On the Aleksandrov problem for mappings preserving fuzzy n -distance in fuzzy n -normed spaces

Abstract

We study the problem of Aleksandrov in fuzzy n-normed spaces and prove that every surjective fuzzy function preserving unit n-distance is affine, and thus is a fuzzy n-isometry. Finally, we show that every fuzzy function preserving two fuzzy unit n-distances confirmers the result of Benz theorem when target space is fuzzy n-strictly convex.

Keywords

Aleksandrov problem 54H20 46L05 11Y50

1 Introduction

Inspired by the basic notions of Katsaras’ paper [1], using the concept of Minkowski functionals of L-fuzzy sets introduced by Höhle [2], and fuzzy metric space by Kaleva and Seikkala [3], in 1988, Morsi [4] introduced a notion of fuzzy (pseudo) normed spaces. Next, by using the notion of random normed spaces introduced by S̆erstnev [5], and studied by Mus̆tari [6], Radu [7], Cheng and Mordeson [8], Rano and Bag [9] and using the method of Shi [10, 11] and Mardones-Pérez [12, 13], Narayanan and Vijayabalaji [14, 15] defined fuzzy n-normed spaces. In this paper, we consider two fuzzy n-normed spaces U and V. It is natural to ask which of the conditions for h : U → V are needed to be sure that h is an isometry. We can consider the following condition: for some fixed number K > 0, suppose that h preserves the distance K, i.e., for every u, v ∈ U with ∥u - v ∥ _α = K, we have ∥h (u) - h (v) ∥ _α = K. Then K is called a conservative distance for the function h. The basic problem of preserved distances is whether the existence of a single conservative distance for some h implies that h is an isometry of U into V. It is called the Aleksandrov problem [17].

In 1993 Rassias and S̆emrl [18] proved a series of results on the UDPP (unit distance preserving property) for normed spaces (see [19 –21]). Baker [22] showed that an isometry from a real normed linear space into a strictly convex real normed linear space is affine. Benz [23] (see also [24]) investigated the case when the mapping preserves distance β, for some β > 0 if the target space is strictly convex, and demonstrated in [23] that this mapping is an affine isometry. Yumei Ma generalized the Aleksandrov–Benz–Rassias problem on n-normed spaces [25]. Huang and Tan gave at first a positive answer to the Aleksandrov problem in n-normed spaces under the surjectivity assumption [26]. In the present paper, we study the Aleksandrov problem in fuzzy n-normed spaces under only the condition of surjectivity. Next, we consider the Benz theorem in this space.

2 Preliminaries

Here, at first, we define fuzzy n-normed linear spaces in the sense of Morsi by using the concept of fuzzy real number. Next, we define fuzzy n-normed linear spaces (U, N, T) and show a relationship between them. A fuzzy real number is a fuzzy set on $ℝ$ i.e., a mapping $η : ℝ \to I$ ( = [0, 1]) associating with each real number t its grade of membership η (t). A fuzzy real number η is convex if η (t) ≥ min(η (s) , η (r)), where s ≤ t ≤ r. The α-level set of a fuzzy real number η, 0 < α ≤ 1, denoted [η] _α, is defined by $[η]_{α} = {t : η (t) \geq α} .$

Note that the α-level set of an upper semicontinuous convex normal fuzzy real number, for each α, 0 < α ≤ 1, is a closed interval [a^α, b^α], where a^α = -∞ and b^α =+ ∞ are also admissible. Let us denote the set of all upper semicontinuous normal convex fuzzy real numbers by $ℝ (I)$ . A fuzzy real number η is called non-negative if η (t) =0 for all t < 0. The set of all non-negative fuzzy real numbers of $ℝ (I)$ is denoted by $ℝ^{*} (I)$ . Arithmetic operations ⊕ and ⊖ on $ℝ (I) \times ℝ (I)$ can be defined by $\begin{matrix} (η \oplus δ) (t) & = & sup_{s \in ℝ} min (η (s), δ (t - s)), t \in ℝ, \\ (η ⊖ δ) (t) & = & sup_{s \in ℝ} min (η (s), δ (s - t)), t \in ℝ . \end{matrix}$

Now, we consider a partial ordering ⪯ in $ℝ (I)$ by η ⪯ δ if and only if $a_{1}^{α} \leq a_{2}^{α}$ and $b_{1}^{α} \leq b_{2}^{α}$ for all α ∈ (0, 1], where $[η]_{α} = [a_{1}^{α}, b_{1}^{α}]$ and $[δ]_{α} = [a_{2}^{α}, b_{2}^{α}]$ (for more details please see [3 , 28]). Now, we define fuzzy n-normed linear spaces in the sense of Morsi by using the concept of fuzzy metric spaces introduced by Kaleva and Seikkala [3].

Definition 2.1. Suppose that U is a linear space (or real vector space). Let $∥ ∥ : U^{n} \to ℝ^{*} (I)$ (dimension U = d ≥ n) and let L, R : I² → I be symmetric, nondecreasing in both arguments and satisfying $L (0, 0) = 0 and R (1, 1) = 1 .$ Write $[∥ u_{1}, \dots, u_{n} ∥]_{α} = [∥ u_{1}, \dots, u_{n} ∥_{α}^{1}, ∥ u_{1}, \dots,$ $u_{n} ∥_{α}^{2}]$ for u₁, …, u_n ∈ U, α ∈ (0, 1] and assume that for every linearly independent vectors u₁, …, u_n ∈ U, there exists α₀ ∈ (0, 1] independent of u₁, …, u_n ∈ U such that for each α ≤ α₀, one has $inf ∥ u_{1}, \dots, u_{n} ∥_{α}^{1} > 0, ∥ u_{1}, \dots, u_{n} ∥_{α}^{2} < \infty .$

The quadruple (Uⁿ, ∥ ∥ , L, R) is called a fuzzy n-normed linear space (in short f-n-NLS) and ∥ ∥ is a fuzzy n-norm if

(F1) $∥ u_{1}, \dots, u_{n} ∥ = \bar{0}$ if and only if u₁, …, u_n are linearly dependent;

(F2) ∥u₁, … , u_n ∥ is invariant under any permutation of u₁, … , u_n ∈ U ;

(F3) ∥ρu₁, … , u_n ∥ = |ρ| ∥ u₁, …, u_n ∥, $ρ \in ℝ;$

(F4) for each u₁, …, u_n ∈ U, $\begin{matrix} ∥ u_{0} + u_{1}, u_{2}, \dots, u_{n} ∥ ⪯ ∥ u_{0}, u_{2}, \dots, u_{n} ∥ \\ \oplus ∥ u_{1}, u_{2}, \dots, u_{n} ∥; \end{matrix}$

(a) whenever $s \leq ∥ u_{0}, u_{2}, \dots, u_{n} ∥_{1}^{1}$ , t ≤ ∥ u₁, u₂, $\dots, u_{n} ∥_{1}^{1}$ and t + s ≤ ∥ u₀ + u₁, u₂, …, $u_{n} ∥_{1}^{1}$ , $\begin{matrix} ∥ u_{0} + u_{1}, u_{2}, \dots, u_{n} ∥ (s + t) \geq L (∥ u_{0}, u_{2}, \\ \dots, u_{n} ∥ (s), ∥ u_{1}, u_{2}, \dots, u_{n} ∥ (t)), \end{matrix}$

