# On the Mazur-Ulam Theorem in Non-Archimedean Fuzzyn-Normed Spaces

Authors: Tian Zhou Xu
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

The motivation of this paper is to present a new notion of non-Archimedean fuzzyn-normed space over a field with valuation. We obtain a Mazur-Ulam theorem  for fuzzy  n-isometry mappings in the strictly convex non-Archimedean fuzzy n-normed spaces. We also prove that the interior preserving mapping carries the barycenter of a triangle to the barycenter point of the corresponding triangle. And then, using this result, we get a Mazur-Ulam theorem for the interior preserving fuzzy n-isometry mappings in non-Archimedean fuzzy n-normed spaces over a linear ordered non-Archimedean field.


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## Body

## 1. Introduction

Let𝕂 be a field. A valuation mapping on 𝕂 is a function |·|:𝕂→ℝ such that for any r,s∈𝕂 the following conditions are satisfied: (i) |r|≥0 and equality holds if and only if r=0; (ii) |rs|=|r|·|s|; (iii) |r+s|≤|r|+|s|.A field endowed with a valuation mapping will be called a valued field. The usual absolute values ofℝ and ℂ are examples of valuations. A trivial example of a non-Archimedean valuation is the function |·| taking everything except for 0 into 1 and |0|=0. In the following, we will assume that |·| is nontrivial; that is, there is an r0∈𝕂 such that |r0|≠0,1.Throughout this paper, we assume that𝕂 is a valued field and n≥2 is a positive integer. We denote the set of all elements of 𝕂 whose norms are 1 by 𝒞; that is, 𝒞={r∈𝕂∣|r|=1}. Moreover, ℕ stands for the set of all positive integers and ℝ (resp., ℂ) denotes the set of all real numbers (resp., complex numbers).If condition (iii) in the definition of a valuation mapping is replaced with a strong triangle inequality (ultrametric),|r+s|≤max{|r|,|s|}, then the valuation |·| is said to be non-Archimedean. In any non-Archimedean field, we have |1|=|-1|=1 and |k|≤1 for all k∈ℕ.Let𝒳 and 𝒴 be metric spaces. A map f:𝒳→𝒴 is called a distance preserving mapping (isometry) if d(f(x),f(y))=d(x,y) for any x,y∈𝒳. Automatically, an isometry is injective. Two metric spaces 𝒳 and 𝒴 are called isometric if there is an isometry from 𝒳 to 𝒴.The classical result of Mazur and Ulam states that if𝒳, 𝒴 are real normed linear spaces and f:𝒳→𝒴 is a surjective isometry, then f is affine; that is, f is a linear mapping up to translation. Numerous generalizations of this fact were presented by many authors (see, e.g., [1–13] and references therein). Unfortunately, the Mazur-Ulam theorem is not true for normed complex vector space. It was a natural step to ask if the theorem holds without the onto assumption. In fact, the onto assumption is essential. Without this assumption, Baker [14] proved that every isometry, not necessary surjective, f:𝒳→𝒴, between real normed linear spaces is affine if 𝒴 is strictly convex. Moslehian and Sadeghi presented a non-Archimedean version of this result [11]; they also noted that a Mazur-Ulam theorem generally fails in a non-Archimedean case. Choy et al. [3] 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. Chu et al. [4] studied the Mazur-Ulam theorem in linear n-normed spaces. Alaca [1] introduced the concepts of 2-isometry, collinearity, and 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. Kubzdela [10] gave some new results for isometries, Mazur-Ulam theorem, and Aleksandrov problem in the framework of non-Archimedean normed spaces. Kang et al. [9] proved that the Mazur-Ulam theorem holds under some conditions in non-Archimedean fuzzy normed space.The motivation of this paper is to introduce the notion of non-Archimedean fuzzyn-normed space over a field with valuation as a generalization of n-normed space [2, 15, 16], non-Archimedean 2-normed space [3], fuzzy n-normed space [17], and non-Archimedean fuzzy normed space [9, 18]. We will prove that the Mazur-Ulam theorem holds in the strictly convex non-Archimedean fuzzy n-normed spaces.


