An introduction to 2-fuzzy n-normed linear spaces and a new perspective to the Mazur-Ulam problem

Abstract

The purpose of this article is to introduce the concept of 2-fuzzy n-normed linear space or fuzzy n-normed linear space of the set of all fuzzy sets of a non-empty set. We define the concepts of n-isometry, n-collinearity n-Lipschitz mapping in this space. Also, we generalize the Mazur-Ulam theorem, that is, when X is a 2-fuzzy n-normed linear space or ℑ(X) is a fuzzy n-normed linear space, the Mazur-Ulam theorem holds. Moreover, it is shown that each n-isometry in 2-fuzzy n-normed linear spaces is affine.

Mathematics Subject Classification (2010): 03E72; 46B20; 51M25; 46B04; 46S40.

1. Introduction

A satisfactory theory of 2-norms and n-norms on a linear space has been introduced and developed by Gähler [1, 2]. Following Misiak [3], Kim and Cho [4], and Malčeski [5] developed the theory of n-normed space. In [6], Gunawan and Mashadi gave a simple way to derive an (n - 1)-norm from the n-norms and realized that any n-normed space is an (n - 1)-normed space. Different authors introduced the definitions of fuzzy norms on a linear space. Cheng and Mordeson [7] and Bag and Samanta [8] introduced a concept of fuzzy norm on a linear space. The concept of fuzzy n-normed linear spaces has been studied by many authors (see [4, 9]).

Recently, Somasundaram and Beaula [10] introduced the concept of 2-fuzzy 2-normed linear space or fuzzy 2-normed linear space of the set of all fuzzy sets of a set. The authors gave the notion of α-2-norm on a linear space corresponding to the 2-fuzzy 2-norm by using some ideas of Bag and Samanta [8] and also gave some fundamental properties of this space.

In 1932, Mazur and Ulam [11] proved the following theorem.

Mazur-Ulam Theorem. Every isometry of a real normed linear space onto a real normed linear space is a linear mapping up to translation.

Baker [12] showed an isometry from a real normed linear space into a strictly convex real normed linear space is affine. Also, Jian [13] investigated the generalizations of the Mazur-Ulam theorem in F*-spaces. Rassias and Wagner [14] described all volume preserving mappings from a real finite dimensional vector space into itself and Väisälä [15] gave a short and simple proof of the Mazur-Ulam theorem. Chu [16] proved that the Mazur-Ulam theorem holds when X is a linear 2-normed space. Chu et al. [17] generalized the Mazur-Ulam theorem when X is a linear n-normed space, that is, the Mazur-Ulam theorem holds, when the n-isometry mapped to a linear n-normed space is affine. They also obtain extensions of Rassias and Šemrl's theorem [18]. Moslehian and Sadeghi [19] investigated the Mazur-Ulam theorem in non-archimedean spaces. Choy et al. [20] proved the Mazur-Ulam theorem for the interior preserving mappings in linear 2-normed spaces. They also proved the theorem on non-Archimedean 2-normed spaces over a linear ordered non-Archimedean field without the strict convexity assumption. Choy and Ku [21] proved that the barycenter of triangle carries the barycenter of corresponding triangle. They showed the Mazur-Ulam problem on non-Archimedean 2-normed spaces using the above statement. Xiaoyun and Meimei [22] introduced the concept of weak n-isometry and then they got under some conditions, a weak n-isometry is also an n-isometry. Cobzaş [23] gave some results of the Mazur-Ulam theorem for the probabilistic normed spaces as defined by Alsina et al. [24]. Cho et al. [25] investigated the Mazur-Ulam theorem on probabilistic 2-normed spaces. Alaca [26] introduced the concepts of 2-isometry, collinearity, 2-Lipschitz mapping in 2-fuzzy 2-normed linear spaces. Also, he gave a new generalization of the Mazur-Ulam theorem when X is a 2-fuzzy 2-normed linear space or ℑ(X) is a fuzzy 2-normed linear space. Kang et al. [27] proved that the Mazur-Ulam theorem holds under some conditions in non-Archimedean fuzzy normed space. Kubzdela [28] gave some new results for isometries, Mazur-Ulam theorem and Aleksandrov problem in the framework of non-Archimedean normed spaces. The Mazur-Ulam theorem has been extensively studied by many authors (see [29, 30]).

In the present article, we introduce the concept of 2-fuzzy n-normed linear space or fuzzy n-normed linear space of the set of all fuzzy sets of a non-empty set. We define the concepts of n-isometry, n-collinearity, n-Lipschitz mapping in this space. Also, we generalize the Mazur-Ulam theorem, that is, when X is a 2-fuzzy n-normed linear space or ℑ(X) is a fuzzy n-normed linear space, the Mazur-Ulam theorem holds. It is moreover shown that each n-isometry in 2-fuzzy n-normed linear spaces is affine.

2. Preliminaries

Definition 2.1([31]) Let n ∈ ℕ and let X be a real vector space of dimension d ≥ n. (Here we allow d to be infinite.) A real-valued function ∥●, ..., ●∥ on $\underset{n}{\underbrace{X\times \cdots \times X}}$ satisfying the following properties

• (1) ∥x₁, x₂, ..., x_n∥ = 0 if and only if x₁, x₂, ..., x _n are linearly dependent,

• (2) ∥x₁, x₂, ..., x _n ∥ is invariant under any permutation,

• (3) ∥x₁, x₂, ..., αx _n ∥ = |α| ∥x₁, x₂, ..., x_n∥ for any α ∈ ℝ,

• (4) ∥x₁, x₂, ..., x_n-1, y + z∥ ≤ ∥x₁, x₂, ..., x_n-1, y∥ + ∥x₁, x₂, ..., x_n-1, z∥, is called an n-norm on X and the pair (X, ∥●, ..., ●∥) is called an n-normed linear space.

Definition 2.2 [9] Let X be a linear space over S (field of real or complex numbers). A fuzzy subset N of Xⁿ× ℝ (ℝ, the set of real numbers) is called a fuzzy n-norm on X if and only if:

(N1) For all t ∈ ℝ with t ≤ 0, N(x₁, x₂, ..., x _n , t) = 0,

(N2) For all t ∈ ℝ with t > 0, N(x₁, x₂, ..., x _n , t) = 1 if and only if x₁, x₂, ..., x _n are linearly dependent,

(N3) N(x₁, x₂, ..., x _n , t) is invariant under any permutation of x₁, x₂, ..., x _n ,

(N4) For all t ∈ ℝ with t > 0, $N\left(x_1,x_2,\dots ,\lambda x_n,t\right)=N\left(x_1,x_2,\dots ,x_n,\frac{t}{\lambda }\right)$, if λ ≠ 0, λ ∈ S,

(N5) For all s, t ∈ ℝ

$$N\left(x_1,x_2,\dots ,x_n+x_n^{\prime },s+t\right)\ge \text{min}\left\{N\left(x_1,x_2,\dots ,x_n,s\right),N\left(x_1,x_2,\dots ,\underset{n}{\overset{\prime }{x}},t\right)\right\},$$

(N6) N(x₁, x₂, ..., x _n , t) is a non-decreasing function of t ∈ ℝ and $\underset{t\to \infty }{\text{lim}}N\left(x_1,x_2,\dots ,x_n,t\right)=1$.

Then (X, N) is called a fuzzy n-normed linear space or in short f-n-NLS.