(b) whenever $s \geq ∥ u_{0}, u_{2}, \dots, u_{n} ∥_{1}^{1}$ , t ≥ ∥ u₁, $u_{2}, \dots, u_{n} ∥_{1}^{1}$ and t + s ≥ ∥ u₀ + u₁, u₂, $\dots, u_{n} ∥_{1}^{1}$ , $\begin{matrix} ∥ u_{0} + u_{1}, u_{2}, \dots, u_{n} ∥ (s + t) \leq R (∥ u_{0}, u_{2}, \\ \dots, u_{n} ∥ (s), ∥ u_{1}, u_{2}, \dots, u_{n} ∥ (t)) . \end{matrix}$

Definition 2.2. [14, 29] Suppose that U is a linear space (or real vector space) and T is a continuous t-norm. Let the fuzzy subset N of Uⁿ × (0, + ∞) (dimension U = d ≥ n) satisfy

(N1) N (u₁, …, u_n, t) =0, for t ≤ 0;

(N2) N (u₁, …, u_n, t) =1 for t ≥ 0 if and only if u₁, …, u_n are linearly dependent;

(N3) N (u₁, …, u_n, t) is invariant under any permutation of u₁, …, u_n ∈ U;

(N4) $N (α u_{1}, \dots, u_{n}, t) = N (u_{1}, \dots, u_{n}, \frac{t}{| α |})$ if α ≠ 0;

(N5) N (u₀ + u₁, u₂, …, u_n, s + t) ≥ T (N (u₀, u₂, …, u_n, s) , N (u₁, u₂, …, u_n, t));

(N6) N (u₁, …, u_n,.) : (0, + ∞) → [0, 1] is left continuous;

(N7) $lim_{t \to + \infty} N (u_{1}, \dots, u_{n}, t) = 1$ .

Then the triple (U, N, T) is said to be a fuzzy n-normed linear space or in short f-n-NLS.

For more details see [30 –36].

Example 2.3. Let (U, || . , …,. ||) be an n-normed space. We define triangular norm T, product t-norm and N (u₁, …, u_n, t) = exp(- ||u₁, …, u_n||/t) for u₁, …, u_n ∈ U and t ∈ (0, + ∞). Then (U, N, T) is an f-n-NLS.

Proof. Clearly exp(- ||u₁, …, u_n||/t) >0; also exp(- ||u₁, …, u_n||/t) =1 if and only if ||u₁, …, u_n||=0 for t > 0; then u₁, …, u_n are linearly dependent, and exp(- ||u₁, …, u_n||/t) is invariant under any permutation of u₁, …, u_n ∈ U.

For t > 0, we have $exp (- ‖ α u_{1}, \dots, u_{n} ‖ / t) = exp (- ‖ u_{1}, \dots, u_{n} ‖ / \frac{t}{| α |})$ and this proves the condition four.

Further we have $\begin{matrix} exp (- \frac{‖ u_{0}, u_{2}, \dots, u_{n} ‖}{t}) \cdot exp (- \frac{‖ u_{1}, u_{2}, \dots, u_{n} ‖}{s}) \\ = exp (- \frac{‖ u_{0}, u_{2}, \dots, u_{n} ‖}{t} - \frac{‖ u_{1}, u_{2}, \dots, u_{n} ‖}{s}) \\ \leq exp (- \frac{‖ u_{0}, u_{2}, \dots, u_{n} ‖}{t + s} - \frac{‖ u_{1}, u_{2}, \dots, u_{n} ‖}{t + s}), \end{matrix}$ and since $\begin{matrix} ‖ u_{0} + u_{1}, u_{2}, \dots, u_{n} ‖ \leq ‖ u_{0}, u_{2}, \dots, u_{n} ‖ \\ + ‖ u_{1}, u_{2}, \dots, u_{n} ‖, \end{matrix}$ then $\begin{matrix} - \frac{‖ u_{0} + u_{1}, u_{2}, \dots, u_{n} ‖}{t + s} \geq - \frac{‖ u_{0}, u_{2}, \dots, u_{n} ‖}{t + s} \\ - \frac{‖ u_{1}, u_{2}, \dots, u_{n} ‖}{t + s}, \end{matrix}$ and we have $\exp (- \frac{| | u_{0}, u_{2}, \dots, u_{n} | |}{t}) \cdot \exp (- \frac{| | u_{1}, u_{2}, \dots, u_{n} | |}{s}) \leq \exp (- \frac{| | u_{0} + u_{1}, u_{2}, \dots, u_{n} | |}{t + s}),$ for t-norm product.

Conditions (N6) and (N7) are evident. □

Note that the above example holds with t-norm mininmum.

Theorem 2.4. [14], see also [12 , 16] Let (U, T, N) be an f-n-NLS in which T = min and

(N8) N (u₁, …, u_n, t) >0 for all t > 0 implies u₁, …, u_n are linearly dependent.

Define $∥ u_{1}, \dots, u_{n} ∥_{α} : = inf [N (u_{1}, \dots, u_{n}, t)]_{α}, α \in (0, 1) .$ Then {∥ • , • , ⋯ , • ∥ _α : α ∈ (0, 1)} is an ascending family of n-norms on U.

These n-norms will be called the α-n-norms on U corresponding to the fuzzy n-norm on U.

Hereinafter, we let T = min and fuzzy n-norm satisfy (N8).

Theorem 2.5. Let (U, ∥ ∥ , 0, max) be an n-f-NLS, in the sense of Morsi, in which $lim_{t \to \infty} ∥ u_{1}, \dots, u_{n} ∥ (t) = 0$ . Let (U, T, N) be an f-n-NLS. Then we have $\begin{matrix} ∥ u_{1}, \dots, u_{n} ∥ (t) = \\ {\begin{matrix} 0, & if t < s = sup {t : N (u_{1}, \dots, u_{n}, t) = 0}, \\ 1 - N (u_{1}, \dots, u_{n}, t), & if t \geq s, \end{matrix} \end{matrix}$ and $\begin{matrix} N (u_{1}, \dots, u_{n}, t) = \\ {\begin{matrix} 0, & if t < ∥ u_{1}, \dots, u_{n} ∥_{1}^{1}, \\ 1 - ∥ u_{1}, \dots, u_{n} ∥ (t), & if t \geq ∥ u_{1}, \dots, u_{n} ∥_{1}^{1} . \end{matrix} \end{matrix}$

Proof. See [3, page 223]. □

Remark 2.6. [37] Let N′ : R × (0, + ∞) → (0, 1] be an Euclidean fuzzy normed and (U, N, T) be an f-n-NLS. Then u₁, u₂, …, u_n ∈ U are linearly independent if and only if N (u₁, u₂, …, u_n, t) = N′ (1, t) for t > 0.

