} - λ a_{i})}{2})^{2}} \leq 1$ , so $\sum_{i = 1}^{n} \frac{(\frac{1}{λ} x_{i} - \frac{b_{i} + a_{i}}{2})^{2}}{(\frac{b_{i} - a_{i})}{2})^{2}} \leq 1$ , i.e., $(\frac{1}{λ} x_{1}, \frac{1}{λ} x_{2}, \dots, \frac{1}{λ} x_{n}) \in {(a_{i}, b_{i})}_{i = 1}^{n}$ . Therefore, we have that $x = (x_{1}, \dots, x_{n}) = λ (\frac{1}{λ} x_{1}, \dots, \frac{1}{λ} x_{n})$ $\in λ {(a_{i}, b_{i})}_{i = 1}^{n}$ . Thus, we know that Equation (3) holds as λ > 0. With the same method, we can show that Equation (3) holds as λ < 0. In addition, it is obvious that Equation (3) holds as λ = 0. The proof of the theorem is completed.

Theorem 3.

u, v ∈ E (Eⁿ) does not necessarily implies u + v ∈ E (Eⁿ), i.e., for fuzzy n-ellipsoid numbers the addition “+” does not preserve the closeness of the operation.

k ∈ R and u ∈ E (Eⁿ) must imply ku ∈ E (Eⁿ), and for any r ∈ [0, 1]

(\underline{(ku)_{i}} (r), \bar{(ku)_{i}} (r)) i = 1 n = {\begin{matrix} (k \underline{u_{i}} (r), k \bar{u_{i}} (r)) i = 1 n, & k \geq 0 \\ (k \bar{u_{i}} (r), k \underline{u_{i}} (r)) i = 1 n, & k < 0 \end{matrix}

(5)

Proof. From [u + v]^r= [u]^r+ [v]^r and [ku]^r= k [u]^r for any u, v ∈ Eⁿ, k ∈ R and r ∈ [0, 1], we can easily see the validity of (1) and (2) of the theorem by the (1) and (2) of Lemma 2, respectively.

Theorem 4. Let u, v ∈ E (Eⁿ). Then ${(\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r))}_{i = 1}^{n}, r \in [0, 1]$ decides a unique fuzzy n-ellipsoid number, i.e., there exists a unique w ∈ E (Eⁿ) such that $[w]^{r} = {(\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r))}_{i = 1}^{n}$ for any r ∈ [0, 1].

Proof. From $\underline{u_{i}} (r)$ , $\bar{u_{i}} (r)$ (i = 1, 2, ⋯ , n) and $\underline{v_{i}} (r)$ , $\bar{v_{i}} (r)$ (i = 1, 2, ⋯ , n) satisfying respectively the conditions (1)-(4) of Theorem 2, we see that $\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r), i = 1, 2, \dots, n$ satisfies the conditions (1)-(4) of Theorem 2. Therefore, by Theorem 2, we know that there exists a unique w ∈ E (Eⁿ) such that $[w]^{r} = {(x_{1}, \dots, x_{n}) \in R^{n} | \sum_{i = 1}^{n} \frac{(x_{i} - \frac{\bar{u_{i}} (r) + \bar{v_{i}} (r) + \underline{u_{i}} (r) - \underline{v_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) + \bar{v_{i}} (r) - \underline{u_{i}} (r) + \underline{v_{i}} (r)}{2})^{2}} \leq 1}$ i.e., [w]^r = $(\underline{u_{i}} (r)$ + $\underline{v_{i}} (r)$ , $\bar{u_{i}} (r)$ + ${\bar{v_{i}} (r))}_{i = 1}^{n}$ for any r ∈ [0, 1].

Definition 1. For any u, v ∈ E (Eⁿ), we define the addition u ⊕ v of u and v is the fuzzy n-ellipsoid number which is decided by ${(\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r))}_{i = 1}^{n}, r \in [0, 1]$

Theorem 5. If u, v ∈ E (Eⁿ), then u ⊕ v ∈ E (Eⁿ), and for any r ∈ [0, 1] $[u \oplus v]^{r} = {(\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r))}_{i = 1}^{n}$ (6)i.e., $\underline{u_{i} \oplus v_{i}} (r) = \underline{u_{i}} (r) + \underline{v_{i}} (r) \bar{u_{i} \oplus v_{i}} (r) = \bar{u_{i}} (r) + \bar{v_{i}} (r)$ for any i = 1, 2, ⋯ , n.

Proof. From the definition of addition u ⊕ v of u and v, the theorem obviously holds.

In the following, we give a result about the relation of fuzzy n-ellipsoid number u ⊕ v and fuzzy subset u + v of Rⁿ.

Theorem 6. Let u, v ∈ E (Eⁿ). Then

u ⊕ v ⊂ u + v;

u ⊕ v = u + v as u + v ∈ E (Eⁿ).

Proof. From u, v ∈ E (Eⁿ), we know that there exist $\underline{u_{i}} (r), \bar{u_{i}} (r), \underline{v_{i}} (r), \bar{v_{i}} (r) \in R$ with $\underline{u_{i}} (r) \leq \bar{u_{i}} (r)$ and $\underline{v_{i}} (r) \leq \bar{v_{i}} (r)$ such that $[u]^{r} = {(\underline{u_{i}} (r), \bar{u_{i}} (r))}_{i = 1}^{n}$ and $[v]^{r} = {(\underline{v_{i}} (r), \bar{v_{i}} (r))}_{i = 1}^{n}$ for any r ∈ [0, 1], i = 1, 2, …, n. By Theorem 5 and Lemma 1, we know that $\begin{matrix} [u \oplus v]^{r} = & {(\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r))}_{i = 1}^{n} \\ \subset & {(\underline{u_{i}} (r), \bar{u_{i}} (r))}_{i = 1}^{n} + {(\underline{v_{i}} (r), \bar{v_{i}} (r))}_{i = 1}^{n} \\ = & [u]^{r} + [v]^{r} \\ = & [u + v]^{r} \end{matrix}$ for any r ∈ [0, 1]. This implies (u ⊕ v) (x) ≤ (u + v) (x) for any x ∈ Rⁿ, i.e., u ⊕ v ⊂ u + v.