Theorem 2.1 [9] Let (X, N) be an f-n-NLS. Assume that

(N7) N(x₁, x₂, ..., x _n ,t) > 0 for all t > 0 implies that x₁, x₂, ..., x _n are linearly dependent.

Define

$$\left|\right|x_1,x_2,\dots ,x_n|{|}_{\alpha }=\text{inf}\left\{t:N\left(x_1,x_2,\dots x_n,t\right)\ge \alpha ,\alpha \in \left(0,1\right)\right\}.$$

Then {∥●, ●, ..., ●∥ _α : α ∈ (0, 1)} is an ascending family of n-norms on X.

We call these n-norms as α-n-norms on X corresponding to the fuzzy n-norm on X.

Definition 2.3 Let X be any non-empty set and ℑ(X) the set of all fuzzy sets on X. For U, V ∈ ℑ(X) and λ ∈ S the field of real numbers, define

$$U+V=\left\{\left(x+y,\nu \wedge \mu \right):\left(x,\nu \right)\in U,\left(y,\mu \right)\in V\right\}$$

and λU = {(λx, ν): (x, ν) ∈ U}.

Definition 2.4 A fuzzy linear space $\hat{X}=X\times \left(0,1\right]$ over the number field S, where the addition and scalar multiplication operation on X are defined by (x, ν) + (y, μ) = (x + y, ν∧μ), λ(x, ν) = (λx, ν) is a fuzzy normed space if to every $\left(x,\nu \right)\in \hat{X}$ there is associated a non-negative real number, ∥(x, ν)∥, called the fuzzy norm of (x, ν), in such away that

• (i) ∥(x, ν)∥ = 0 iff x = 0 the zero element of X, ν ∈ (0, 1],

• (ii) ∥λ(x, ν)∥ = |λ| ∥(x, ν)∥ for all $\left(x,\nu \right)\in \hat{X}$ and all λ ∈ S,

• (iii) ∥(x, ν) + (y, μ) || ≤ ∥(x, ν ∧ μ)∥ + ∥(y, ν ∧ μ)∥ for all $\left(x,\nu \right),\left(y,\mu \right)\in \hat{X}$,

• (iv) ∥(x, ∨ _t ν _t )∥ = ∧ _t ∥(x, ν _t )∥ for all ν _t ∈ (0, 1].

3. 2-fuzzy n-normed linear spaces

In this section, we define the concepts of 2-fuzzy n-normed linear spaces and α-n-norms on the set of all fuzzy sets of a non-empty set.

Definition 3.1 Let X be a non-empty and ℑ(X) be the set of all fuzzy sets in X. If f ∈ ℑ(X) then f = {(x, μ): x ∈ X and μ ∈ (0, 1]}. Clearly f is bounded function for |f(x)| ≤ 1. Let S be the space of real numbers, then ℑ(X) is a linear space over the field S where the addition and scalar multiplication are defined by

$$f+g=\left\{\left(x,\mu \right)+\left(y,\eta \right)\right\}=\left\{\left(x+y,\mu \wedge \eta \right):\left(x,\mu \right)\in f\text{and}\left(y,\eta \right)\in g\right\}$$

and

$$\lambda f=\left\{\left(\lambda x,\mu \right):\left(x,\mu \right)\in f\right\}$$

where λ ∈ S.

The linear space ℑ(X) is said to be normed linear space if, for every f ∈ ℑ(X), there exists an associated non-negative real number ∥f∥ (called the norm of f) which satisfies

• (i) ∥f∥ = 0 if and only if f = 0. For

  $$\begin{array}{l}\hfill \left|\right|f\left|\right|=0\\ \iff \left\{\left|\right|\left(x,\mu \right)\left|\right|:\left(x,\mu \right)\in f\right\}=0\\ \iff x=0,\mu \in \left(0,1\right]\iff f=0.\end{array}$$

• (ii) ∥λf∥ = |λ| ∥f∥, λ ∈ S. For

  $$\begin{array}{ll}\hfill \left|\right|\lambda f\left|\right|& =\left\{\left|\right|\lambda \left(x,\mu \right)\left|\right|:\left(x,\mu \right)\in f,\lambda \in S\right\}\\ =\left\{\left|\lambda \right|\left|\right|\left(x,\mu \right)\left|\right|:\left(x,\mu \right)\in f\right\}=\left|\lambda \right|\left|\right|f\left|\right|.\end{array}$$

• (iii) ∥f + g∥ ≤ ∥f∥ + ∥g∥ for every f, g ∈ ℑ(X). For

  $$\begin{array}{ll}\hfill \left|\right|f+g\left|\right|& =\left\{\left|\right|\left(x,\mu \right)+\left(y,\eta \right)\left|\right|:x,y\in X,\mu ,\eta \in \left(0,1\right]\right\}\\ =\left\{\left|\right|\left(x+y\right),\left(\mu \wedge \eta \right)\left|\right|:x,y\in X,\mu ,\eta \in \left(0,1\right]\right\}\\ =\left\{\left|\right|\left(x,\mu \wedge \eta \right)\left|\right|+\left|\right|\left(y,\mu \wedge \eta \right)\left|\right|:\left(x,\mu \right)\in f,\left(y,\eta \right)\in g\right\}\\ =\left|\right|f\left|\right|+\left|\right|g\left|\right|.\end{array}$$

Then (ℑ(X),∥●∥) is a normed linear space.

Definition 3.2 A 2-fuzzy set on X is a fuzzy set on ℑ(X).

Definition 3.3 Let X be a real vector space of dimension d ≥ n (n ∈ ℕ) and ℑ(X) be the set of all fuzzy sets in X. Here we allow d to be infinite. Assume that a [0, 1]-valued function ∥●, ..., ●∥ on $\underset{n}{\underbrace{\Im \left(X\right)\times \cdots \times \Im \left(X\right)}}$ satisfies the following properties

• (1) ∥f₁, f₂, ..., f _n ∥ = 0 if and only if f₁, f₂, ..., f _n are linearly dependent,

• (2) ∥f₁, f₂, ..., f _n ∥ is invariant under any permutation,

• (3) ∥f₁, f₂, ..., λf _n ∥ = |λ| ∥f₁, f₂, ..., f _n ∥ for any λ ∈ S,

• (4) ∥f₁, f₂, ..., f_n-1, y + z∥ ≤ ∥f₁, f₂, ..., f_n-1, y∥ + ∥f₁, f₂, ..., f_n-1, z∥.

Then (ℑ(X),∥●,...,●∥) is an n-normed linear space or (X, ∥●, ..., ●∥) is a 2-n-normed linear space.