By the above remark, we have that, u₁, u₂, …, u_n ∈ U are linearly independent if and only if $\begin{matrix} ∥ u_{1}, \dots, u_{n} ∥_{α} \\ = & inf {t : N (u_{1}, \dots, u_{n}, t) \geq α, α \in (0, 1)} \\ = & inf {t : N^{'} (1, t) \geq α, α \in (0, 1)} \\ = & | 1 |_{α} . \end{matrix}$

Definition 2.7. Let U and V be f-n-NLS and let h be a function from U to V. Then,

h is called a fuzzy n-isometry if it satisfies $\begin{matrix} N (h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}), t) \\ = N (u_{1} - v_{1}, \dots, u_{n} - v_{n}, t) \end{matrix}$ for all u₁, …, u_n, v₁, …, v_n ∈ U and t ∈ (0, + ∞). In particular, if v₁ = v₂ = ⋯ = v_n, then h is called a weak fuzzy n-isometry.

If N (u₁ - v₁, …, u_n - v_n, t) = N′ (1, t) imp-lies that $N (h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}), t) = N^{'} (1, t)$ for all u₁, …, u_n, v₁, …, v_n ∈ U, t∈ (0, + ∞), then we say that h has the fuzzy unit n-distance preserving property (f-n-UDPP). In particular, if v₁ = v₂ = ⋯ = v_n, then we say that h has the weak fuzzy unit n-distance preserving property (w-f-n-UDPP), with attention to define N, if t→ + ∞, then h has always w-f-n-UDPP. Let u₁, …, u_n, v₁, …, v_n ∈ U, t ∈ (0, + ∞) and $N (h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}), t) = N^{'} (1, t)$ if and only if N (u₁ - v₁, …, u_n - v_n, t) = N′ (1, t), then we say that h has the strong fuzzy unit n-distance preserving property (s-f-n-UDPP).

If N (u₁ - v₁, …, u_n - v_n, t) = N′ (β, t) implies $N (h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}), t) = N^{'} (β, t)$ for all u₁, …, u_n, v₁, …, v_n ∈ U and t ∈ (0, + ∞), then h is said to preserve fuzzy β-n-distance for some β > 0. In particular, if v₁ = v₂ = ⋯ = v_n, then h is said to preserve w-f-β-n-distance.

Corollary 2.8. Let (U, ∥ • , • , ⋯ , • ∥ _α) and (V, ∥ • , •, ⋯ , • ∥ _α) be f-n-NLS corresponding to the fuzzy n-norm N on U and V receptively; let also $(ℝ, | • |_{α})$ correspond to the Euclidean fuzzy norm (see Remark 2.6). Now, by Definition 2.7, for the function h : U → V we say that:

it is called a fuzzy n-isometry if it satisfies $\begin{matrix} ∥ h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}) ∥_{α} \\ = ∥ u_{1} - v_{1}, \dots, u_{n} - v_{n} ∥_{α} \end{matrix}$ for all u₁, …, u_n, v₁, …, v_n ∈ U and α ∈ (0, 1). In particular, if v₁ = v₂ = ⋯ = v_n, then h is called a weak fuzzy n-isometry.

If ∥u₁ - v₁, …, u_n - v_n ∥ _α = |1|_α implies that $∥ h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}) ∥_{α} = | 1 |_{α}$ for all u₁, …, u_n, v₁, …, v_n ∈ U, α ∈ (0, 1), then we say that h has the fuzzy unit n-distance preserving property (f-n-UDPP). In particular, if v₁ = v₂ = ⋯ = v_n, then h has the weak fuzzy unit n-distance preserving property (w-f-n-UDPP). Let u₁, …, u_n, v₁, …, v_n ∈ U, t ∈ (0, + ∞) and $∥ h (u_{1}) - h (v_{1}), \dots, h (u_{n}) - h (v_{n}) ∥_{α} = | 1 |_{α}$ if and only if ∥u₁ - v₁, …, u_n - v_n ∥ _α = |1|_α, then we say that h has the strong fuzzy unit n-distance preserving property (s-f-n-UDPP).

If ∥u₁ - v₁, …, u_n - v_n ∥ _α = |β|_α implies ∥h (u₁) - h (v₁) , …, h (u_n) - h (v_n) ∥ _α = |β|_α for all u₁, …, u_n, v₁, …, v_n ∈ U and α ∈ (0, 1), then h is said to preserve fuzzy β-n-distance for some β > 0. In particular, if v₁ = v₂ = ⋯ = v_n, then h is said to preserve w-f-β-n-distance.

Lemma 2.9. Let (U, ∥ • , • , ⋯ , • ∥ _α) be an f-n-NLS, u₁, …, u_n ∈ U and let v be a linear combination of u₂, …, u_n ∈ U. Then, $∥ u_{1}, u_{2}, \dots, u_{n} ∥_{α} = ∥ u_{1} + v, u_{2}, \dots, u_{n} ∥_{α} .$

Proof. Since v is a linear combination of u₂, …, u_n, then ∥v, u₂, …, u_n ∥ _α = 0. But $\begin{matrix} ∥ u_{1} + v, u_{2}, \dots, u_{n} ∥_{α} \leq ∥ u_{1}, \\ \dots, u_{n} ∥_{α} + ∥ v, u_{2}, \dots, u_{n} ∥_{α}, \end{matrix}$ thus $∥ u_{1} + v, u_{2}, \dots, u_{n} ∥_{α} \leq ∥ u_{1}, \dots, u_{n} ∥_{α} .$ (2.1)

On the other hand, $\begin{matrix} ∥ u_{1}, \dots, u_{n} ∥_{α} \\ = & ∥ u_{1} + v - v, u_{2}, \dots, u_{n} ∥_{α} \\ \leq & ∥ u_{1} + v, u_{2}, \dots, u_{n} ∥_{α} + ∥ - v, u_{2}, \dots, u_{n} ∥_{α} \\ = & ∥ u_{1} + v, u_{2}, \dots, u_{n} ∥_{α} + | - 1 | ∥ v, u_{2}, \dots, u_{n} ∥_{α} \\ = & ∥ u_{1} + v, u_{2}, \dots, u_{n} ∥_{α} . \end{matrix}$ (2.2)

Now, from 2.1 and 2.2, we conclude that ∥u₁, …, u_n ∥ _α = ∥ v + u₁, …, u_n ∥ _α. □

3 Isometry in f-n-NLS

Now we study the Aleksandrov problem in f-n-NLS. Of course, we solve it under a weaker hypotheses in f-n-NLS. For some real number m, we say that u, v, w of U are 2-collinear if v - w = m (u - w). Also, if u_j - u_i, are linearly dependent for some i, in which 0 ≤ j ≠ i ≤ n, then we say that u₀, u₁, …, u_n of U are n-collinear.

Definition 3.1. Let U and V be f-n-NLS and let h be a function from U to V.

When u, v, w ∈ U are f-2-collinear implies that h (u), h (v), h (w) are f-2-collinear, then we say that h preserves f-2-collinearity. Also, h preserves f-2-collinearity for the midpoint of a segment if $w = \frac{u + v}{2}$ .

When u₀, u₁, …, u_n of U are f-n-collinear implies that h (u₀) , h (u₁) , …, h (u_n) are f-n-collinear, then we say that h preserves f-n-collinearity. It means that h preserves w-f-0-n-distance, i.e., if $N (u_{1} - u_{0}, u_{2} - u_{0}, \dots, u_{n} - u_{0}, t) = 1,$ then $\begin{matrix} N (h (u_{1}) - h (u_{0}), h (u_{2}) - h (u_{0}), \\ \dots, h (u_{n}) - h (u_{0}), t) = 1, \end{matrix}$ for each u₀, u₁, …, u_n ∈ U, t > 0.