If u + v ∈ E (Eⁿ), then for any r ∈ [0, 1], [u + v]^r is always a closed n-ellipsoid. Pay attention to the fact that the two closed n-ellipsoids [u ⊕ v]^r and [u + v]^r possess the same vertexes

$\begin{matrix} (\underline{u_{1}} (r) + \underline{v_{1}} (r), 0, \dots, 0) \\ (\bar{u_{1}} (r) + \bar{v_{1}} (r), 0, \dots, 0) \\ ⋮ \\ (0, 0, \dots, \underline{u_{n}} (r) + \underline{v_{n}} (r)) \\ (0, 0, \dots, \bar{u_{n}} (r) + \bar{v_{n}} (r)) \end{matrix}$ we know [u ⊕ v]^r= [u + v]^r, r ∈ [0, 1]. Thus, we see u ⊕ v = u + v. The proof is completed.

Remark 1. For u, v ∈ E (Eⁿ), by Lemma 1, we know that ${(\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r))}_{i = 1}^{n}$ is the best approximation n-ellipsoid of ${(\underline{u_{i}} (r), \bar{u_{i}} (r))}_{i = 1}^{n} + {(\underline{v_{i}} (r), \bar{v_{i}} (r))}_{i = 1}^{n}$ (i.e., [u + v]^r= [u]^r+ [v]^r) in the all n-ellipsoids which are contained in $\prod_{i = 1}^{n} [\underline{u_{i}} (r) + \underline{v_{i}} (r), \bar{u_{i}} (r) + \bar{v_{i}} (r)]$ . So the addition operation “⊕” defined in this paper is the best approximation of the general addition operation “+” in the all addition operations which preserve the closeness of the operation for fuzzy n-ellipsoid numbers.

4 Joint membership function and edge membership functions

For any u ∈ E (Eⁿ), by the definition of fuzzy n-ellipsoid numbers, we know that for any r ∈ [0, 1], [u]^r is a n-ellipsoid, i.e., exist $\underline{u_{i}} (r), \bar{u_{i}} (r) \in R$ with $\underline{u_{i}} (r) \leq \bar{u_{i}} (r)$ , i = 1, 2, …, n, such that $[u]^{r} = {(x_{1}, x_{2}, \dots, x_{n}) \in R^{n} | \sum_{i = 1}^{n} \frac{(x_{i} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} \leq 1}$ , i.e., $[u]^{r} = {(\underline{u_{i}} (r), \bar{u_{i}} (r))}_{i = 1}^{n}$ . By Theorem 2, we know that $\underline{u_{i}} (r)$ , $\bar{u_{i}} (r)$ , i = 1, 2, ⋯ , n satisfy the conditions (1)-(4) in Theorem 2. Furthermore, by the Theorem 3.1 (there n = 1) in [36], we see that for each pair (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ, there exists a unique u_i ∈ E such that $[u_{i}]^{r} = [\underline{u_{i}} (r), \bar{u_{i}} (r)]$ for any r ∈ [0, 1]. Therefore, for any u ∈ E (Eⁿ), there exist n unique one-dimensional fuzzy numbers u₁, u₂, ⋯ , u_n such that $[u_{i}]^{r} = [\underline{u_{i}} (r), \bar{u_{i}} (r)]$ for any r ∈ [0, 1] and i = 1, 2, ⋯ , n. According to this corresponding relation, each u ∈ E (Eⁿ) determines a unique ordered one-dimension fuzzy number class u₁, u₂, ⋯ , u_n ∈ E, i.e., determines a unique n-dimension fuzzy vector (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ.

Conversely, for any (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ, by the Theorem 3.1 (there n = 1) in [36], we know that $\underline{u_{i}} (r)$ , $\bar{u_{i}} (r)$ , i = 1, 2, ⋯ , n satisfy the conditions (1)-(4) in Theorem 2 since u_i ∈ E (i = 1, 2, ⋯ , n), so by Theorem 2, we see that there exists a unique u ∈ E (Eⁿ) such that $[u]^{r} = {(\underline{u_{i}} (r), \bar{u_{i}} (r))}_{i = 1}^{n}$ for any r ∈ [0, 1].

Therefore, we have the following result:

Theorem 7. For any u ∈ E (Eⁿ) with $[u]^{r} = (\underline{u_{i}} (r),$ $\bar{u_{i}} (r))_{i = 1}^{n}, r \in [0, 1]$ , there exists a unique (u₁, ⋯ , u_n) ∈ (E)ⁿ such that $[u_{i}]^{r} = [\underline{u_{i}} (r), \bar{u_{i}} (r)], i = 1, 2, \dots, n, r \in [0, 1]$

Conversely, for any (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ with $[u_{i}]^{r} = [\underline{u_{i}} (r), \bar{u_{i}} (r)]$ (i = 1, 2, ⋯ , n, r ∈ [0, 1]), there exists a unique u ∈ E (Eⁿ) such that $[u]^{r} = {(\underline{u_{i}} (r), \bar{u_{i}} (r))}_{i = 1}^{n}, r \in [0, 1]$

Remark 2. Theorem 7 tell us that according to this corresponding relation (denote it as I), fuzzy n-ellipsoid numbers and n-dimensional fuzzy vectors can uniquely determine each other. So, if u ∈ E (Eⁿ) and (u₁, ⋯ , u_n) ∈ (E)ⁿ determines uniquely each other, i.e., I (u) = (u₁, ⋯ , u_n) or I^-1(u₁, ⋯ , u_n) = u, we can use the each one of u and (u₁, ⋯ , u_n) to represent the other one, i.e., we can directly denote u = (u₁, ⋯ , u_n) for the sake of simplicity.