## 2. Preliminaries

In 1897, Hensel discovered thep-adic numbers as a number-theoretical analogue of power series in complex analysis. Let p be a prime number. For any nonzero rational number a, there exists a unique integer r such that a=prm/k, where m and k are integers not divisible by p. Then |a|p:=p-r defines a non-Archimedean norm on ℚ. The completion of ℚ with respect to the metric d(a,b)=|a-b|p is denoted by ℚp which is called the p-adic number field. Note that if p>2, then |2k|p=1 for each integer k but |2|2<1.During the last three decades,p-adic numbers have gained the interest of physicists for their research, in particular, in problems derived from quantum physics, p-adic strings, and superstrings (see, e.g., [19]).Definition 1.
Let𝒳 be a linear space over a field 𝕂 with a non-Archimedean valuation |·|. A function ∥·∥:𝒳→[0,∞) is said to be a non-Archimedean norm if it satisfies the following conditions:(i)
∥x∥=0 if and only if x=0,(ii)
∥rx∥=|r|∥x∥,    r∈𝕂,  x∈𝒳,(iii)
the strong triangle inequality:∥x+y∥≤max{∥x∥,∥y∥}(x,y∈𝒳).
Then(𝒳,∥·∥) is called a non-Archimedean normed space. By a complete non-Archimedean normed space, we mean one in which every Cauchy sequence is convergent.Definition 2.
Let𝒳 be a linear space over a valued field 𝕂. A function N:𝒳n×ℝ→[0,1] is called a non-Archimedean fuzzy n-norm on 𝒳 if the following conditions hold for all x,y,x1,…,xn∈𝒳 and all s,t∈ℝ:(nN1)
N(x1,…,xn,t)=0 if t≤0,(nN2)
for allt>0, N(x1,…,xn,t)=1 if and only if x1,…,xn are linearly dependent,(nN3)
N(x1,…,xn,t) is invariant under any permutation of x1,…,xn,(nN4)
N(cx1,…,xn,t)=N(x1,…,xn,t/|c|) for all t>0 if c∈𝕂,c≠0,(nN5)
N(x+y,x2,…,xn,max{s,t})≥min{N(x,x2,…,xn,s),N(y,x2,…,xn,t)} for all s,t>0,(nN6)
limt→∞N(x1,…,xn,t)=1.
IfN is a non-Archimedean fuzzy n-norm on 𝒳, then (𝒳,N) is called a non-Archimedean fuzzy n-normed space. It should be noticed that from the condition (nN5) it follows that
(1)N(x1,…,xn,t)≥min{N(0,…,0,t),N(x1,…,xn,s)}=N(x1,…,xn,s)
for every t>s>0 and x1,…,xn∈𝒳; that is, N(x1,…,xn,·) is nondecreasing for every x1,…,xn∈𝒳. This implies that
(2)N(x1,…,xn,s+t)≥N(x1,…,xn,max{s,t}),s,t>0.
If (nN5) holds, then so is (nN7)
N(x+y,x2,…,xn,s+t)≥min{N(x,x2,…,xn,s),N(y,x2,…,xn,t)}.Example 3.
Let(𝒳,∥·,…,·∥) be an n-normed space (see [2]). For each k∈ℕ, consider
(3)Nk(x1,…,xn,t)={tt+k∥x1,…,xn∥,t>0,0,t≤0.