Definition 3.4 Let ℑ(X) be a linear space over the real field S. A fuzzy subset N of $\underset{n}{\underbrace{\Im \left(X\right)\times \cdots \times \Im \left(X\right)}}\times ℝ$ is called a 2-fuzzy n-norm on X (or fuzzy n-norm on ℑ(X)) if and only if

(2-N1) for all t ∈ ℝ with t ≤ 0, N(f₁, f₂, ..., f _n , t) = 0,

(2-N2) for all t ∈ ℝ with t > 0, N(f₁, f₂, ..., f _n , t) = 1 if and only if f₁, f₂, ..., f _n are linearly dependent,

(2-N3) N(f₁, f₂, ..., f _n , t) is invariant under any permutation of f₁, f₂, ..., f _n ,

(2-N4) for all t ∈ ℝ with t > 0, N(f₁, f₂, ..., λf _n , t) = N(f₁, f₂, ..., f _n , t/|λ|), if λ ≠ 0, λ ∈ S,

(2-N5) for all s, t ∈ ℝ,

$$N\left(f_1,f_2,\dots ,f_n+f_n^{\prime },s+t\right)\ge \text{min}\left\{N\left(f_1,f_2,\dots ,f_n,s\right),N\left(f_1,f_2,\dots ,\underset{n}{\overset{\prime }{f}},t\right)\right\},$$

(2-N6) N(f₁, f₂, ..., f _n , ·): (0, ∞) → [0, 1] is continuous,

(2-N7) $\underset{t\to \infty }{\text{lim}}N\left(f_1,f_2,\dots ,f_{n,}t\right)=1$.

Then ℑ(X), N) is a fuzzy n-normed linear space or (X, N) is a 2-fuzzy n-normed linear space.

Remark 3.1 In a 2-fuzzy n-normed linear space (X, N), N(f₁, f₂, ..., f _n , ·) is a non-decreasing function of ℝ for all f₁, f₂,...,f _n ∈ ℑ(X).

Remark 3.2 From (2-N4) and (2-N5), it follows that in a 2-fuzzy n-normed linear space,

(2-N4) for all t ∈ ℝ with t > 0, $N\left(f_1,f_2,\dots ,\lambda f_i,\dots ,f_n,t\right)=N\left(f_1,f_2,\dots ,f_i,\dots ,f_n,\frac{t}{\left|\lambda \right|}\right),$ if λ ≠ 0, λ ∈ S,

(2-N5) for all s, t ∈ ℝ,

$$\begin{array}{c}N\left(f_1,f_2,\dots ,f_i+{f^{\prime }}_i,\dots ,f_n,s+t\right)\\ \ge \text{min}\left\{N\left(f_1,f_2,\dots ,f_i,\dots ,f_n,s\right),N\left(f_1,f_2,\dots ,{f^{\prime }}_i,\dots ,f_n,t\right)\right\}.\end{array}$$

The following example agrees with our notion of 2-fuzzy n-normed linear space.

Example 3.1 Let (ℑ(X),∥●,●,...,●∥) be an n-normed linear space as in Definition 3.3. Define

$$N\left(f_1,f_2,\dots ,f_{n,}t\right)=\left\{\begin{array}{c}\frac{t}{t+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}ift>0,t\in ℝ,\hfill \\ 0\hfill & ift\le 0\hfill \end{array}\right.$$

for all $\left(f_1,f_2,\dots ,f_n\right)\in \underset{n}{\underbrace{\Im \left(X\right)\times \cdots \times \Im \left(X\right)}}$. Then (X, N) is a 2-fuzzy n-normed linear space.

Solution. (2-N1) For all t ∈ ℝ with t ≤ 0, by definition, we have N(f₁, f₂, ..., f _n , t) = 0.

(2-N2) For all t ∈ ℝ with t > 0,

$$\begin{array}{ll}\hfill N\left(f_1,f_2,\dots ,f_n,t\right)& =1\iff \frac{t}{t+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}=1\\ \iff t=t+\left|\right|f_1,f_2,\dots ,f_n\left|\right|\\ \iff \left|\right|f_1,f_2,\dots ,f_n\left|\right|=0\\ \iff f_1,f_2,\dots ,f_n\text{ are linearly dependent}\text{.}\end{array}$$

(2-N3) For all t ∈ ℝ with t > 0,

$$\begin{array}{ll}\hfill N\left(f_1,f_2,\dots ,f_n,t\right)& =\frac{t}{t+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}=\frac{t}{t+\left|\right|f_1,f_2,\dots ,f_n,f_{n-1}\left|\right|}\\ =N\left(f_1,f_2,\dots ,f_n,f_{n-1},t\right)=\cdots .\end{array}$$

(2-N4) For all t ∈ ℝ with t > 0 and λ ∈ F, λ ≠ 0,

$$\begin{array}{l}N(f_1,f_2,\dots ,f_n,t/|\lambda |)=\frac{t/\left|\lambda \right|}{t/\left|\lambda \right|+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}\\ =\frac{t/\left|\lambda \right|}{(t+|\lambda \left|\right||f_1,f_2,\dots ,f_n||)/\left|\lambda \right|}\\ =\frac{t}{t+\left|\lambda \right|\left|\right|f_1,f_2,\dots ,f_n\left|\right|}\\ =\frac{t}{(t+||f_1,f_2,\dots ,\lambda f_n||}=N(f_1,f_2,\dots ,\lambda f_n,t).\end{array}$$

(2-N5) We have to prove

$$N\left(f_1,f_2,\dots ,f_n+f_n^{\prime },s+t\right)\ge \text{min}\left\{f\left(x_1,f_2,\dots ,f_n,s\right),N\left(f_1,f_2,\dots ,\underset{n}{\overset{\prime }{f}},t\right)\right\}.$$

• (i) s + t < 0,

• (ii) s = t = 0,

• (iii) s + t > 0; s > 0, t < 0; s < 0, t > 0, then the above relation is obvious. If

• (iv) s > 0, t > 0, s + t > 0, then

  $$N(f_1,f_2,\dots ,f_n+{f^{\prime }}_n,s+t=\frac{s+t}{s+t+\left|\right|f_1,f_2,\dots ,f_n+{f^{\prime }}_n\left|\right|}.$$

If

$$\begin{array}{ll}\hfill \frac{s}{s+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}& \ge \frac{t}{t+\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}\Rightarrow \frac{\left|\right|f_1,f_2,\dots ,f_n\left|\right|}{s}\le \frac{\left|\right|x_1,x_2,\dots ,{x^{\prime }}_n\left|\right|}{t}\\ \Rightarrow \frac{\left|\right|f_1,f_2,\dots ,f_n\left|\right|}{s}+\frac{\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}{s}\le \frac{\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}{t}\\ +\frac{\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}{s}\\ \Rightarrow \frac{\left|\right|f_1,f_2,\dots ,f_n+{f^{\prime }}_n\left|\right|}{s}\le \left(\frac{s+t}{s\cdot t}\right)\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|\\ \Rightarrow \frac{\left|\right|f_1,f_2,\dots ,f_n+{f^{\prime }}_n\left|\right|}{s+t}\le \frac{\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}{t}\\ \Rightarrow \frac{s+t+\left|\right|f_1,f_2,\dots ,f_n+{f^{\prime }}_n\left|\right|}{s+t}\le \frac{t+\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}{t}\\ \Rightarrow \frac{s+t}{s+t+\left|\right|f_1,f_2,\dots ,f_n+{f^{\prime }}_n\left|\right|}\ge \frac{t}{t+\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}\\ \Rightarrow N\left(f_1,f_2,\dots ,f_n+{f^{\prime }}_n,s+t\right)\ge N\left(f_1,f_2,\dots ,{f^{\prime }}_n,t\right).\end{array}$$

Similarly, if

$$\begin{array}{c}\frac{t}{t+\left|\right|f_1,f_2,\dots ,{f^{\prime }}_n\left|\right|}\ge \frac{s}{s+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}\\ \Rightarrow N\left(f_1,f_2,\dots ,f_n+{f^{\prime }}_n,s+t\right)\ge N\left(f_1,f_2,\dots ,f_n,t\right).\end{array}$$

Thus

$$N\left(f_1,f_2,\dots ,f_n+f_n^{\prime },s+t\right)\ge \text{min}\left\{N\left(f_1,f_2,\dots ,f_n,s\right),N\left(f_1,f_2,\dots ,\underset{n}{\overset{\prime }{f}},t\right)\right\}.$$

(2-N6) It is clear that N(f₁, f₂, ..., f _n , ·): (0, ∞) → [0, 1] is continuous.