By the definition of the addition “⊕” and scalar product rule of fuzzy n-ellipsoid numbers and the addition “+” rule of 1-dimensional fuzzy numbers, we can obtained directly the following result.

Proposition 1. Let u, v ∈ E (Eⁿ), (u₁, u₂, ⋯ , u_n), (v₁, v₂, ⋯ , v_n) ∈ (E)ⁿ with u = (u₁, u₂, ⋯ , u_n) and u = (v₁, v₂, ⋯ , v_n), and k ∈ R. Then

ku = (ku₁, ku₂, ⋯ , ku_n);

u ⊕ v = (u₁ + v₁, u₂ + v₂, ⋯ , u_n + v_n).

Definition 2. Let u ∈ E (Eⁿ) and (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ with u = (u₁, u₂, ⋯ , u_n). We call u : Rⁿ→ [0, 1] the joint membership function (with respect to corresponding relation I) of u₁, u₂, ⋯ , u_n, and call u_i : R → [0, 1] the ith edge membership functionof u.

In the following, we study the direct relationship between the joint membership function and the edge membership functions.

Theorem 8. Let u ∈ E (Eⁿ) and (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ with u = (u₁, u₂, ⋯ , u_n). Then for any $x_{i}^{0} \in R$ , i = 1, 2, ⋯ , n, we have

$u_{i} (x_{i}^{0}) = max_{\begin{matrix} x_{j} \in R, j = 1, \dots, \\ i - 1, i + 1, \dots, n \end{matrix}} u (x_{1}, \dots, x_{i - 1}, x_{i}^{0}, x_{i + 1}, \dots, x_{n})$

Proof. For any i = 1, 2, ⋯ , n and $x_{i}^{0} \in R$ , from

$\begin{matrix} {(x_{1}, x_{2}, \dots, x_{n}) \in R^{n} : u (x_{1}, x_{2}, \dots, x_{n}) \geq r} \\ = [u]^{r} \\ = {(x_{1}, \dots, x_{n}) \in R^{n} : \sum_{j = 1}^{n} \frac{(x_{j} - \frac{\bar{u_{j}} (r) + \underline{u_{j}} (r)}{2})^{2}}{(\frac{\bar{u_{j}} (r) - \underline{u_{j}} (r)}{2})^{2}} \leq 1} \\ \subset {(x_{1}, \dots, x_{n}) \in R^{n} : \underline{u_{i}} (r) \leq x_{i} \leq \bar{u_{i}} (r), x_{1}, \dots, \\ x_{i - 1}, x_{i + 1}, \dots, x_{n} \in R} \\ = {(x_{1}, \dots, x_{n}) \in R^{n} : x_{i} \in [u_{i}]^{r}, x_{1}, \dots, \\ x_{i - 1}, x_{i + 1}, \dots, x_{n} \in R} \\ = {(x_{1}, \dots, x_{n}) \in R^{n} : u_{i} (x_{i}) \geq r, x_{1}, \dots, \\ x_{i - 1}, x_{i + 1}, \dots, x_{n} \in R} \end{matrix}$ for any r ∈ [0, 1], we know that for any r ∈ [0, 1] and x₁, ⋯, x_i-1, x_i+1, ⋯, x_n ∈ R, if u (x₁, ⋯, x_i-1, $x_{i}^{0},$ x_i+1, ⋯, x_n) ≥ r, then $(x_{1}, \dots, x_{i - 1}, x_{i}^{0}, x_{i + 1}, \dots, x_{n})$ ∈ {(x₁, ⋯, x_n) ∈ Rⁿ: u_i (x_i) ≥ r, x_j ∈ R, j = 1, ⋯, i - 1, i + 1, ⋯, n}, so $u_{i} (x_{i}^{0})$ ≥ r holds. This implies that $u_{i} (x_{i}^{0})$ ≥ u (x₁, ⋯, x_i-1, $x_{i}^{0}$ , x_i+1, ⋯, x_n) for any x₁, ⋯, x_i-1, x_i+1, ⋯, x_n ∈ R, so we have $u_{i} (x_{i}^{0}) \geq max_{\begin{matrix} x_{j} \in R, j = 1, \dots, \\ i - 1, i + 1, \dots, n \end{matrix}} u (x_{1}, \dots, x_{i - 1}, x_{i}^{0}, x_{i + 1}, \dots, x_{n})$ (7)