Note that, by Definition 3.1, when u₀, u₁, …, u_n of U are f-n-collinear implies that h (u₀) , h (u₁) , …, h (u_n) are f-n-collinear, then we say that h preserves f-n-collinearity. It means that h preserves w-f-0-n-distance, i.e., if $∥ u_{1} - u_{0}, u_{2} - u_{0}, \dots, u_{n} - u_{0} ∥_{α} = 0,$ then $\begin{matrix} ∥ h (u_{1}) - h (u_{0}), h (u_{2}) - h (u_{0}), \dots, h (u_{n}) \\ - h (u_{0} ∥_{α} = 0, \forall u_{0}, u_{1}, \dots, u_{n} \in U, α \in (0, 1) . \end{matrix}$

Now, we show that a function h : U → V where U and V are two f-n-NLS, which preserves w-f-n-distance and f-2-collinearity for the midpoint of a segment, satisfies the Jensen equality $h (\frac{u + v}{2}) = \frac{h (u) + h (v)}{2} \forall u, v \in U .$

Lemma 3.2. Let U and V be f-n-NLS, and let h : U → V preserve weak f-β-n-distance for some β > 0. Then h is injective. Also if for the midpoint of a segment, h preserves f-2-collinearity, then the function g : U → V defined by g (u) = h (u) - h (0) for each u ∈ U is additive.

Proof. Let (U, ∥ • , • , ⋯ , • ∥ _α) and (V, ∥ • , • , ⋯, • ∥ _α) be f-n-NLS corresponding to the fuzzy n-norm N on U and V receptivity, and let $(ℝ, | • |_{α})$ corresponding to the Euclidean fuzzy norm (see Remark 2.6). Let u, v ∈ U, u ≠ v, and dim U ≥ n, so we can find u₂, u₃, …, u_n ∈ U such that $∥ v - u, u_{2} - u, \dots, u_{n} - u ∥_{α} = | β |_{α} .$

Since the mapping h preserves w-f-β-n-distance, we have that $∥ h (v) - h (u), h (u_{2}) - h (u), \dots, h (u_{n}) - h (u) ∥_{α} = | β |_{α},$ which implies h (u) ≠ h (v), thus h is injective. Now, for each u, v ∈ U, we show that $h (\frac{u + v}{2}) = \frac{h (u) + h (v)}{2} .$ (3.1)

Set $k = \frac{u + v}{2}$ for distinct u, v ∈ U. Choose v₂, v₃, …, v_n ∈ U such that $\begin{matrix} ∥ v - k, v_{2} - k, \dots, v_{n} - k ∥_{α} = ∥ u - k, v_{2} - k, \\ \dots, v_{n} - k ∥_{α} = | β |_{α}, \end{matrix}$ then $\begin{matrix} ∥ h (v) - h (k), h (v_{2}) - h (k), \dots, h (v_{n}) - h (k) ∥_{α} \\ = ∥ h (u) - h (k), h (v_{2}) - h (k), \dots, h (v_{n}) - h (k) ∥_{α} \end{matrix}$ (3.2) $= | β |_{α} .$

Since for the midpoint of a segment, h preserves f-2-collinearity, there exist $r \in ℝ$ such that $h (v) - h (k) = r (h (u) - h (k)) .$ Now (3.2) and injectivity of h imply that r = -1. Hence, $h (\frac{u + v}{2}) = h (k) = \frac{h (u) + h (v)}{2} .$ □

Remark 3.3. Note that a mapping h is said to preserve f-2-collinearity if 2-collinearity of points u, v, w i.e., if N (u - w, v - w, t) =1 for t > 0, implies N (h (u) - h (w) , h (v) - h (w) , t) =1. In fact, we do not need surjectivity, but need only preservation of w-f-n-UDPP and of 2-collinearity.

Lemma 3.4. Consider two f-n-NLS U and V and a mapping h : U → V preserving weak f-β-n-distance, for some β > 0. Then the below properties are equivalent.

h preserves f-n-collinearity.

h preserves f-2-collinearity.

For the midpoint of a segment, h preserves f-2-collinearity.

Proof. Let (U, ∥ • , • , ⋯ , • ∥ _α) and (V, ∥ • , • , ⋯, • ∥ _α) be f-n-NLS corresponding to the fuzzy n-norm N on U and V respectively, and let $(ℝ, | • |_{α})$ corresponding to the Euclidean fuzzy norm (see Remark 2.6).

(a) ⇒ (b). Let u₀, u₁, u₂ ∈ U be f-2-collinear such that h (u₁) - h (u₀) and h (u₂) - h (u₀) are linearly independent. Note that u₀ ≠ u₁ and h preserves w-f-β-n-distance. We can choose v₂, …, v_n ∈ U such that $\begin{matrix} ∥ h (u_{1}) - h (u_{0}), h (v_{2}) - h (u_{0}), \dots, h (v_{n}) - h (u_{0}) ∥_{α} \\ = ∥ u_{1} - u_{0}, v_{2} - u_{0}, \dots, v_{n} - u_{0} ∥_{α} \\ = & | β |_{α}, \end{matrix}$ for α ∈ (0, 1). Then there are n linearly independent vectors in A : = {h (u) - h (u₀) : u ∈ U} and so we can find u₃, ⋯ , u_n in U such that $\begin{matrix} ∥ h (u_{1}) - h (u_{0}), h (u_{2}) - h (u_{0}), \dots, h (u_{n}) \\ - h (u_{0} ∥_{α} = | β |_{α} . \end{matrix}$

Assume that h preserves f-n-collinearity; then ∥u₁ - u₀, u₂ - u₀, …, u_n - u₀ ∥ _α = 0, i.e., dependency of {u₁ - u₀, …, u_n - u₀} implies a contradiction $∥ h (u_{1}) - h (u_{0}), \dots, h (u_{n}) - h (u_{0}) ∥ = 0 .$

So, h preserves f-2-collinearity.

By Definition 3.1(a), (b) ⇒ (c) is obvious.