Then (𝒳,Nk) is a non-Archimedean fuzzy n-normed space.Definition 4.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces, and let f:𝒳→𝒴 be a mapping. We call f a fuzzy n-isometry if
(4)N(x1-x0,…,xn-x0,t)=N(f(x1)-f(x0),…,f(xn)-f(x0),t)
for all x0,x1,…,xn∈𝒳 and all t>0.For given pointsx,y, and z in 𝒳, Δxyz denotes the triangle determined by x,y, and z. A point (x+y+z)/3 is called a barycenter of Δxyz. If p is a point of a set {t1x+t2y+t3z∣t1+t2+t3=1,ti∈𝕂,ti>0,i=1,2,3}, then p is called an interior point of Δxyz. Define a mapping f between linear n-normed spaces to be an interior preserving mapping of the triangle if f carries an interior point in a triangle to an interior point in the corresponding triangle.Remark 5.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces and f:𝒳→𝒴 be a mapping. Then f is a fuzzy n-isometry if and only if f satisfies the following property:
(5)|(f(x1′)-f(x0′),…,f(xn′)-f(x0′),t)N(x1-x0,…,xn-x0,t)-N(x1′-x0′,…,xn′-x0′,t)|=|N(f(x1′)-f(x0′),…,f(xn′)-f(x0′),t)N(f(x1)-f(x0),…,f(xn)-f(x0),t)-N(f(x1′)-f(x0′),…,f(xn′)-f(x0′),t)|
for all x0,x1,…,xn,x0′,x1′,…,xn′∈𝒳 and all t>0.Definition 6.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces and f:𝒳→𝒴 be a mapping. Then f is called a weak fuzzy n-isometry if for every ϵ>0, there exists positive real number δ such that
(6)|N(x1-x0,…,xn-x0,t)-N(x1′-x0′,…,xn′-x0′,t)|<δimplies|-f(x0′),t)N(f(x1)-f(x0),…,f(xn)-f(x0),t)-N(f(x1′)-f(x0′),…,f(xn′)-f(x0′),t)|<ϵ
for all x0,x1,…,xn,x0′,x1′,…,xn′∈𝒳 and all t>0.Definition 7.
Let𝒳 be a non-Archimedean fuzzy n-normed space. The points x0,x1,…,xn are said to be n-collinear if for every i∈{0,1,…,n}, {xj-xi∣0≤j≠i≤n} is linearly dependent.Remark 8.
Let𝒳 be a non-Archimedean fuzzy n-normed space over a valued field 𝕂 and x0,x1, x2 mutually disjoint elements of 𝒳. Then x0,x1, and x2 are said to be 2-collinear if and only if x2-x0=t(x1-x0) for some t∈𝕂.Now we define the concept ofn-Lipschitz mapping.Definition 9.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces, and let f:𝒳→𝒴 be a mapping. Then f is called a fuzzy n-Lipschitz mapping if there is a L≥0 such that
(7)N(f(x1)-f(x0),…,f(xn)-f(x0),Lt)≥N(x1-x0,…,xn-x0,t)
for all x0,x1,…,xn∈𝒳 and all t>0. The smallest such L is called the n-Lipschitz constant.Definition 10.
A non-Archimedean fuzzyn-normed space 𝒳 over a valued field 𝕂 is called strictly convex, if for each x1,…,xm,w2,…,wn∈𝒳 and s1,…,sm>0,
(8)N(∑i=1m‍xi,w2,…,wn,max{s1,…,sm})=min{N(x1,w2,…,wn,s1),…,N(xm,w2,…,wn,sm)},N(x1,w2,…,wn,s1)=⋯=N(xm,w2,…,wn,sm)
implies that x1=⋯=xm and s1=⋯=sm.


## 3. On the Mazur-Ulam Problem

Lemma 11.
Let𝒳 be a non-Archimedean fuzzy n-normed space over a valued field 𝕂, x1,…,xn∈𝒳 and all t>0. Then
(9)N(x1,…,xi,…,xj,…,xn,t)=N(x1,…,xi,…,xj-1,xj+αxi,xj+1,…,xn,t)
for all α∈𝕂.Proof.
Letx1,…,xn∈𝒳,t>0, and α∈𝕂, then
(10)N(x1,…,xi,…,xj-1,xj+αxi,xj+1,…,xn,t)≥min{(x1,…,xi,…,xj-1,xi,xj+1,…,xn,t|α|)N(x1,…,xi,…,xj,…,xn,t),N(x1,…,xi,…,xj-1,xi,xj+1,…,xn,t|α|)}≥N(x1,…,xi,…,xj,…,xn,t),N(x1,…,xi,…,xj,…,xn,t)=N(x1,…,xi,…,xj-1,xj+αxi-αxi,xj+1,…,xn,t)≥min{N(x1,…,xi,…,xj-1,xj+αxi,xj+1,…,xn,t),N(x1,…,xi,…,xj-1,-αxi,xj+1,…,xn,t)}=N(x1,…,xi,…,xj-1,xj+αxi,xj+1,…,xn,t).