On the other hand, for any fixed i = 1, 2, ⋯ , n and $x_{i}^{0} \in R$ , we denote $u_{i} (x_{i}^{0})$ by r_{i
₀}, so we have $x_{i}^{0} \in [u_{i}]^{r_{i_{0}}}$ , i.e., $\underline{u_{i}} (r_{i_{0}}) \leq x_{i}^{0} \leq \bar{u_{i}} (r_{i_{0}})$ . It implies $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{i_{0}}) + \underline{u_{i}} (r_{i_{0}})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{i_{0}}) - \underline{u_{i}} (r_{i_{0}})}{2})^{2}} \leq 1$ , so $\begin{matrix} \sum_{j = 1}^{i - 1} \frac{(\frac{\bar{u_{j}} (r_{i_{0}}) + \underline{u_{j}} (r_{i_{0}})}{2} - \frac{\bar{u_{j}} (r_{i_{0}}) + \underline{u_{j}} (r_{i_{0}})}{2})^{2}}{(\frac{\bar{u_{j}} (r_{i_{0}}) - \underline{u_{j}} (r_{i_{0}})}{2})^{2}} \\ + \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{i_{0}}) + \underline{u_{i}} (r_{i_{0}})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{i_{0}}) - \underline{u_{i}} (r_{i_{0}})}{2})^{2}} \\ + \sum_{j = i + 1}^{n} \frac{(\frac{\bar{u_{j}} (r_{i_{0}}) + \underline{u_{j}} (r_{i_{0}})}{2} - \frac{\bar{u_{j}} (r_{i_{0}}) + \underline{u_{j}} (r_{i_{0}})}{2})^{2}}{(\frac{\bar{u_{j}} (r_{i_{0}}) - \underline{u_{j}} (r_{i_{0}})}{2})^{2}} \\ \leq 1 \end{matrix}$ So we see $(\frac{\bar{u_{1}} (r_{i_{0}}) + \underline{u_{1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{i - 1}} (r_{i_{0}}) + \underline{u_{i - 1}} (r_{i_{0}})}{2},$ $x_{i}^{0}, \frac{\bar{u_{i + 1}} (r_{i_{0}}) + \underline{u_{i + 1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{n}} (r_{i_{0}}) + \underline{u_{n}} (r_{i_{0}})}{2}) \in [u]^{r_{i_{0}}}$ , i.e., $u (\frac{\bar{u_{1}} (r_{i_{0}}) + \underline{u_{1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{i - 1}} (r_{i_{0}}) + \underline{u_{i - 1}} (r_{i_{0}})}{2}, x_{i}^{0},$ $\frac{\bar{u_{i + 1}} (r_{i_{0}}) + \underline{u_{i + 1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{n}} (r_{i_{0}}) + \underline{u_{n}} (r_{i_{0}})}{2}) \geq r_{i_{0}} = u_{i} (x_{i}^{0})$ . From this, we have $\begin{matrix} u_{i} (x_{i}^{0}) \\ \leq u (\frac{\bar{u_{1}} (r_{i_{0}}) + \underline{u_{1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{i - 1}} (r_{i_{0}}) + \underline{u_{i - 1}} (r_{i_{0}})}{2}, x_{i}^{0}, \\ \frac{\bar{u_{i + 1}} (r_{i_{0}}) + \underline{u_{i + 1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{n}} (r_{i_{0}}) + \underline{u_{n}} (r_{i_{0}})}{2}) \\ \underset{\neq}{\subset} \begin{matrix} x_{j} \in R, j = 1, \dots, \\ i - 1, i + 1, \dots, n \end{matrix} sup u (x_{1}, \dots, x_{i - 1}, x_{i}^{0}, x_{i + 1}, \dots, x_{n}) \\ \leq u_{i} (x_{i}^{0}) \end{matrix}$ so $\begin{matrix} u_{i} (x_{i}^{0}) \\ = u (\frac{\bar{u_{1}} (r_{i_{0}}) + \underline{u_{1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{i - 1}} (r_{i_{0}}) + \underline{u_{i - 1}} (r_{i_{0}})}{2}, x_{i}^{0}, \\ \frac{\bar{u_{i + 1}} (r_{i_{0}}) + \underline{u_{i + 1}} (r_{i_{0}})}{2}, \dots, \frac{\bar{u_{n}} (r_{i_{0}}) + \underline{u_{n}} (r_{i_{0}})}{2}) \\ = \underset{\neq}{\subset} \begin{matrix} x_{j} \in R, j = 1, \dots, \\ i - 1, i + 1, \dots, n \end{matrix} sup u (x_{1}, \dots, x_{i - 1}, x_{i}^{0}, x_{i + 1}, \dots, x_{n}) \end{matrix}$ go a step further, we have $u_{i} (x_{i}^{0}) = max_{\begin{matrix} x_{j} \in R, j = 1, \dots, \\ i - 1, i + 1, \dots, n \end{matrix}} u (x_{1}, \dots, x_{i - 1}, x_{i}^{0}, x_{i + 1}, \dots, x_{n})$

Remark 3. In [13] (see the page 365 in [13]), Full $\overset{´}{e}$ r and Majlander gave the concept of the joint possibility distribution of a set of fuzzy numbers: A fuzzy set B in R^m is said to be a joint possibility distribution of fuzzy numbers A_i ∈ E, i = 1, 2, ⋯ , m, if it satisfies the relationship $max_{x_{j} \in R, j \neq i} B (x_{1}, \dots, x_{m}) = A_{i} (x_{i}), \forall x_{i} \in R, i = 1, 2, \dots, m$ . By Theorem 8 in this paper and Theorem 3.2 in [30], we see that both the fuzzy m-ellipsoid number and the fuzzy m-cell number which’s components are A_i ∈ E, i = 1, 2, ⋯ , m satisfy the condition of the definition of the joint possibility distribution, so the fuzzy m-ellipsoid number and the fuzzy m-cell number are all the joint possibility distribution of fuzzy numbers A_i ∈ E, i = 1, 2, ⋯ , m. Therefore, according to Full $\overset{´}{e}$ r and Majlander’s definition of the joint possibility distribution, for a given set of fuzzy numbers A_i ∈ E, i = 1, 2, ⋯ , m, their joint possibility distribution is not unique.