(c) ⇒ (a). So h has w-f-n-UDPP. Suppose that g (u) = h (u) - h (0) for each u ∈ U. Lemma 3.2 implies that g (u) is additive and hence g is Q-linear. Also, for each u ∈ U, g (u) - g (u₀) = h (u) - h (u₀) and g (0) =0, so we have

$\begin{matrix} ∥ h (u_{1}) - h (u_{0}), \dots, h (u_{n}) - h (u_{0}) ∥_{α} \\ = ∥ g (u_{1}) - g (u_{0}), \dots, g (u_{n}) - g (u_{0}) ∥_{α}, \end{matrix}$ for all u₀, u₁, …, u_n ∈ U and α ∈ (0, 1). Then g also has w-f-n-UDPP. Thus for all u₁, …, u_n ∈ U, $m, k \in ℕ$ and $∥ u_{1}, \dots, u_{n} ∥_{α} = | \frac{m}{k} |_{α}$ , we have $∥ \frac{k}{m} u_{1}, u_{2}, \dots, u_{n} ∥_{α} = | 1 |_{α}$ and since g has w-f-n-UDPP, we conclude that $∥ g (\frac{k}{m} u_{1}), g (u_{2}), \dots, g (u_{n}) ∥_{α} = | 1 |_{α}$ and since g is Q-linear, then $\begin{matrix} ∥ \frac{k}{m} g (u_{1}), \dots, g (u_{n}) ∥_{α} \\ = | \frac{k}{m} | {∥ g (u_{1}), g (u_{2}), \dots, g (u_{n}) ∥}_{α} = | 1 |_{α} . \end{matrix}$ So, ${∥ g (u_{1}), g (u_{2}), \dots, g (u_{n}) ∥}_{α} = | \frac{m}{k} |_{α}$ which implies that g preserves $\frac{m}{k}$ for all $m, k \in ℕ$ .

It remains to show that ∥g (u₁) , g (u₂) , …, g (u_n) ∥ _α = 0 for all nonzero u₁, …, u_n ∈ U satisfying ∥u₁, …, u_n ∥ _α = 0

Since ∥u₁, …, u_n ∥ _α = 0, we know that u₁, …, u_n ∈ U are linearly dependent. Select v_k+1, …, v_n ∈ U such that ∥u₁, …, u_k, v_k+1, …, v_n ∥ _α = |1|_α for α ∈ (0, 1). From Lemma 3.2, we conclude that u_j is a linear combination of u₁, …, u_k for each k + 1 ≤ j ≤ n, and so $\begin{matrix} ∥ u_{1}, \dots, u_{k}, u_{k + 1} + \frac{1}{m} v_{k + 1}, \dots, u_{n} + \frac{1}{m} v_{n} ∥_{α} \\ = & ∥ u_{1}, \dots, u_{k}, \frac{1}{m} v_{k + 1}, \dots, \frac{1}{m} v_{n} ∥_{α} \\ = & | \frac{1}{m^{n - k}} | ∥ u_{1}, \dots, u_{k}, v_{k + 1}, \dots, v_{n} ∥_{α} \\ = & | \frac{1}{m^{n - k}} | | 1 |_{α} \\ = & | \frac{1}{m^{n - k}} |_{α} . \end{matrix}$

Since g (u) is additive and preserves w-f-n-distance, we conclude that $\begin{matrix} ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), \dots, \\ g (u_{n}) + \frac{1}{m} g (v_{n}) ∥_{α} = | \frac{1}{m^{n - k}} |_{α} . \end{matrix}$

Also, ∥ g (u₁) , …, g (u_k) , g (u_k+1) , …, g (u_n) ∥ _α $= ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}) - \frac{1}{m} g (v_{k + 1}), g (u_{k + 2}) \dots, g (u_{n}) ∥_{α}$ $\leq ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), g (u_{k + 2}), \dots, g (u_{n}) ∥_{α}$ $+ ∥ g (u_{1}), \dots, g (u_{k}), - \frac{1}{m} g (v_{k + 1}), g (u_{k + 2}),$ … , g (u_n) ∥ _α $\leq ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), g (u_{k + 2})$ $+ \frac{1}{m} g (v_{k + 2}), g (u_{k + 3}), g (u_{k + 4}), \dots, g (u_{n}) ∥_{α}$ $+ ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), - \frac{1}{m} g (v_{k + 2}), g (u_{k + 3}), \dots, g (u_{n}) ∥_{α}$ $+ ∥ g (u_{1}), \dots, g (u_{k}), - \frac{1}{m} g (v_{k + 1}), g (u_{k + 2}), \dots, g (u_{n}) ∥_{α} \leq \dots$ $\leq ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), g (u_{k + 2}) + \frac{1}{m} g (v_{k + 2}), \dots, g (u_{n}) + \frac{1}{m} g (v_{n}) ∥_{α}$ $+ ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), \dots, g (u_{n - 1}) + \frac{1}{m} g (v_{n - 1}), - \frac{1}{m} g (v_{n}) ∥_{α}$ $+ ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), \dots, - \frac{1}{m} g (v_{n - 1}), g (u_{n}) ∥_{α} + \dots$ $+ ∥ g (u_{1}), \dots, g (u_{k}), - \frac{1}{m} g (v_{k + 1}), g (u_{k + 2}), \dots, g (u_{n}) ∥_{α}$ $= | \frac{1}{m^{n - k}} | + \frac{1}{m} ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), g (u_{k + 2})$ $+ \frac{1}{m} g (v_{k + 2}), \dots, g (u_{n - 1}) + \frac{1}{m} g (v_{n - 1}), g (v_{n}) ∥_{α}$ $+ \frac{1}{m} ∥ g (u_{1}), \dots, g (u_{k}), g (u_{k + 1}) + \frac{1}{m} g (v_{k + 1}), \dots, g (u_{n - 2}) + \frac{1}{m} g (v_{n - 2}), g (v_{n - 1}), g (u_{n}) ∥_{α}$ $+ \dots + \frac{1}{m} ∥ g (u_{1}), \dots, g (u_{k}), g (v_{k + 1}), g (u_{k + 2}), \dots, g (u_{n}) ∥_{α}$ , which implies that u_j, like above, is a linear combination of u₁, …, u_k, v for each k + 1 ≤ j ≤ n - 1, so $\begin{array}{l} ‖ g (u_{1}), \dots, g (u_{k}), g (υ_{n}), g (u_{k + 1}) + \frac{1}{m} g (υ_{k + 1}), \\ {\dots, g (u_{n - 1}) + \frac{1}{m} g (υ_{n - 1}) ‖}_{α} = | \frac{1}{m^{n - k - 1}} |_{α} \end{array}$ wherefrom $\begin{array}{l} ‖ g (u_{1}), \dots, g (u_{k}), g (υ_{n - 1}), g (u_{k + 1}) + \frac{1}{m} g (υ_{k + 1}), \\ {\dots, g (u_{n - 2}) + \frac{1}{m} g (υ_{n - 2}), g (u_{n}) ‖}_{α} = 0, \end{array}$ because g is additive and u_n is a linear combination of u₁, …, u_k and v_n-1, and so ∥g (u₁) , …, g (u_k) , …, g (v_j) , …, g (u_n) ∥ _α = 0 for all k + 1 ≤ j ≤ n - 1.

Therefore ∥g (u₁) , …, g (u_k) , …, g (u_n) ∥ _α ≤ 2 |1/m^n-k|_α, and letting m→ + ∞, we conclude that ∥g (u₁) , …, g (u_k) , …, g (u_n) ∥ =0, which implies that g (u₁) , …, g (u_n) are linearly dependent.□

Proposition 3.5. Consider two f-n-NLS U and V and h : U → V preserving weak f-β-n-distance for some β > 0. Then, for the midpoint of a segment, h preserves f-2-collinearity, so it is an affine n-isometry.