(2-N7) For all t ∈ ℝ with t > 0,

$$\begin{array}{ll}\hfill \underset{t\to \infty }{\text{lim}}N\left(f_1,f_2,\dots ,f_n,t\right)& =\underset{t\to \infty }{\text{lim}}\frac{t}{t+\left|\right|f_1,f_2,\dots ,f_n\left|\right|}\\ =\underset{t\to \infty }{\text{lim}}\frac{t}{t\left(1+\left(1/t\right)\left|\right|f_1,f_2,\dots ,f_n\left|\right|\right)}=1,\end{array}$$

as desired.

As a consequence of Theorem 3.2 in [10], we introduce an interesting notion of ascending family of α-n-norms corresponding to the fuzzy n-norms in the following theorem.

Theorem 3.1 Let (ℑ(X), N) is a fuzzy n-normed linear space. Assume that

(2-N8) N(f₁, f₂, ..., f _n , t) > 0 for all t > 0 implies f₁, f₂, ..., f _n are linearly dependent.

Define

$$\left|\right|f_1,f_2,\dots ,f_n|{|}_{\alpha }=\text{inf}\left(t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge \alpha ,\alpha \in \left(0,1\right)\right\}.$$

Then {∥●, ●, ..., ●∥ _α : α ∈ (0, 1)} is an ascending family of n-norms on ℑ(X).

These n-norms are called α-n-norms on ℑ(X) corresponding to the 2-fuzzy n-norm on X.

Proof. (i) Let ∥f₁, ..., f _n ∥ _α = 0. This implies that inf {t : N(f₁, ..., f _n , t) ≥ α}. Then, N(f₁, f₂, ..., f _n , t) ≥ α > 0, for all t > 0, α ∈ (0, 1), which implies that f₁, f₂, ..., f _n are linearly dependent, by (2-N8).

Conversely, assume f₁, f₂, ..., f _n are linearly dependent. This implies that N(f₁, f₂, ..., f _n , t) = 1 for all t > 0. For all α ∈ (0, 1), inf {t : N(f₁, f₂, ..., f _n , t) ≥ α}, which implies that ∥f₁, f₂, ..., f _n ∥ _α = 0.

• (ii) Since N(f₁, f₂, ..., f _n , t) is invariant under any permutation, ∥f₁, f₂, ..., f _n ∥ _α = 0 under any permutation.

• (iii) If λ ≠ 0, then

  $$\begin{array}{l}\left|\right|f_1,f_2,\dots ,\lambda f_n|{|}_{\alpha }=\inf \{s:N(f_1,f_2,\dots ,f_n,s)\ge \alpha \}\\ =\inf \{s:N(f_1,f_2,\dots ,f_n,\frac{s}{\left|\lambda \right|}\ge \alpha \}.\end{array}$$

Let $t=\frac{s}{\left|\lambda \right|}$, then

$$\begin{array}{c}\left|\right|f_1,f_2,\dots ,\lambda f_n|{|}_{\alpha }=\text{inf}\left\{\left|\lambda \right|t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge \alpha \right\}\\ =\left|\lambda \right|\text{inf}\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge \alpha \right\}=\left|\lambda \right|\left|\right|f_1,f_2,\dots ,f_n\left|\right|\alpha .\end{array}$$

If λ = 0, then

$$\begin{array}{l}\left|\right|f_1,f_2,\dots ,\lambda f_n|{|}_{\alpha }=||f_1,f_2,\dots ,0|{|}_{\alpha }=0=0\left|\right|f_1,f_2,\dots ,f_n\left|\right|\alpha \\ =\left|\lambda \right|\left|\right|f_1,f_2,\dots ,f_n\left|\right|\alpha ,\forall \lambda \in S(\text{field}).\end{array}$$

(iv)

$$\begin{array}{c}{∥f_1,f_2,\dots ,f_n∥}_{\alpha }+{∥f_1,f_2,\dots ,{f^{\prime }}_n∥∥}_{\alpha }\\ =\text{inf}\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge \alpha \right\}+\text{inf}\left(s:N\left(f_1,f_2,\dots ,{f^{\prime }}_n,s\right)\ge \alpha \right\}\\ =\text{inf}\left\{t+s:N\left(f_1,f_2,\dots ,f_n,t\right)\ge \alpha ,N\left(f_1,f_2,\dots ,{f^{\prime }}_n,s\right)\ge \alpha \right\}\\ \ge \text{inf}\left\{t+s:N\left(f_1,f_2,\dots ,f_n+{f^{\prime }}_n,t+s\right)\ge \alpha \right\},\\ \ge \text{inf}\left\{r:N\left(f_1,f_2,\dots ,f_n+{f^{\prime }}_n,r\right)\ge \alpha \right\},r=t+s\\ ={∥f_1,f_2,\dots ,f_n+{f^{\prime }}_n∥}_{\alpha }.\end{array}$$

Hence

$$\left|\right|f_1,f_2,\dots ,f_n+f_n^{\prime }|{|}_{\alpha }\le ||f_1,f_2,\dots ,f_n|{|}_{\alpha }+\left|\right|f_1,f_2,\dots ,f_n^{\prime }|{|}_{\alpha }.$$

Thus {∥●, ●, ..., ●∥ _α : α ∈ (0, 1)} is an α-n-norm on X.

Let 0 < α₁ < α₂. Then,

$$\begin{array}{c}{∥f_1,f_2,\dots ,f_n∥}_{{\alpha }_1}=\text{inf}\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge {\alpha }_1\right\},\\ {∥f_1,f_2,\dots ,f_n∥}_{{\alpha }_2}=\text{inf}\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge {\alpha }_2\right\}.\end{array}$$

As α₁ < α₂,

$$\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge {\alpha }_2\right\}\subset \left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge {\alpha }_1\right\}$$

implies that

$$\text{inf}\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge {\alpha }_2\right\}\ge \text{inf}\left\{t:N\left(f_1,f_2,\dots ,f_n,t\right)\ge {\alpha }_1\right\}$$

which implies that

$$\left|\right|f_1,f_2,\dots ,f_n|{|}_{{\alpha }_2}\ge ||f_1,f_2,\dots ,f_n|{|}_{{\alpha }_1}.$$

Hence {∥●, ●, ..., ●∥ _α : α ∈ (0, 1)} is an ascending family of α-n-norms on x corresponding to the 2-fuzzy n-norm on X.

4. On the Mazur-Ulam problem

In this section, we give a new generalization of the Mazur-Ulam theorem when X is a 2-fuzzy n-normed linear space or ℑ(X) is a fuzzy n-normed linear space. Hereafter, we use the notion of fuzzy n-normed linear space on ℑ(X) instead of 2-fuzzy n-normed linear space on X.