Theorem 9. Let u ∈ E (Eⁿ) and (u₁, u₂, ⋯ , u_n) ∈ (E)ⁿ with u = (u₁, ⋯ , u_n), and denote $[\underline{u_{i}} (r), \bar{u_{i}} (r)]$ = [u_i]^r for any i = 1, 2, ⋯ , n and r ∈ [0, 1]. Then for any $x_{0} = (x_{1}^{0}, x_{2}^{0}, \dots, x_{n}^{0}) \in supp u \ [u]^{1}$ (= {x ∈ Rⁿ: 0 < u (x) <1}), there exists a unique r₀ ∈ (0, 1) such that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{0}) + \underline{u_{i}} (r_{0})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{0}) - \underline{u_{i}} (r_{0})}{2})^{2}} = 1$ or there exist r₁, r₂ ∈ (0, 1) with r₁ < r₂ such that

$\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} = 1$ for anyr ∈ (r₁, r₂);

$\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} < 1$ for anyr ∈ [0, r₁);

$\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ for anyr ∈ (r₂, 1].

Proof. From $x_{0} = (x_{1}^{0}, \dots, x_{n}^{0}) \in supp u \ [u]^{1}$ , we know $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (0) + \underline{u_{i}} (0)}{2})^{2}}{(\frac{\bar{u_{i}} (0) - \underline{u_{i}} (0)}{2})^{2}} < 1$ and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (1) + \underline{u_{i}} (1)}{2})^{2}}{(\frac{\bar{u_{i}} (1) - \underline{u_{i}} (1)}{2})^{2}} > 1$ .

Firstly, we show that there exists s₀ ∈ (0, 1) such that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{0}) + \underline{u_{i}} (s_{0})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{0}) - \underline{u_{i}} (s_{0})}{2})^{2}} = 1$ (8) Denoting a₁ = 0, b₁ = 1, taking λ ∈ (0, 1), and we denote s₁ = λa₁ + (1 - λ) b₁, then a₁ < b₁, s₁ ∈ (a₁, b₁), $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{1}) + \underline{u_{i}} (a_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{1}) - \underline{u_{i}} (a_{1})}{2})^{2}} < 1, \sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{1}) + \underline{u_{i}} (b_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{1}) - \underline{u_{i}} (b_{1})}{2})^{2}} > 1$ . If $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}} = 1$ , we take s₀ = s₁, then Equality (8) holds; If $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}} > 1$ , we denote a₂ = a₁, b₂ = s₁ and s₂ = λa₂ + (1 - λ) b₂; If $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}} < 1$ , we denote a₂ = s₁, b₂ = 1 and s₂ = λa₂ + (1 - λ) b₂. Therefore, if $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}} \neq 1$ , we can obtain real numbers a₂, b₂ and s₂ with a₂ < b₂, s₂ ∈ (a₂, b₂), b₂ - a₂ ≤ max {λ, (1 - λ)}, [a₂, b₂] ⊂ [a₁, b₁] (= [0, 1]) and $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{2}) + \underline{u_{i}} (a_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{2}) - \underline{u_{i}} (a_{2})}{2})^{2}} < 1,$ $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{2}) + \underline{u_{i}} (b_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{2}) - \underline{u_{i}} (b_{2})}{2})^{2}} > 1$ ; If $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{2}) + \underline{u_{i}} (s_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{2}) - \underline{u_{i}} (s_{2})}{2})^{2}}$ = 1, we take s₀ = s₂, then Equality (8) holds; If $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{2}) + \underline{u_{i}} (s_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{2}) - \underline{u_{i}} (s_{2})}{2})^{2}} > 1$ , we denote a₃ = a₂, b₃ = s₂ and s₃ = λa₃ + (1 - λ) b₃; If $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{2}) + \underline{u_{i}} (s_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{2}) - \underline{u_{i}} (s_{2})}{2})^{2}}$ <1, we denote a₃ = s₂, b₃ = b₂ and s₃ = λa₃ + (1 - λ) b₃. Therefore, if $\sum_{i = 1}^{n}$ $\frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{2}) + \underline{u_{i}} (s_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{2}) - \underline{u_{i}} (s_{2})}{2})^{2}} \neq 1$ , we can obtain real numbers a₃, b₃ and s₃ with a₃ < b₃, s₃ ∈ (a₃, b₃) , b₃ - a₃ ≤ (max {λ, (1 - λ)})², [a₃, b₃] ⊂ [a₂, b₂] and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{3}) + \underline{u_{i}} (a_{3})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{3}) - \underline{u_{i}} (a_{3})}{2})^{2}} < 1, \sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{3}) + \underline{u_{i}} (b_{3})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{3}) - \underline{u_{i}} (b_{3})}{2})^{2}} > 1$ , ⋯. Keep the discussing like this, after the kth discussing, we can obtain s₀ = s_k with Equality (8), or real numbers a_k+1, b_k+1 and s_k+1 with a_k+1 < b_k+1, s_k+1 ∈ (a_k+1, b_k+1) , b_k+1 - a_k+1 ≤ (max {λ, (1 - λ)})^k, [a_k+1, b_k+1] ⊂ [a_k, b_k] and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{k + 1}) + \underline{u_{i}} (a_{k + 1})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{k + 1}) - \underline{u_{i}} (a_{k + 1})}{2})^{2}} < 1$ and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{k + 1}) + \underline{u_{i}} (b_{k + 1})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{k + 1}) - \underline{u_{i}} (b_{k + 1})}{2})^{2}} > 1$ Therefore, if there is no a natural number k such that s₀ = s_k with Equality (8), then for any natural number k, there are real numbers a_k+1, b_k+1 and s_k+1 with a_k+1 < b_k+1, s_k+1 ∈ (a_k+1, b_k+1), b_k+1 - a_k+1 ≤ (max {λ, (1 - λ)}))^k, [a_k+1, b_k+1] ⊂ [a_k, b_k] and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{k + 1}) + \underline{u_{i}} (a_{k + 1})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{k + 1}) - \underline{u_{i}} (a_{k + 1})}{2})^{2}} < 1$ and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{k + 1}) + \underline{u_{i}} (b_{k + 1})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{k + 1}) - \underline{u_{i}} (b_{k + 1})}{2})^{2}} > 1$ so by Theorem of nested closed interval and Squeeze Theorem, we know there exists real number $\tilde{s}$ such that $lim_{k \to \infty} a_{k} = lim_{k \to \infty} s_{k} = lim_{k \to \infty} b_{k} = \tilde{s}$ . Taking $s_{0} = \tilde{s}$ , we can see that s₀ ∈ (0, 1) and Equality (8) hold. Thus, we have shown that there exists s₀ ∈ (0, 1) such that Equality (8) holds.