It follows that (9) holds.Lemma 12.
Let𝒳 be a strictly convex non-Archimedean fuzzy n-normed space over a linear ordered non-Archimedean field 𝕂 with 𝒞={2n∣n∈ℤ}, and let x,y∈𝒳 and t>0. Then u=(x+y)/2 is the unique element of 𝒳 satisfying
(11)N(x-u,x2-u,…,xn-u,t)=N(y-u,x2-u,…,xn-u,t)=N(x-y,x2-x,…,xn-x,t)
for all x2,…,xn∈𝒳 and x,y, and u are 2-collinear.Proof.
Letu:=(x+y)/2,x2,…,xn∈𝒳 and t>0. By Lemma 11, we have
(12)N(x-u,x2-u,…,xn-u,t)=N(x-x+y2,x2-x+y2,…,xn-x+y2,t)=N(x-x+y2,x2-x,…,xn-x,t)=N(y-x,x2-x,…,xn-x,|2|t),N(y-u,x2-u,…,xn-u,t)=N(y-x+y2,x2-x+y2,…,xn-x+y2,t)=N(y-x+y2,x2-y,…,xn-y,t)=N(y-x,x2-y,…,xn-y,|2|t)=N(y-x,x2-x,…,xn-x,|2|t).
Hence, the existence holds. For the uniqueness of u, we may assume that there is another v∈𝒳, collinear with x,y such that
(13)N(x-v,x2-v,…,xn-v,t)=N(y-v,x2-v,…,xn-v,t)=N(x-y,x2-x,…,xn-x,t).
Since x,y, and v are collinear, v=sx+(1-s)y for some s∈𝕂. We may assume s≠0 and s≠1. Then
(14)N(x-y,x2-x,…,xn-x,t)=N(x-v,x2-v,…,xn-v,t)=N(x-v,x2-x,…,xn-x,t)=N(x-sx-(1-s)y,x2-x,…,xn-x,t)=N(x-y,x2-x,…,xn-x,t|1-s|),N(x-y,x2-x,…,xn-x,t)=N(y-v,x2-v,…,xn-v,t)=N(y-v,x2-y,…,xn-y,t)=N(y-sx-(1-s)y,x2-y,…,xn-y,t)=N(x-y,x2-y,…,xn-y,t|s|)=N(x-y,x2-x,…,xn-x,t|s|).
Then
(15)N(x-y+x-y,x2-x,…,xn-x,max{t|s|,t|1-s|})=N(x-y,x2-x,…,xn-x,1|2|max{t|s|,t|1-s|})=N(x-y,x2-x,…,xn-x,t)=min{N(x-y,x2-x,…,xn-x,t|s|),N(x-y,x2-x,…,xn-x,t|1-s|)},N(x-y+x-y,x2-x,…,xn-x,max{t,t|s|})=N(x-y,x2-x,…,xn-x,1|2|max{t,t|s|})=N(x-y,x2-x,…,xn-x,t)=min{(x-y,x2-x,…,xn-x,t|s|)N(x-y,x2-x,…,xn-x,t),N(x-y,x2-x,…,xn-x,t|s|)}.
By the strict convexity of 𝒳, we have |1-s|=|s|=1. Then there exist two integers k1,k2 such that 1-s=2k1 and s=2k2. Since 2k1+2k2=1, we know that k1<0,k2<0. Without loss of generality, we let 1-s=2-n1 and s=2-n2 with n1≥n2. If n1>n2, then 1=2-n1+2-n2=2-n1(1+2n1-n2); that is, 2n1=1+2n1-n2. This is a contradiction. Thus, n1=n2; that is, s=1/2. This completes the proof.Lemma 13.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces over a valued field 𝕂. If f:𝒳→𝒴 is a fuzzy n-isometry and x0,x1, and x2 are 2-collinear, then f(x0),f(x1), and f(x2) are 2-collinear.Proof.