Definition 4.1 Let ℑ(X) and ℑ(X) be fuzzy n-normed linear spaces and Ψ : ℑ(X) → ℑ(Y) a mapping. We call Ψ an n-isometry if

$${∥f_1-f_0,\dots ,f_n-f_0∥}_{\alpha }={∥\Psi \left(f_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)∥}_{\beta }$$

for all f₀, f₁, f₂,...,f _n ∈ ℑ(X) and α, β ∈ (0, 1).

For a mapping Ψ, consider the following condition which is called the n-distance one preserving property (n DOPP).

$$\left(n\text{DOPP}\right)\text{Let}{f}_0,f_1,f_2,\dots ,f_n\in \Im \left(X\right)\text{with}{∥f_1-f_0,\dots ,f_n-f_0∥}_{\alpha }=1.$$

Then ∥Ψ(f₁) - Ψ(f₀), ..., Ψ(f _n ) - Ψ(f₀)∥ _β = 1.

Lemma 4.1 Let f₁, f₂,...,f _n ∈ ℑ(X), α ∈ (0, 1) and ħ ∈ ℝ. Then,

$$\left|\right|f_1,\dots ,f_i,\dots ,f_j,\dots ,f_n|{|}_{\alpha }=||f_1,\dots ,f_i,\dots ,f_j+\hslash f_i,\dots ,f_n|{|}_{\alpha }$$

for all 1 ≤ i ≠ j ≤ n.

Proof. It is obviously true.

Lemma 4.2 For $f_0,f_0^{\prime }\in \Im \left(X\right)$, if f₀ and $f_0^{\prime }$ are linearly dependent with some direction, that is, $f_0^{\prime }=t{f}_0$ for some t > 0, then

$$\left|\right|f_0+f_0^{\prime },f_1,\dots ,f_n|{|}_{\alpha }=||f_0,f_1,\dots ,f_n|{|}_{\alpha }+\left|\right|f_0^{\prime },f_1,\dots ,f_n|{|}_{\alpha }$$

for all f₁, f₂,...,f _n ∈ ℑ(X) and α ∈ (0, 1).

Proof. Let $f_0^{\prime }=t{f}_0$ for some t > 0. Then we have

$$\begin{array}{ll}\hfill \left|\right|f_0+{f^{\prime }}_0,f_1,\dots ,f_n|{|}_{\alpha }& =\left|\right|f_0+t{f}_0,f_1,\dots ,f_n|{|}_{\alpha }\\ =\left(1+t\right)\left|\right|f_0,f_1,\dots ,f_n|{|}_{\alpha }\\ =\left|\right|f_0,f_1,\dots ,f_n|{|}_{\alpha }+t||f_0,f_1,\dots ,f_n|{|}_{\alpha }\\ =\left|\right|f_0,f_1,\dots ,f_n|{|}_{\alpha }+||{f^{\prime }}_0,f_1,\dots ,f_n|{|}_{\alpha }\end{array}$$

for all f₁, f₂,...,f _n ∈ ℑ(X) and α ∈ (0, 1).

Definition 4.2 The elements f₀, f₁, f₂, ..., f _n of ℑ(X) are said to be n-collinear if for every i, {f _j - f _i : 0 ≤ j ≠ i ≤ n} is linearly dependent.

Remark 4.1 The elements f₀, f₁, and f₂ are said to be 2-collinear if and only if f₂ - f₀ = r(f₁ - f₀) for some real number r.

Now we define the concept of n-Lipschitz mapping.

Definition 4.3 We call Ψ an n-Lipschitz mapping if there is a κ ≥ 0 such that

$$\left|\right|\Psi \left(f_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\le k||f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }$$

for all f₀, f₁, f₂,...,f _n ∈ ℑ(X) and α, β ∈ (0, 1). The smallest such κ is called the n-Lipschitz constant.

Lemma 4.3 Assume that if f₀, f₁, and f₂ are 2 -collinear then Ψ(f₀), Ψ(f₁) and Ψ(f₂) are 2-collinear, and that Ψ satisfies (n DOPP). Then Ψ preserves the n-distance k for each k ∈ ℕ.

Proof. Suppose that there exist f₀, f₁ ∈ ℑ(X) with f₀ ≠ f₁ such that Ψ(f₀) = Ψ(f₁). Since dimℑ(X) ≥ n, there are f₂,...,f _n ∈ ℑ(X) such that f₁ - f₀, f₂ - f₀, ..., f _n - f₀ are linearly independent. Since ∥f₁ - f₀, f₂ - f₀, ..., f _n - f₀∥ _α ≠ 0, we can set

$$z_2=f_0+\frac{f_2-f_0}{\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }}.$$

Then we have

$$\begin{array}{c}\left|\right|f_1-f_0,z_2-f_0,f_3-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ ={∥f_1-f_0,\frac{f_2-f_0}{\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }},f_3-f_0,\dots ,f_n-f_0∥}_{\alpha }=1.\end{array}$$

Since Ψ preserves the unit n-distance,

$$\left|\right|\Psi \left(f_1\right)-\Psi \left(f_0\right),\Psi \left(z_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }=1.$$

But it follows from Ψ (f₀) = Ψ (f₁) that

$$\left|\right|\Psi \left(f_1\right)-\Psi \left(f_0\right),\Psi \left(z_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }=0,$$

which is a contradiction. Hence, Ψ is injective.

Let f₀, f₁, f₂, ..., f _n be elements of ℑ(X), k ∈ ℕ and

$$\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }=k.$$

We put

$$g_i=f_0+\frac{i}{k}\left(f_1-f_0\right),i=0,1,\dots ,k.$$

Then

$$\begin{array}{c}\left|\right|g_{i+1}-g_i,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ ={∥f_0+\frac{i+1}{k}\left(f_1-f_0\right)-\left(f_0+\frac{i}{k}\left(f_1-f_0\right)\right),f_2-f_0,\dots ,f_n-f_0∥}_{\alpha }\\ ={∥\frac{1}{k}\left(f_1-f_0\right),f_2-f_0,\dots ,f_n-f_0∥}_{\alpha }\\ =\frac{i}{k}\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }=\frac{k}{k}=1\end{array}$$

for all i = 0, 1, ..., k - 1. Since Ψ satisfies (n DOPP),

$$\left|\right|\Psi \left(g_{i+1}\right)-\Psi \left(g_i\right),\Psi \left(f_2\right)-\Psi \left(f_0\right)\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }=1$$

(4.1)

for all i = 0, 1, ..., k - 1. Since g₀, g₁, and g₂ are 2-collinear, Ψ (g₀), Ψ(g₁) and Ψ(g₂) are also 2-collinear. Thus there is a real number r₀ such that Ψ(g₂) - Ψ(g₁) = r₀ (Ψ(g₁) - Ψ(g₀)). It follows from (4.1) that

$$\begin{array}{c}\left|\right|\Psi \left(g_1\right)-\Psi \left(g_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\left|\right|\Psi \left(g_2\right)-\Psi \left(g_1\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\left|\right|r_0\left(\Psi \left(g_1\right)-\Psi \left(g_2\right)\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\left|r_0\right|\left|\right|\left(\Psi \left(g_1\right)-\Psi \left(g_0\right)\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }.\end{array}$$