If s₀ is unique number in (0, 1) which satisfy Equality (8), we take r₀ = s₀, then the theorem holds.

If s₀ is not unique number in (0, 1) which satisfy Equality (8), in the following, we show the existence of r₁ and s₂ which satisfy the conclusions of the theorem. Denoting a₁ = s₀, b₁ = 1, and taking λ ∈ (0, 1), we denote s₁ = λa₁ + (1 - λ) b₁, then $a_{1} < b_{1}, s_{1} \in (a_{1}, b_{1}), b_{1} - a_{1} \leq s_{0}, \sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{1}) + \underline{u_{i}} (a_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{1}) - \underline{u_{i}} (a_{1})}{2})^{2}} = 1,$ $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{1}) + \underline{u_{i}} (b_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{1}) - \underline{u_{i}} (b_{1})}{2})^{2}} > 1$ . If $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}}$ = 1, then we take a₂ = s₁, b₂ = b₁, and denote s₂ = λa₂ + (1 - λ) b₂; If $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}} \neq 1$ , then $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (s_{1}) + \underline{u_{i}} (s_{1})}{2})^{2}}{(\frac{\bar{u_{i}} (s_{1}) - \underline{u_{i}} (s_{1})}{2})^{2}} > 1$ (since s₁ > a₁), we take a₂ = a₁, b₂ = s₁, and denote s₂ = λa₂ + (1 - λ) b₂. Thus we can obtain a₂, b₂ and s₂ with a₂ < b₂, s₂ ∈ (a₂, b₂) , b₂ - a₂ ≤ max {λ, (1 - λ)} s₀, [a₂, b₂] ⊂ [a₁, b₁] and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{2}) + \underline{u_{i}} (a_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{2}) - \underline{u_{i}} (a_{2})}{2})^{2}} = 1, \sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{2}) + \underline{u_{i}} (b_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{2}) - \underline{u_{i}} (b_{2})}{2})^{2}}$ >1, ⋯. Keep the discussing like this, we can obtain a_k+1, b_k+1 and s_k+1 with a_k+1 < b_k+1, s_k+1 ∈ (a_k+1, b_k+1) , b_k+1 - a_k+1 ≤ (max {λ, (1 - λ)})^ks₀, [a_k+1, b_k+1] ⊂ [a_k, b_k] and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (a_{k + 1}) + \underline{u_{i}} (a_{k + 1})}{2})^{2}}{(\frac{\bar{u_{i}} (a_{k + 1}) - \underline{u_{i}} (a_{k + 1})}{2})^{2}} = 1$ and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (b_{k + 1}) + \underline{u_{i}} (b_{k + 1})}{2})^{2}}{(\frac{\bar{u_{i}} (b_{k + 1}) - \underline{u_{i}} (b_{k + 1})}{2})^{2}} > 1$ for any k = 1, 2, ⋯. By Theorem of nested closed interval and Squeeze Theorem, we know there exists real number $\tilde{s}$ such that $lim_{k \to \infty} a_{k} = lim_{k \to \infty} s_{k} = lim_{k \to \infty} b_{k} = \tilde{s}$ . Take $r_{2} = \tilde{s}$ . If $\tilde{s} = s_{0}$ , then we can see that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} = 1$ as r = r₂ ( = s₀), and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ as r ∈ (r₂, 1]; If $\tilde{s} \neq s_{0}$ , then we can see that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} = 1$ as r ∈ [s₀, r₂), and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ as r ∈ (r₂, 1].

Likewise, we can show there exists $r_{1} = \overset{ˇ}{s}$ such that if $\overset{ˇ}{s} = s_{0}$ , then $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} = 1$ as r = r₁ (= s₀), and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ as r ∈ [0, r₁), if $\overset{ˇ}{s} \neq s_{0}$ , then $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} = 1$ as r ∈ (r₁, s₀], and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ as r ∈ [0, r₁). Thus, the proof is completed.

Theorem 10. Let u ∈ E (Eⁿ) and (u₁, ⋯ , u_n) ∈ (E)ⁿ with u = (u₁, ⋯ , u_n), and denote $[\underline{u_{i}} (r), \bar{u_{i}} (r)] = [u_{i}]^{r}$ for any i = 1, 2, ⋯ , n and r ∈ [0, 1]. Then u (x₀) = r₀ or u (x₀) = r₂ for any $x_{0} = (x_{1}^{0}, \dots, x_{n}^{0})$ ∈suppu \ [u]¹ (= {x ∈ Rⁿ: 0 < u (x) <1}), where r₀ and r₂ are respectively the r₀ and r₂ in Theorem 9.