Proof. Let (U, ∥ • , • , ⋯ , • ∥ _α) and (V, ∥ • , • , ⋯, • ∥ _α) be f-n-NLS corresponding to the fuzzy n-norm N on U and V receptively; Let also $(ℝ, | • |_{α})$ corresponding to the Euclidean fuzzy norm (see Remark 2.6). First, we prove that the function g : U → V defined by g (u) = h (u) - h (0) is linear, which will imply that h is affine. From Lemmas 3.2 and 3.4, we can conclude that g is injective, additive and that preserves f-2-collinearity. Assume that 0 ≠ u ∈ U and that α ≠ 0, 1 is a real number. By collinearity of 0, u, αu, we can find a unique $t \in ℝ$ such that g (αu) = tg (u). Consider the real function K given by K (α) = t. We have $g (α u) = K (α) g (u), \forall α \in ℝ,$ which implies the injectivity and additivity of K and K (0) =0, K (1) =1. Now choose v ∈ U such that u and v are linearly independent and suppose that K₁ is a real function on $ℝ$ such that $g (α v) = K_{1} (α) g (v), \forall α \in ℝ .$

Collinearity of 0, u + v, α (u + v) implies the collinearity of $0, g (u) + g (v), K (α) g (u) + K_{1} (α) g (v) .$

Also, linear independency of g (u) and g (v) implies that K (α) = K₁ (α) and hence there are u₃, u₄, …, u_n ∈ U such that for β > 0, $\begin{matrix} ∥ g (u), g (v), g (u_{3}), \dots, g (u_{n}) ∥_{α} \\ = ∥ u, v, u_{3}, \dots, u_{n} ∥_{α} = | β |_{α}, \end{matrix}$ which implies linear independency of g (u) and g (v). For n = 2, choose a real number $0 \neq m \in ℝ$ , such that for β > 0, $∥ g (u), g (mv) ∥_{α} = ∥ u, mv ∥_{α} = | β |_{α},$ which implies linear independency of g (u) and g (mv), and $\begin{matrix} ∥ g (u), g (mv) ∥_{α} = ∥ g (u), K (a) g (v) ∥_{α} \\ = | K (a) | ∥ g (u), g (v) ∥_{α} = | K (a) β |_{α}, \end{matrix}$ which implies linear independency of g (u) and g (v). Now, we show that K is an endomorphism. Any real numbers a, b, 0, u + bv, au + abv are collinear and therefore so are $0, g (u) + K (b) g (v), K (a) g (u) + K (ab) g (v) .$

It follows that K (ab) = K (a) K (b) for any $a, b \in ℝ$ . The additivity of K confirms the claim. The fact K (α) = α for each real numbers α implies that for every u ∈ U, g (αu) = αg (u) and hence g is linear and h is affine. Also $∥ g (u_{1}), \dots, g (u_{n}) ∥_{α} = ∥ h (u_{1}) - h (0), \dots, h (u_{n}) - h (0) ∥_{α}$ $= ∥ u_{1} - 0, \dots, u_{n} - 0 ∥_{α} = ∥ u_{1}, \dots, u_{n} ∥_{α},$ shows that g is an n-isometrically and hence so is h.□

We study main theorem this paper. In a real f-n-NLS U, the line joining two points a and b in U will be denoted by $\bar{ab}$ and the affine subspace generated by K ⊂ U will be (affine) (K).

Theorem 3.6. Consider two f-n-NLS U and V. If a surjective function h from U to V has f-n-UDPP, then h is an affine f-n-isometry.

Proof. Let (U, ∥ • , • , ⋯ , • ∥ _α) and (V, ∥ • , •, ⋯ , • ∥ _α) be f-n-NLS corresponding to the fuzzy n-norm N on U and V receptively; let also $(ℝ, | • |_{α})$ corresponding to the Euclidean fuzzy norm (see Remark 2.6). Let h (0) =0. Since h is injective and surjective, then there exist h^-1. We show that if u, v, w ∈ U are not collinear, then h (u) , h (v) , h (w) are not collinear which implies that h^-1 preserves 2-collinearity. Select u₃, …, u_n ∈ U such that, $∥ u - w, v - w, u_{3} - w, \dots, u_{n} - w ∥_{α} = | β |_{α} .$

Set $z : = w + \frac{(u - w) + (v - w)}{| β |_{α}} .$

By Lemma 2.9, we have $\begin{matrix} ∥ u - w, z - w, u_{3} - w, \dots, u_{n} - w ∥_{α} \\ = ∥ u - w, \frac{(u - w) + (v - w)}{| β |_{α}}, u_{3} - w, \dots, u_{n} - w, t ∥ \\ = \frac{1}{| β |_{α}} ∥ u - w, (u - w) + (v - w), u_{3} - w, \dots, u_{n} - w ∥_{α} \\ = \frac{1}{| β |_{α}} ∥ u - w, v - w, u_{3} - w, \dots, u_{n} - w ∥ \\ = \frac{1}{| β |_{α}} ∥ v - w, u - w, u_{3} - w, \dots, u_{n} - w ∥_{α} \\ = \frac{1}{| β |_{α}} ∥ v - w, (u - w) + (v - w), u_{3} - w, \dots, u_{n} - w ∥_{α} \\ = ∥ v - w, \frac{(u - w) + (v - w)}{| β |_{α}}, u_{3} - w, \dots, u_{n} - w ∥_{α} \\ = ∥ v - w, z - w, u_{3} - w, \dots, u_{n} - w ∥_{α} . \end{matrix}$

Since h has f-n-UDPP, one has $\begin{matrix} ∥ h (u) - h (w), h (z) - h (w), h (u_{3}) - h (w), \\ \dots, h (u_{n}) - h (w) ∥_{α} \end{matrix}$ $\begin{matrix} = ∥ h (v) - h (w), h (z) - h (w), h (u_{3}) - h (w), \\ \dots, h (u_{n}) - h (w) ∥_{α} = | 1 |_{α} \end{matrix}$

Let s be a real number such that h (u) - h (w) = s (h (v) - h (w)). The injectivity of h implies that $h (w) = \frac{h (u) + h (v)}{2}$ . Similarly, $h (u) = \frac{h (w) + h (v)}{2}$ . It follows that h (u) = h (v) = h (w), which is impossible. Consequently h^-1 preserves collinearity. By using the previous proposition, for the midpoint of a segment, we prove that h preserves f-2-collinearity. On the other hand, there is u ≠ v ∈ U with $w = \frac{u + v}{2}$ such that h (u) , h (v) , h (w) are not collinear.

Now let z be in U such that $h (z) = \frac{h (u) + h (v)}{2}$ . Since h^-1 preserves f-2-collinearity, there is a scalar s ≠ 0 such that v - z = s (u - z). We can choose u₂, …, u_n ∈ U, satisfying ∥v - z, u₂, …, u_n ∥ _α = |1|_α for α ∈ (0, 1), so that $\bar{0 u_{2}}$ cuts $\bar{uv}$ only in one point u₀. We claim that for $h (\bar{0 u_{2}})$ and $\bar{0 h (u_{2})}$ , we get $h (\bar{0 u_{2}}) \subset \bar{0 h (u_{2})} \subset V .$

Otherwise, there are $x, y \in \bar{0 u_{2}}$ such that h (x) , h (y) , h (u₀) are not collinear. To check the claim, set $\begin{matrix} E : = affine (h (x), h (u_{0}), h (y)) and \\ F : = affine (h (u), h (u_{0}), h (v), h (w)) . \end{matrix}$

Since h^-1 preserves 2-collinearity, we have $h^{- 1} (E) \subset \bar{0 u_{2}}$ and $h^{- 1} (F) \subset \bar{uv}$ .