Thus, we have r₀ = 1 or -1. If r₀ = -1, Ψ(g₂) - Ψ(g₁) = -Ψ(g₁) + Ψ(g₀), that is, Ψ(g₂) = Ψ(g₀). Since Ψ is injective, g₂ = g₀, which is a contradiction. Thus r₀ = 1. Then we have Ψ(g₂) - Ψ(g₁) = Ψ(g₁) - Ψ(g₀). Similarly, one can obtain that Ψ(g_i+1) - Ψ(g _i ) = Ψ(g _i ) - Ψ(g_i-1) for all i = 0, 1, ..., k - 1. Thus Ψ(g_i+1) - Ψ(g _i ) = Ψ(g₁) - Ψ(g₀) for all i = 0, 1, ..., k - 1. Hence

$$\begin{array}{c}\Psi \left(f_1\right)-\Psi \left(f_0\right)=\Psi \left(g_k\right)-\Psi \left(g_0\right)\\ =\Psi \left(g_k\right)-\Psi \left(g_{k-1}\right)+\Psi \left(g_{k-1}\right)-\Psi \left(g_{k-2}\right)+\cdots +\Psi \left(g_1\right)-\Psi \left(g_0\right)\\ =k\left(\Psi \left(g_1\right)-\Psi \left(g_0\right)\right).\end{array}$$

Hence

$$\begin{array}{c}\left|\right|\Psi \left(f_1\right)-\Psi \left(f_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\left|\right|k\left(\Psi \left(g_1\right)-\Psi \left(g_0\right)\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\left|\right|k\Psi \left(g_1\right)-\Psi \left(g_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }=k.\end{array}$$

This completes the proof.

Lemma 4.4 Let h, f₀, f₁, ..., f _n be elements of ℑ(X) and let h, f₀, f₁ be 2-collinear. Then

$$\left|\right|f_1-h,f_2-h,\dots ,f_n-h|{|}_{\alpha }=||f_1-h,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }.$$

Proof. Since h, f₀, f₁ are 2-collinear, there exists a real number r such that f₁ - h = r(f₀ - h). It follows from Lemma 4.1 that

$$\begin{array}{ll}\hfill \left|\right|f_1-h,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }& =\left|\right|r\left(f_0-h\right),f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ =\left|r\right|\left|\right|f_0-h,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ =\left|r\right|\left|\right|f_0-h,f_2-h,\dots ,f_n-h|{|}_{\alpha }\\ =\left|\right|r\left(f_0-h\right),f_2-h,\dots ,f_n-h|{|}_{\alpha }\\ =\left|\right|f_1-h,f_2-h,\dots ,f_n-h|{|}_{\alpha }.\end{array}$$

This completes the proof.

Theorem 4.1 Let Ψ be an n-Lipschitz mapping with the n-Lipschitz constant κ ≤ 1. Assume that if f₀, f₁, ..., f _n are m-collinear then Ψ(f₀), Ψ(f₁), ..., Ψ(f _m ) are m-collinear, m = 2, n, and that Ψ satisfies (n DOPP), then Ψ is an n-isometry.

Proof. It follows from Lemma 4.3 that Ψ preserves n-distance k for all k ∈ ℕ. For f₀, f₁, ..., f _n ∈ X, there are two cases depending upon whether ∥f₁ - f₀, ..., f _n - f₀∥ _α = 0 or not. In the case ∥f₁ - f₀, ..., f _n - f₀∥ _α = 0, f₁ - f₀, ..., f _n - f₀ are linearly dependent, that is, n-collinear. Thus f₁ - f₀, ..., f _n - f₀ are linearly dependent. Thus ∥Ψ(f₁) - Ψ(f₀), ..., Ψ(f _n ) - Ψ(f₀)∥ _β = 0.

In the case ∥f₁ - f₀, ..., f _n - f₀∥ _α > 0, there exists an n₀ ∈ ℕ such that

$$n_0>\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }.$$

Assume that

$$\left|\right|\Psi \left(f_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }<||f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }.$$

We can set

$$h=f_0+\frac{n_0}{\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }}\left(f_1-f_0\right).$$

Then we get

$$\begin{array}{c}\left|\right|h-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ ={∥f_0+\frac{n_0}{\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }}\left(f_1-f_0\right)-f_0,\dots ,f_n-f_0∥}_{\alpha }\\ =\frac{n_0}{\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }}\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }=n_0.\end{array}$$

It follows from Lemma 4.3 that

$$\left|\right|\Psi \left(h\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }=n_0.$$

By the definition of h,

$$h-f_1=\left(\frac{n_0}{\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }}-1\right)\left(f_1-f_0\right).$$

Since

$$\frac{n_0}{\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }}>1,$$

h - f₁ and f₁ - f₀ have the same direction. It follows from Lemma 4.2 that

$$\begin{array}{c}\left|\right|h-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ =\left|\right|h-f_1,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }+||f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }.\end{array}$$

Since Ψ(h), Ψ(f₁), Ψ(f₂) are 2-collinear, we have

$$\begin{array}{c}\left|\right|\Psi \left(h\right)-\Psi \left(f_1\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\left|\right|\Psi \left(f_1\right)-\Psi \left(h\right),\Psi \left(f_2\right)-\Psi \left(h\right),\dots ,\Psi \left(f_n\right)-\Psi \left(h\right)|{|}_{\beta }\\ \le \left|\right|f_1-h,f_2-h,\dots ,f_n-h|{|}_{\alpha }\\ =\left|\right|f_1-h,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ =n_0-\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\end{array}$$

by Lemma 4.4. By the assumption,

$$\begin{array}{ll}\hfill n_0& =\left|\right|\Psi \left(h\right)-\Psi \left(f_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ \le \left|\right|\Psi \left(h\right)-\Psi \left(f_1\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ +\left|\right|\Psi \left(f_1\right)-\Psi \left(f_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ <n_0-\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0|{|}_{\alpha }\\ +\left|\right|f_1-f_0,f_2-f_0,\dots ,f_n-f_0\left|\right|\alpha \\ =n_0,\end{array}$$

which is a contradiction. Hence Ψ is an n-isometry.

Lemma 4.5 Let g₀, g₁ be elements of ℑ(X). Then $v=\frac{g_0+g_1}{2}$ is the unique element of ℑ(X) satisfying

$$\begin{array}{c}\frac{1}{2}\left|\right|g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|\right|g_1-v,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|\right|g_0-g_n,g_0-v,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\end{array}$$

for some g₂,...,g _n ∈ ℑ(X) with ∥g₀ - g _n , g₁ - g _n , g₂ - g _n , ..., g_n-1- g _n ∥ _α ≠ 0 and v, g₀, g₁ 2-collinear.

Proof. Let ∥g₀ - g _n , g₁ - g _n , g₂ - g _n , ..., g_n-1- g _n ∥ _α ≠ 0 and $\left|\right|f_0-f_n,f_1-f_n,f_2-f_n,\dots ,f_{n-1}-f_n|{|}_{\alpha }\ne 0.$.

Then v, g₀, g₁ are 2-collinear. It follows from Lemma 4.1 and g _n - g ₀ = g₁ - g₀ - (g₁ - g _n ) that

$$\begin{array}{c}\left|\right|g_1-v,g_1-g_n,g_2-g_n,\dots g_{n-1}-g_n|{|}_{\alpha }\\ ={∥g_1\frac{g_0+g_1}{2},g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n∥}_{\alpha }\\ =\frac{1}{2}\left|\right|g_1-g_0,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\frac{1}{2}\left|\right|g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n\left|\right|\alpha \end{array}$$

and similarly

$$\left|\right|g_0-g_n,g_0-v,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }=\frac{1}{2}||g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }.$$

Now we prove the uniqueness.