Proof. By Theorem 9, we see that for any $x_{0} = (x_{1}^{0}, x_{2}^{0}, \dots, x_{n}^{0}) \in supp u \ [u]^{1}$ , there exists a unique r₀ ∈ (0, 1) such that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{0}) + \underline{u_{i}} (r_{0})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{0}) - \underline{u_{i}} (r_{0})}{2})^{2}} = 1$ or there exist r₁, r₂ ∈ (0, 1) with r₁ < r₂ such that the (1)-(3) in Theorem 9 hold.

If there exists a unique r₀ ∈ (0, 1) such that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{0}) + \underline{u_{i}} (r_{0})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{0}) - \underline{u_{i}} (r_{0})}{2})^{2}} = 1$ From the uniqueness of r₀, we can show that $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} < 1$ for any r ∈ [0, r₀) and $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ for any r ∈ (r₀, 1]. From $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{0}) + \underline{u_{i}} (r_{0})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{0}) - \underline{u_{i}} (r_{0})}{2})^{2}} = 1$ we know $x_{0} \in {(\underline{u_{i}} (r_{0}), \bar{u_{i}} (r_{0}))}_{i = 1}^{n} = [u]^{r_{0}}$ i.e., u (x₀) ≥ r₀. In the following, we prove u (x₀) = r₀ by reduction to absurdity. Assume $u (x_{0}) = u (x_{1}^{0}, x_{2}^{0},$ $\dots, x_{n}^{0}) \neq r_{0}$ . Then, we have u (x₀) > r₀, so there exist ${\overset{`}{r}}_{0} \in [0, 1]$ such that $u (x_{0}) > {\overset{`}{r}}_{0} > r_{0}$ . From $u (x_{0}) > {\overset{`}{r}}_{0}$ , we see $x_{0} \in [u]^{{\overset{`}{r}}_{0}}$ , it implies $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} ({\overset{`}{r}}_{0}) + \underline{u_{i}} ({\overset{`}{r}}_{0})}{2})^{2}}{(\frac{\bar{u_{i}} ({\overset{`}{r}}_{0}) - \underline{u_{i}} ({\overset{`}{r}}_{0})}{2})^{2}} \leq 1$ which is in contradiction to $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r) + \underline{u_{i}} (r)}{2})^{2}}{(\frac{\bar{u_{i}} (r) - \underline{u_{i}} (r)}{2})^{2}} > 1$ for any r ∈ (r₀, 1]. Thus, we have u (x₀) = r₀.

If there exist r₁, r₂ ∈ (0, 1) with r₁ < r₂ such that the (1)-(3) in Theorem 9 hold. Taking a non-decreasing sequence ${r_{m}^{'}} \subset (r_{1}, r_{2})$ with $r_{m}^{'} \to r_{2}, m \to \infty$ , from $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{m}^{'}) + \underline{u_{i}} (r_{m}^{'})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{m}^{'}) - \underline{u_{i}} (r_{m}^{'})}{2})^{2}} = 1$ and the left continuity (see Theorem 2) of $\underline{u_{i}} (r)$ and $\bar{u_{i}} (r)$ , i = 1, 2, ⋯ , n, we have $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (r_{2}) + \underline{u_{i}} (r_{2})}{2})^{2}}{(\frac{\bar{u_{i}} (r_{2}) - \underline{u_{i}} (r_{2})}{2})^{2}} = 1$ It implies $x_{0} \in {(\underline{u_{i}} (r_{2}), \bar{u_{i}} (r_{2}))}_{i = 1}^{n} = [u]^{r_{2}}$ i.e., u (x₀) ≥ r₂. In the following, we prove u (x₀) = r₂ by reduction to absurdity. Assume $u (x_{0}) = u (x_{1}^{0}, x_{2}^{0},$ $\dots, x_{n}^{0}) \neq r_{2}$ . Then, we have u (x₀) > r₂, so there exist ${\overset{`}{r}}_{0} \in [0, 1]$ such that $u (x_{0}) > {\overset{`}{r}}_{0} > r_{2}$ . From $u (x_{0}) > {\overset{`}{r}}_{0}$ , we see $x_{0} \in [u]^{{\overset{`}{r}}_{0}}$ , it implies $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} ({\overset{`}{r}}_{0}) + \underline{u_{i}} ({\overset{`}{r}}_{0})}{2})^{2}}{(\frac{\bar{u_{i}} ({\overset{`}{r}}_{0}) - \underline{u_{i}} ({\overset{`}{r}}_{0})}{2})^{2}} \leq 1$ which is in contradiction to the (3) in Theorem 9. So, we have u (x₀) = r₂. Thus, the proof is completed.

Remark 4. Theorem 9 can be also proved by other method different from our method in this paper. However, the proof method used in this paper can present a iterative algorithm for the value of joint membership function as the edge membership functions are given.

Let $x_{0} = (x_{1}^{0}, x_{2}^{0}, \dots, x_{n}^{0}) \in R^{n}$ , δ₀ > 0, and u₁, u₂, ⋯ , u_n ∈ E be known. Then we can work out $\underline{u_{i}} (r), \bar{u_{i}} (r)$ with $[u_{i}]^{r} = [\underline{u_{i}} (r), \bar{u_{i}} (r)]$ for any i = 1, 2, ⋯ , n and r ∈ [0, 1]. In order to obtain the iterative approximate value (the error is less than the given δ₀) of the joint membership function value u (x₀) of u₁, u₂, ⋯ , u_n as soon as possible, we can take λ = 0.618 (in the proof of Theorem 9) in accordance by the golden section method.