Evidently that h (u₀) ∈ E ∩ F and since any multiple of h (u₀), is in E ∩ F, then |E∩ F| = ∞ and so $h^{- 1} (E \cap F) \subset h^{- 1} (E) \cap h^{- 1} (F) \subset \bar{0 u_{2}} \cap \bar{uv} = u_{0} .$

Observe that the result is incorrect, because h is injective. Now, there are $λ_{1}, λ_{2} \in ℝ$ such that h (su₂) = λ₁h (u₂) and h (- su₂) = λ₂h (u₂). Since h has f-n-UDPP, we have $\begin{matrix} ∥ h (v) - h (z), h (u_{2}), h (u_{3}), \dots, h (u_{n}) ∥_{α} \\ = \frac{1}{2} ∥ h (u) - h (v), h (u_{2}), \dots, h (u_{n}) ∥_{α} = | 1 |_{α}, \\ ∥ h (u) - h (z), h ({su}_{2}), \dots, h (u_{n}) ∥_{α} \\ = \frac{1}{2} ∥ h (u) - h (v), λ_{1} h (u_{2}), \dots, h (u_{n}) ∥_{α} = | 1 |_{α}, \end{matrix}$ and $\begin{matrix} ∥ h (u) - h (z), h (- {su}_{2}), \dots, h (u_{n}) ∥_{α} \\ = \frac{1}{2} ∥ h (u) - h (v), λ_{2} h (u_{2}), \dots, h (u_{n}) ∥_{α} = | 1 |_{α} . \end{matrix}$

It follows that |λ_i|=1 for i = 1, 2. Since h is injective and s ≠ 1, the only possibility is λ₁ = -1 and λ₂ = 1, thus s = -1. (Note that |h (su₂) | = |λ₁h (u₂) | = |λ₂h (u₂) | = |h (- su₂) |, h injective and s ≠ 1, then λ₁ = -1 and λ₂ = 1, and s = -1, because h (su₂) = - h (- su₂)). Therefore $z = \frac{u + v}{2}$ . This contradiction means that, for the midpoint of a segment, h preserves f-2-collinearity. □

Now, we study Benz’ theorem [23].

Definition 3.7. Let (U, ∥ • , • , ⋯ , • ∥ _α) be the f-n-NLS corresponding to the fuzzy n-norm N on U. An f-n-NLS, (U, N, T), is an f-n-strictly convex space if for every u₀, u₁, …, u_n ∈ U, such that u₂, u₃, …, u_n ∉ span {u₀, u₁}, $\begin{matrix} ∥ u_{0} + u_{1}, u_{2}, \dots, u_{n} ∥_{α} = ∥ u_{0}, u_{2}, \dots, u_{n} ∥_{α} \\ + ∥ u_{1}, u_{2}, \dots, u_{n} ∥_{α}, \end{matrix}$ implies that u₀ = Ku₁, for some K ≥ 0.

Theorem 3.8. Consider two f-n-NLS U and V, where V is f-n-strictly convex. If h : U → V preserves f-β-n-distance, for β > 0, then it is an affine f-n-isometry.

Proof. Let (U, ∥ • , • , ⋯ , • ∥ _α) and (V, ∥ • , •, ⋯ , • ∥ _α) be f-n-NLS corresponding to the fuzzy n-norm N on U and V receptively, also let $(ℝ, | • |_{α})$ corresponding to the Euclidean fuzzy norm (see Remark 2.6). The proof goes through three steps.

We show that h preserves f-2β-n-distance. Let u₁, u₂, …, u_n, v₁, v₂, …, v_n be in U such that ∥u₁ - v₁, u₂ - v₂, …, u_n - v_n ∥ _α = 2|β|_α, for α ∈ (0, 1). Set $x_{i} = v_{1} + i (\frac{u_{1} - v_{1}}{2})$ , for each $i \in ℕ \cup {0}$ . Observe that x₀ = v₁ and x₂ = u₁ and $x_{i} - x_{i - 1} = \frac{u_{1} - v_{1}}{2}$ , for each i ∈ N. It follows that $\begin{matrix} ∥ x_{i} - x_{i - 1}, u_{2} - v_{2}, \dots, u_{n} - v_{n} ∥_{α} \\ = \frac{1}{2} ∥ u_{1} - v_{1}, u_{2} - v_{2}, \dots, u_{n} - v_{n} ∥_{α} \\ = | β |_{α}, \forall i \in ℕ . \end{matrix}$ Since h preserves f-β-n-distance and V is f-n-strictly convex, we have $\begin{matrix} ∥ h (u_{1}) - h (v_{1}), h (u_{2}) - h (v_{2}), \\ \dots, h (u_{n}) - h (v_{n}) ∥_{α} \\ = ∥ h (u_{1}) - h (x_{1}) + h (x_{1}) - h (v_{1}), \\ ah (u_{2}) - h (v_{2}), \dots, h (u_{n}) - h (v_{n}) ∥_{α} \\ = ∥ h (u_{1}) - h (x_{1}), h (u_{2}) - h (v_{2}), \\ \dots, h (u_{n}) - h (v_{n}) ∥_{α} \\ + ∥ h (x_{1}) - h (v_{1}), h (u_{2}) - h (v_{2}), \\ \dots, h (u_{n}) - h (v_{n}) ∥_{α} \\ = | β |_{α} + | β |_{α} \\ = | 2 β |_{α}, for α \in (0, 1) . \end{matrix}$

Let $w = \frac{u + v}{2}$ for any u ≠ v ∈ U and g (u) = h (u) - h (0), which implies that g preserves both f-β-n-distances and f-2β-n-distances. Let also h (0) be zero. We show that there are u₂, u₃, …, u_n ∈ U such that ∥v - w, u₂, …, u_n ∥ _α = |β|_α, for α ∈ (0, 1), and h (u_i) ∉ span {h (v) - h (w) , h (u) - h (w)} for i ∈ {2, 3, … n}.