Let u be an element of ℑ(X) satisfying the above properties. Since u, g₀, g₁ are 2-collinear, there exists a real number t such that u = tg₀ + (1 - t)g₁. It follows from Lemma 4.1 that

$$\begin{array}{c}\frac{1}{2}\left|\right|g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|\right|g_1-u,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|\right|g_1-\left(t{g}_0+\left(1-t\right)g_1\right),g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|t\right|\left|\right|g_1-g_0,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|t\right|\left|\right|g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\end{array}$$

and

$$\begin{array}{c}\frac{1}{2}\left|\right|g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|\right|g_0-g_n,g_0-u,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =\left|\right|g_0-g_n,g_0-\left(t{g}_0+\left(1-t\right)g_1\right),g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =|1-t|\left|\right|g_0-g_n,g_0-g_1,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }\\ =|1-t|\left|\right|g_0-g_n,g_1-g_n,g_2-g_n,\dots ,g_{n-1}-g_n|{|}_{\alpha }.\end{array}$$

Since ∥g₀ - g _n , g₁ - g _n , g₂ - g _n , ..., g_n-1- g _n ∥ _α ≠ 0, we have $\frac{1}{2}=|1-t|=\left|t\right|$. Therefore, we get $t=\frac{1}{2}$ and hence v = u.

Lemma 4.6 If Ψ is an n-isometry and f₀, f₁, f₂ are 2-collinear then Ψ(f₀), Ψ(f₁), Ψ(f₂) are 2-collinear.

Proof. Since dimℑ(X) ≥ n, for any f₀ ∈ ℑ(X), there exist g₁,...,g _n ∈ ℑ(X) such that g₁ - f₀, ..., g _n - f₀ are linearly independent. Then

$$\left|\right|g_1-f_0,\dots ,g_n-f_0|{|}_{\alpha }=||\Psi \left(g_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(g_n\right)-\Psi \left(f_0\right)|{|}_{\beta }\ne 0$$

and hence, the set A = {Ψ(f) -Ψ(f₀) : f ∈ ℑ(X)} contains n linearly independent vectors.

Assume that f₀, f₁, f₂ are 2-collinear. Then, for any f₃,...,f _n ∈ ℑ(X),

$$\left|\right|f_1-f_0,\dots ,f_n-f_0|{|}_{\alpha }=||\Psi \left(f_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_n\right)-\Psi \left(f_0\right)|{|}_{\beta }=0,$$

i.e. Ψ(f₁) - Ψ(f₀), ..., Ψ(f _n ) - Ψ(f₀) are linearly dependent.

If there exist f₃, ..., f_n-1such that Ψ(f₁) - Ψ(f₀), ..., Ψ(f_{n- 1}) - Ψ(f₀) are linearly independent, then

$$A=\left\{\Psi \left(f_n\right)-\Psi \left(f_0\right):f_n\in \Im \left(X\right)\right\}\subset \text{span }\left\{\Psi \left(f_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_0\right)\right\},$$

which contradicts the fact that A contains n linearly independent vectors.

Then, for any f₃, ..., f_n-1, Ψ(f₁) - Ψ(f₀), ..., Ψ(f_n-1) - Ψ(f₀) are linearly dependent.

If there exist f₃, ..., f_n-2such that Ψ(f₁) - Ψ(f₀), ..., Ψ(f_n-2) - Ψ(f₀) are linearly independent, then

$$A=\left\{\Psi \left(f_{n-1}\right)-\Psi \left(f_0\right):f_{n-1}\in \Im \left(X\right)\right\}\subset \text{span }\left\{\Psi \left(f_1\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_{n-2}\right)-\Psi \left(f_0\right)\right\},$$

which contradicts the fact that A contains n linearly independent vectors.

And so on, Ψ(f₁) - Ψ(f₀), Ψ(f₂) - Ψ(f₀) are linearly dependent. Thus Ψ(f₀), Ψ(f₁), and Ψ(f₂) are 2-collinear.

Theorem 4.2 Every n-isometry mapping is affine.

Proof. Let Ψ be an n-isometry and Φ(f) = Ψ(f) - Ψ(0). Then Φ is an n-isometry and Φ(0) = 0. Thus we may assume that Ψ(0) = 0. Hence it suffices to show that Ψ is linear.

Let f₀, f₁ ∈ ℑ(X) with f₀ ≠ f₁. Since dimℑ(X) ≥ n, there exist f₂,...,f _n ∈ ℑ(X) such that

$$\left|\right|f_0-f_n,f_1-f_n,f_2-f_n,\dots ,f_{n-1}-f_n|{|}_{\alpha }\ne 0.$$

Since Ψ is an n-isometry, we have

$$\left|\right|\Psi \left(f_0\right)-\Psi \left(f_n\right),\Psi \left(f_1\right)-\Psi \left(f_n\right),\Psi \left(f_2\right)-\Psi \left(f_n\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_n\right)|{|}_{\beta }\ne 0.$$

It follows from Lemma 4.1 that

$$\begin{array}{c}{∥\Psi \left(f_0\right)-\Psi \left(f_n\right),\Psi \left(f_0\right)-\Psi \left(\frac{f_0+f_1}{2}\right),\Psi \left(f_2\right)-\Psi \left(f_n\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_n\right)∥}_{\beta }\\ ={∥\Psi \left(f_n\right)-\Psi \left(f_0\right),\Psi \left(\frac{f_0+f_1}{2}\right)-\Psi \left(f_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_0\right)∥}_{\beta }\\ ={∥f_n-f_0,\frac{f_0+f_1}{2}-f_0,f_2-f_0,\dots ,f_{n-1}-f_0∥}_{\alpha }\\ =\frac{1}{2}\left|\right|f_n-f_0,f_1-f_0,f_2-f_0,\dots ,f_{n-1}-f_0|{|}_{\alpha }\\ =\frac{1}{2}\left|\right|\Psi \left(f_n\right)-\Psi \left(f_0\right),\Psi \left(f_1\right)-\Psi \left(f_0\right),\Psi \left(f_2\right)-\Psi \left(f_0\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_0\right)|{|}_{\beta }\\ =\frac{1}{2}\left|\right|\Psi \left(f_0\right)-\Psi \left(f_n\right),\Psi \left(f_1\right)-\Psi \left(f_n\right),\Psi \left(f_2\right)-\Psi \left(f_n\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_n\right)|{|}_{\beta }.\end{array}$$

And we get

$$\begin{array}{c}{∥\Psi \left(f_1\right)-\Psi \left(\frac{f_0+f_1}{2}\right),\Psi \left(f_1\right)-\Psi \left(f_n\right),\Psi \left(f_2\right)-\Psi \left(f_n\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_n\right)∥}_{\beta }\\ ={∥\Psi \left(\frac{f_0+f_1}{2}\right)-\Psi \left(f_1\right),\Psi \left(f_n\right)-\Psi \left(f_1\right),\Psi \left(f_2\right)-\Psi \left(f_1\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_1\right)∥}_{\beta }\\ ={∥\frac{f_0+f_1}{2}-f_1,f_n-f_1,f_2-f_1,\dots ,f_{n-1}-f_1∥}_{\alpha }\\ =\frac{1}{2}\left|\right|f_0-f_1,f_n-f_1,f_2-f_1,\dots ,f_{n-1}-f_1|{|}_{\alpha }\\ =\frac{1}{2}\left|\right|\Psi \left(f_0\right)-\Psi \left(f_1\right),\Psi \left(f_n\right)-\Psi \left(f_1\right),\Psi \left(f_2\right)-\Psi \left(f_1\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_1\right)|{|}_{\beta }\\ =\frac{1}{2}\left|\right|\Psi \left(f_0\right)-\Psi \left(f_n\right),\Psi \left(f_1\right)-\Psi \left(f_n\right),\Psi \left(f_2\right)-\Psi \left(f_n\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(f_n\right)|{|}_{\beta }.\end{array}$$