Sincedim𝒳≥n, for any x0∈𝒳, there exist y1,…,yn∈𝒳 such that y1-x0,…,yn-x0 are linearly independent. Then
(16)N(y1-x0,…,yn-x0,t)=N(f(y1)-f(x0),…,f(yn)-f(x0),t)≠1,
and hence, the set {f(x)-f(x0):x∈𝒳} contains n linearly independent vectors. Assume that x0,x1, and x2 are 2-collinear. Then, for any x3,…,xn∈𝒳,
(17)N(x1-x0,…,xn-x0,t)=N(f(x1)-f(x0),…,f(xn)-f(x0),t)=1.
By (nN2), it follows that f(x1)-f(x0),…,f(xn)-f(x0) are linearly dependent. If there exist x3,…,xn-1 such that f(x1)-f(x0),…,f(xn-1)-f(x0) are linearly independent, then
(18){f(x)-f(x0):x∈𝒳}⊂span{f(x1)-f(x0),…,f(xn-1)-f(x0)}
which contradicts the fact that {f(x)-f(x0):x∈𝒳} contains n linearly independent vectors. Hence, for any x3,…,xn-1∈𝒳, f(x1)-f(x0),…,f(xn-1)-f(x0) are linearly dependent. If there exist x3,…,xn-2 such that f(x1)-f(x0),…,f(xn-2)-f(x0) are linearly independent, then
(19){f(x)-f(x0):x∈𝒳}⊂span{f(x1)-f(x0),…,f(xn-2)-f(x0)}
which contradicts the fact that {f(x)-f(x0):x∈𝒳} contains n linearly independent vectors. And thus, f(x1)-f(x0),f(x2)-f(x0) are linearly dependent. Therefore, f(x0),f(x1), and f(x2) are 2-collinear.Theorem 14.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces over a linear ordered non-Archimedean field 𝕂 with 𝒞={2n∣n∈ℤ} such that 𝒴 is strictly convex. If f:𝒳→𝒴 is a fuzzy n-isometry, then f(x)-f(0) is additive.Proof.
Letg(x)=f(x)-f(0) for x∈𝒳. Then g is a fuzzy n-isometry and g(0)=0. For each x,y,x2,…,xn∈𝒳. Since g is a fuzzy n-isometry, we have
(20)N(g(x)-g(x+y2),g(x2)-g(x+y2),…,g(xn)-g(x+y2),t)=N(x-x+y2,x2-x+y2,…,xn-x+y2,t)=N(x-x+y2,x2-x,…,xn-x,t)=N(x-y,x2-x,…,xn-x,t)=N(g(x)-g(y),g(x2)-g(x),…,g(xn)-g(x),t),N(g(y)-g(x+y2),g(x2)-g(x+y2),…,g(xn)-g(x+y2),t)=N(y-x+y2,x2-x+y2,…,xn-x+y2,t)=N(y-x+y2,x2-y,…,xn-y,t)=N(x-y,x2-y,…,xn-y,t)=N(x-y,x2-x,…,xn-x,t)=N(g(x)-g(y),g(x2)-g(x),…,g(xn)-g(x),t).
Since (x+y)/2, x, and y are collinear, by Lemma 13, g((x+y)/2), g(x), and g(y) are also collinear. It follows from Lemma 12 that
(21)g(x+y2)=g(x)+g(y)2,
for all x,y∈𝒳. Hence, g(x)=f(x)-f(0) is additive since g(0)=0.In the following, we prove that the interior preserving mapping carries the barycenter of a triangle to the barycenter point of the corresponding triangle. And then, using this result, we get a Mazur-Ulam theorem on non-Archimedean fuzzyn-normed spaces over a linear ordered non-Archimedean field 𝕂 with 𝒞={3n∣n∈ℤ}.Lemma 15.