Since dim U ≥ n, select v₂, v₃, …, v_n ∈ U such that $∥ v - w, v_{2}, v_{3}, \dots, v_{n} ∥_{α} = | β |_{α} .$

Now consider M₂ : = {y ∈ U : ∥ v - w, y, v₃, …, v_n ∥ = |β|_α}. We can choose u₂ ∈ M₂ such that h (u₂) ∉ span {h (v) - h (w) , h (u) - h (w)}. Let for each y ∈ M₂ there exist $0 \neq λ, μ \in ℝ$ such that $h (y) = λ (h (v) - h (w)) + μ (h (u) - h (w)) .$ Since ∥v - w, y, v₃, …, v_n ∥ _α = |β|_α, we get ∥u - w, y, v₃, …, v_n ∥ _α = |β|_α. Then h preserves f-β-n-distance and $∥ h (v) - h (w), h (y), h (v_{3}), \dots, h (v_{n}) ∥_{α} = | β |_{α},$ $∥ h (u) - h (w), h (y), h (v_{3}), \dots, h (v_{n}) ∥_{α} = | β |_{α} .$ Now using Lemma 2.9, we get $\begin{matrix} | μ | ∥ h (v) - h (w), h (u) - h (w), h (v_{3}), \\ \dots, h (v_{n}) ∥_{α} = | β |_{α}, \\ | λ | ∥ h (u) - h (w), h (v) - h (w), h (v_{3}), \\ \dots, h (v_{n}) ∥_{α} = | β |_{α} . \end{matrix}$ Then $\begin{matrix} | μ | ∥ h (v) - h (w), h (u) - h (w), \\ \dots, h (u_{n}) ∥_{α} \\ = | λ | ∥ h (v) - h (w), h (u) - h (w), \\ \dots, h (u_{n}) ∥_{α} . \end{matrix}$ This yields |μ| = |λ|. Since h is injective, for every real number r, y_r : = v₂ + r (v - w) belongs to M₂. So |h (M₂) |≤4, which is a contradiction. So there exists u₂ ∈ M₂ such that h (u₂) ∉ span {h (v) - h (w) , h (u) - h (w)}.

Next, set $\begin{matrix} M_{3} : = {y \in U : ∥ v - w, u_{2}, y, v_{4}, \\ \dots, v_{n} ∥_{α} = | β |_{α}} . \end{matrix}$ By the same method as above, we can choose u₃ ∈ M₃ such that $h (u_{3}) \notin span {h (v) - h (w), h (u) - h (w)} .$ We can find u₂, u₃, …, u_n ∈ U such that ∥v - w, u₂, …, u_n ∥ _α = |β|_α and h (u_i) ∉ span {h (v) - h (w) , h (u) - h (w)} for i = 2, 3, … n.

Let $w = \frac{u + v}{2}$ for any u ≠ v ∈ U, and let u₂, u₃, …, u_n be in U such that $∥ v - w, u_{2}, \dots, u_{n} ∥_{α} = | β |_{α},$ and h (u_i) ∉ span {h (v) - h (w) , h (u) - h (w)} for i = 2, 3, … n. Then, h preserves both f-β-n-distances and f-2β-n-distances, since, $\begin{matrix} | β |_{α} = ∥ v - w, u_{2}, \dots, u_{n} ∥_{α} \\ = \frac{1}{2} ∥ v - u, u_{2}, \dots, u_{n} ∥_{α}, \end{matrix}$ and so ∥v - u, u₂, …, u_n ∥ _α = |2β|_α. Hence, $\begin{matrix} ∥ h (v) - h (u), h (u_{2}), \dots, h (u_{n}) ∥_{α} \\ = | 2 β |_{α} = | β |_{α} + | β |_{α} \\ = ∥ h (v) - h (w), h (u_{2}), \dots, h (u_{n}) ∥_{α} \\ + ∥ h (u) - h (w), h (u_{2}), \dots, h (u_{n}) ∥_{α}, \end{matrix}$ for α ∈ (0, 1). The n-strict convexity of V implies that there exist a scalar K > 0 such that (v) - h (w) = K (h (w) - h (u)).

This completes the proof. □

As an example, we present a function that is continuous, one-to-one, surjective and has the s-f-n-UDPP property, but is not isometry.

Example 3.9. Consider $h : ℝ \to ℝ$ such that h (u) = [u] + (u - [u]) ², where [u] denotes the integer part of u. Since h (u) = [u] + u² - 2u [u] + [u] ² is polynomial, so h is continuous on $ℝ$ .

Now we show that h is one-to-one. Let for any u and v in $ℝ$ , h (u) = h (v) and put u - [u] = {u} for any $u \in ℝ$ , then [u] + {u} ² = [v] + {v} ² and we have [u] - [v] = {v} ² - {u} ² then {v} ² - {u} ² is an integer but -1 < {v} ² - {u} ² < 1, so {v} ² - {u} ² = 0, hence [u] = [v], and since ({v} - {u}) ({v} + {u}) =0, then either {v} = {u} =0 that u = [u] = [v] = v, or 0 < {v} = {u} <1 that v - [v] = u - [u] and again since [u] = [v] we have u = v. Then h is one-to-one.

Also, since $ℝ$ is connected set and h is continuous, by definition of h, h is surjective. Now, if |u - v|=1 then | [u] + {u} - [v] - {v} |=1 therefore $| [u] - [v] + {u} - {v} | = 1, ([u] - [v]) \in ℤ$ then $({u} - {v}) \in ℤ$ whereas -1 < {u} - {v} <1 then {u} - {v} =0 and therefore | [u] - [v] |=1 and |h (u) - h (v) | = | [u] - [v] + {u} ² - {v} ²|=1 and on the other hand, if |h (u) - h (v) |=1 then {u} ² - {v} ² = 0 that if {u} + {v} =0 then {u} = {v} =0 therefore u = [u] and v = [v] that consequently |u - v|=1 and if {u} - {v} =0 then u - [u] = v - [v] that is u - v = [u] - [v] then |u - v| = | [u] - [v] |=1. Then h has s-f-n-UDPP.

Now, suppose $h : ℝ \to ℝ$ and define $∥ • ∥_{α} : ℝ \to [0, 1]$ such that $∥ u ∥_{α} = inf [\frac{t}{| u | + t}]_{α}$ . We know that $(ℝ, ∥ • ∥_{α})$ is a fuzzy normed linear space. Now, suppose ||u - v||_α = |1|_α meaning $inf [\frac{t}{| u - v | + t}]_{α} = inf [\frac{t}{1 + t}]_{α}$ that is |u - v|=1 if and only if ∥h (u) - h (v) ∥ _α = ∥ [u] + {u} ² - [v] - {v} ² ∥ _α = $inf [\frac{t}{| [u] + {u}^{2} - [v] - {v}^{2} | + t}]_{α} = inf [\frac{t}{1 + t}]_{α} = | 1 |_{α}$ . Then h has s-f-n-UDPP, but it is not isometry, because if ∥u - v ∥ _α = ∥ h (u) - h (v) ∥ _α, then $inf [\frac{t}{| u - v | + t}]_{α} = inf [\frac{t}{| [u] - [v] + {u}^{2} - {v}^{2} | + t}]_{α}$ and so |u - v| = | [u] - [v] + {u} ² - {v} ²|, so that {u} + {v} =1, which is not always true, for example for $u = \frac{10}{3}$ and $v = \frac{23}{7}$ .

4 Conclusion

In this paper, we defined a fuzzy n-normed linear space in the sense of Morsi by using the concept of fuzzy real number and showed the relationship to fuzzy n-normed linear space (U, N, T). Next, we considered the Aleksandrov problem in fuzzy n-normed spaces, and show that every surjective fuzzy function preserving unit n-distance is affine, and thus is a fuzzy n-isometry. Finally, we proved the Benz theorem when the target space is fuzzy n-strictly convex.

Footnotes

Acknowledgments

The authors are thankful to the three anonymous referees for giving valuable comments and suggestions which helped to improve the final version of this paper. Also, we wish to thank the area Editor for useful comments that improved the presentation of the paper.

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