The iterative algorithm

If $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (0) + \underline{u_{i}} (0)}{2})^{2}}{(\frac{\bar{u_{i}} (0) - \underline{u_{i}} (0)}{2})^{2}} \geq 1$ , then u (x₀) =0, and the algorithm end;

If $\sum_{i = 1}^{n} \frac{(x_{i}^{1} - \frac{\bar{u_{i}} (1) + \underline{u_{i}} (1)}{2})^{2}}{(\frac{\bar{u_{i}} (1) - \underline{u_{i}} (1)}{2})^{2}} \leq 1$ , then u (x₀) =1, and the algorithm end;

If $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\bar{u_{i}} (0) + \underline{u_{i}} (0)}{2})^{2}}{(\frac{\bar{u_{i}} (0) - \underline{u_{i}} (0)}{2})^{2}} < 1$ , $\sum_{i = 1}^{n} \frac{(x_{i}^{1} - \frac{\bar{u_{i}} (1) + \underline{u_{i}} (1)}{2})^{2}}{(\frac{\bar{u_{i}} (1) - \underline{u_{i}} (1)}{2})^{2}}$ >1, then take a₁ = 0, b₁ = 1.

Take s₁ = 0.382a₁ + 0.618b₁ = 0.618.

If b₁ - a₁ ≤ δ₀, then u (x₀) = s₁, and the algorithm end;

If b₁ - a₁ > δ₀, then

If $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\underline{u_{i}} (s_{1}) + \bar{u_{i}} (s_{1})}{2})^{2}}{(\frac{\underline{u_{i}} (s_{1}) - \bar{u_{i}} (s_{1})}{2})^{2}} \leq 1$ , then take a₂ = s₁, b₂ = b₁;

Take s₂ = 0.382a₂ + 0.618b₂.

If b₂ - a₂ ≤ δ₀, then u (x₀) = s₂, and the algorithm end;

If b₂ - a₂ > δ₀, then

If $\sum_{i = 1}^{n} \frac{(x_{i}^{0} - \frac{\underline{u_{i}} (s_{2}) + \bar{u_{i}} (s_{2})}{2})^{2}}{(\frac{\underline{u_{i}} (s_{2}) - \bar{u_{i}} (s_{2})}{2})^{2}} \leq 1$ , then take a₃ = s₂, b₃ = b₂;

Take s₃ = 0.382a₃ + 0.618b₃.

...... and so on until the end.

Example 1. Let u₁, u₂ ∈ E be defined respectively by

$u_{1} (x) = {\begin{matrix} \frac{1}{2} x, & x \in [0, 1) \\ 1, & x \in [1, 2] \\ 3 - x, & x \in (2, 3] \\ 0, & x \bar{\in} [0, 3] \end{matrix}$ $u_{2} (x) = {\begin{matrix} \frac{1}{2}, & x \in [0, 1] \cup [3, 4] \\ 1 - \frac{1}{2} (x - 2)^{2}, & x \in (1, 3) \\ 0, & x \bar{\in} [0, 4] \end{matrix}$ Then $\underline{u_{1}} (r) = {\begin{matrix} 2 r, & r \in [0, \frac{1}{2}) \\ 1, & r \in [\frac{1}{2}, 1] \end{matrix}, \bar{u_{1}} (r) = 3 - r$ $\underline{u_{2}} (r) = {\begin{matrix} 0, & r \in [0, \frac{1}{2}] \\ 2 - \sqrt{2 (1 - r)}, & r \in (\frac{1}{2}, 1] \end{matrix}$ $\bar{u_{2}} (r) = {\begin{matrix} 4, & r \in [0, \frac{1}{2}] \\ 2 + \sqrt{2 (1 - r)}, & r \in (\frac{1}{2}, 1] \end{matrix}$ By the iterative algorithm, we can obtain the joint membership function value $u (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ of u₁, u₂ at $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ (i.e., the membership degree of the 2-ellipsoid number u with u = (u₁, u₂) at $(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$ is $u (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) = 0.213779$ (where the allowable error δ₀ = 0.000001).

5 Conclusion

In this paper, we obtained two results (Lemmas 1 and 2) of scalar multiplication and addition of n-ellipsoids. By the two results, we showed that the general scalar multiplication (defined by Zadeh’s expansion) preserves the closeness of the operation, but the general addition (defined by Zadeh’s expansion) does not preserve the closeness of the operation (Theorem 3). Then, by Theorem 4, we defined a new addition “⊕” of fuzzy n-ellipsoid numbers (Definition 1), obtained its algorithm (Theorem 5), and showed that the new addition preserves the closeness of the operation (Theorem 5), and is the best approximation of the general addition operation in the all addition operations which preserve the closeness of the operation for fuzzy ellipsoid numbers (Theorem 6 and Remark 1). And then, we established a bijection between fuzzy n-ellipsoid number space and n-dimension fuzzy vector space (Theorem 7), obtained a formula which are used to express the edge membership functions with the joint membership function (Theorem 8), presented an iterative algorithm to calculate the joint membership function value as the edge membership functions are known, and proved the convergence of the iterative algorithm (Theorems 9 and 10). The theory (about fuzzy ellipsoid numbers) established by us can be used to identify, classify and rank (and so on) objects which are characterized by uncertain or imprecise multi-channel digital information. The specific approach is: From the data set (imprecise or uncertain multi-channel digital signals) which characterize objects to be dealt with, we firstly construct fuzzy ellipsoid numbers to represent these objects to be dealt with. Secondly, by the theory established by us, we can set up some corresponding aggregation operators in fuzzy ellipsoid number space. And finally, we can use these aggregation operators to realize the identification, classification and ranking for these objects to be dealt with.

Footnotes

Acknowledgments

This work is partially supported by the Nature Science Foundation of China (61273077, 61573112), Joint Key Grant by National Nature Science Foundation of China and Zhejiang Province Government (U1509217), and the Australian Research Council (DP140102180, LP140100471, LE150100079).

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