By Lemma 4.6, we obtain that $\Psi \left(\frac{f_0+f_1}{2}\right)$, Ψ(f₀), and Ψ(f₁) are 2-collinear. By Lemma 4.5, we get $\Psi \left(\frac{f_0+f_1}{2}\right)=\frac{\Psi \left(f_0\right)+\Psi \left(f_1\right)}{2}$ for all f, g ∈ ℑ(X) and α, β ∈ (0, 1). Since Ψ(0) = 0, we can easily show that Ψ is additive. It follows that Ψ is $\mathbb{Q}-\text{linear}$-linear.

Let r ∈ ℝ⁺ with r ≠ 1 and f ∈ ℑ(X). By Lemma 4.6, Ψ(0), Ψ(f) and Ψ(rf) are also 2-collinear. It follows from Ψ(0) = 0 that there exists a real number k such that Ψ(rf) = k Ψ(f). Since dimℑ(X) ≥ n, there exist f₁, ..., f_n-1 ∈ ℑ(X) such that ∥f, f₁, f₂, ..., f_n-1∥ _α ≠ 0. Since Ψ(0) = 0, for every f₀, f₁, f₂, ..., f_n-1 ∈ ℑ(X),

$$\begin{array}{c}\left|\right|f_0,f_1,f_2,\dots ,f_{n-1}|{|}_{\alpha }\\ =\left|\right|f_0-0,f_1-0,f_2-0,\dots ,f_{n-1}-0|{|}_{\alpha }\\ =\left|\right|\Psi \left(f_0\right)-\Psi \left(0\right),\Psi \left(f_1\right)-\Psi \left(0\right),\Psi \left(f_2\right)-\Psi \left(0\right),\dots ,\Psi \left(f_{n-1}\right)-\Psi \left(0\right)|{|}_{\beta }\\ =\left|\right|\Psi \left(f_0\right),\Psi \left(f_1\right),\Psi \left(f_2\right),\dots ,\Psi \left(f_{n-1}\right)|{|}_{\beta }.\end{array}$$

Thus we have

$$\begin{array}{ll}\hfill r\left|\right|f,f_1,f_2,\dots ,f_{n-1}|{|}_{\alpha }& =\left|\right|rf,f_1,f_2,\dots ,f_{n-1}|{|}_{\alpha }\\ =\left|\right|\Psi \left(rf\right),\Psi \left(f_1\right),\Psi \left(f_2\right),\dots ,\Psi \left(f_{n-1}\right)|{|}_{\beta }\\ =\left|\right|k\Psi \left(f\right),\Psi \left(f_1\right),\Psi \left(f_2\right),\dots ,\Psi \left(f_{n-1}\right)|{|}_{\beta }\\ =\left|k\right|\left|\right|\Psi \left(f\right),\Psi \left(f_1\right),\Psi \left(f_2\right),\dots ,\Psi \left(f_{n-1}\right)|{|}_{\beta }\\ =\left|k\right|\left|\right|f,f_1,f_2,\dots ,f_{n-1}|{|}_{\alpha }.\end{array}$$

Since ∥f, f₁, f₂, ..., f_n-1∥ _α ≠ 0, |k| = r. Then Ψ(rf) = r Ψ(f) or Ψ(rf) = -r Ψ(f). First of all, assume that k = -r, that is, Ψ(rf) = -r Ψ(f). Then there exist positive rational numbers q₁, q₂ such that o < q₁ < r < q₂. Since dimℑ(X) ≥ n, there exist h₁, ..., h_n-1 ∈ ℑ(X) such that

$$\left|\right|rf-q_{2}f,h_1-q_{2}f,h_2-q_{2}f,\dots ,h_{n-1}-q_{2}f|{|}_{\alpha }\ne 0.$$

Then we have

$$\begin{array}{c}\left(q_2+r\right)\left|\right|\Psi \left(f\right),\Psi \left(h_1\right)-\Psi \left(q_{2}f\right),\Psi \left(h_2\right)-\Psi \left(q_{2}f\right),\dots ,\Psi \left(h_{n-1}\right)-\Psi \left(q_{2}f\right)|{|}_{\beta }\\ =\left|\right|\left(q_2+r\right)\Psi \left(f\right),\Psi \left(h_1\right)-\Psi \left(q_{2}f\right),\Psi \left(h_2\right)-\Psi \left(q_{2}f\right),\dots ,\Psi \left(h_{n-1}\right)-\Psi \left(q_{2}f\right)|{|}_{\beta }\\ =\left|\right|q_2\Psi \left(f\right)-\left(-r\Psi \left(f\right)\right),\Psi \left(h_1\right)-\Psi \left(q_{2}f\right),\Psi \left(h_2\right)-\Psi \left(q_{2}f\right),\dots ,\Psi \left(h_{n-1}\right)-\Psi \left(q_{2}f\right)|{|}_{\beta }\\ =\left|\right|\Psi \left(rf\right)-\Psi \left(q_{2}f\right),\Psi \left(h_1\right)-\Psi \left(q_{2}f\right),\Psi \left(h_2\right)-\Psi \left(q_{2}f\right),\dots ,\Psi \left(h_{n-1}\right)-\Psi \left(q_{2}f\right)|{|}_{\beta }\\ =\left|\right|rf-q_{2}f,h_1-q_{2}f,h_2-q_{2}f,\dots ,h_{n-1}-q_{2}f|{|}_{\alpha }\\ =\left(q_2-r\right)\left|\right|f,h_1-q_{2}f,h_2-q_{2}f,\dots ,h_{n-1}-q_{2}f|{|}_{\alpha }\\ \le \left(q_2-q_1\right)\left|\right|f,h_1-q_{2}f,h_2-q_{2}f,\dots ,h_{n-1}-q_{2}f|{|}_{\alpha }\\ =\left|\right|q_{1}f-q_{2}f,h_1-q_{2}f,h_2-q_{2}f,\dots ,h_{n-1}-q_{2}f|{|}_{\alpha }\\ =\left|\right|\Psi \left(q_{1}f\right)-\Psi \left(q_{2}f\right),\Psi \left(h_1\right)-\Psi \left(q_{2}f\right),\Psi \left(h_2\right)-\Psi \left(q_{2}f\right),\dots ,\Psi \left(h_{n-1}\right)-\Psi \left(q_{2}f\right)|{|}_{\beta }.\end{array}$$

And also we have

$$\begin{array}{c}\left|\right|rf-q_{2}f,h_1-q_{2}f,h_2-q_{2}f,\dots ,h_{n-1}-q_{2}f|{|}_{\alpha }\\ =\left|\right|\Psi \left(rf\right)-\Psi \left(q_{2}f\right),\Psi \left(h_1\right)-\Psi \left(q_{2}f\right),\Psi \end{array}$$