Let𝒳 be a strictly convex non-Archimedean fuzzy n-normed space over a linear ordered non-Archimedean field 𝕂 with 𝒞={3n∣n∈ℤ}, and let x,y,z∈𝒳 and t>0. Then u:=(x+y+z)/3 is the unique element of 𝒳 satisfying
(22)N(x-u,y-u,x3-u,…,xn-u,t)=N(y-u,z-u,x3-u,…,xn-u,t)=N(x-u,z-u,x3-u,…,xn-u,t)=N(x-y,x-z,x3-x,…,xn-x,t)
for all x3,…,xn∈𝒳 and u∈{t1x+t2y+t3z∣t1+t2+t3=1,ti∈𝕂,ti>0,i=1,2,3}.Proof.
Letu:=(x+y+z)/3,x3,…,xn∈𝒳 and t>0. By Lemma 11, we have
(23)N(x-u,y-u,x3-u,…,xn-u,t)=N(x-u,y-x,x3-x,…,xn-x,t)=N(x-x+y+z3,y-x,x3-x,…,xn-x,t)=N(2x-y-z,y-x,x3-x,…,xn-x,|3|t)=N(y-z,y-x,x3-x,…,xn-x,t)=N(y-x,z-x,x3-x,…,xn-x,t)=N(x-y,x-z,x3-x,…,xn-x,t).
Similarly, we have(24)N(y-u,z-u,x3-u,…,xn-u,t)=N(x-y,x-z,x3-x,…,xn-x,t),N(x-u,z-u,x3-u,…,xn-u,t)=N(x-y,x-z,x3-x,…,xn-x,t).
Hence, the existence holds. For the uniqueness ofu, we may assume that there is another v∈𝒳 satisfying (22). Since v∈{t1x+t2y+t3z∣t1+t2+t3=1,ti∈𝕂,ti>0,i=1,2,3}, v=s1x+s2y+s3z, s1+s2+s3=1 for some si∈𝕂,si>0,i=1,2,3. Then
(25)N(x-y,z-x,x3-x,…,xn-x,t)=N(x-v,y-v,x3-v,…,xn-v,t)=N(x-v,y-x,x3-x,…,xn-x,t)=N((1-s1)x-s2y-(1-s1-s2)z,y-x,x3-x,…,xn-x,t(1-s1)x-s2y-(1-s1-s2))=N((1-s1-s2)x-(1-s1-s2)z,y-x,x3-x,…,xn-x,t((1-s1)x-s2y-(1-s1-s2)))=N(x-z,y-x,x3-x,…,xn-x,t|1-s1-s2|)=N(x-y,z-x,x3-x,…,xn-x,t|1-s1-s2|),N(x-y,z-x,x3-x,…,xn-x,t)=N(x-v,z-v,x3-v,…,xn-v,t)=N(x-v,z-x,x3-x,…,xn-x,t)=N((1-s1)x-s2y-(1-s1-s2)z,z-x,x3-x,…,xn-x,t(1-s1)x-s2y-(1-s1-s2)z)=N(s2x-s2y,z-x,x3-x,…,xn-x,t)=N(x-z,y-x,x3-x,…,xn-x,t|s2|),N(x-y,z-x,x3-x,…,xn-x,t)=N(y-v,z-v,x3-v,…,xn-v,t)=N(y-v,z-y,x3-y,…,xn-y,t)=N((1-s2)y-s1x-(1-s1-s2)z,z-y,x3-y,…,xn-y,t(1-s2)y-s1x-(1-s1-s2)z)=N(-s1x+s1z,z-y,x3-y,…,xn-y,t)=N(x-z,z-y,x3-y,…,xn-y,t|s1|)=N(x-y,z-x,x3-x,…,xn-x,t|s1|).
Hence,(26)N({t|s1|,t|s2|,t|1-s1-s2|}x-y+x-y+x-y,z-x,x3-x,…,xn-x,max{t|s1|,t|s2|,t|1-s1-s2|})=N({t|s1|,t|s2|,t|1-s1-s2|}x-y,z-x,x3-x,…,xn-x,1|3|max{t|s1|,t|s2|,t|1-s1-s2|})=N(x-y,z-x,x3-x,…,xn-x,t)=min{N(x-y,z-x,x3-x,…,xn-x,t|s1|),N(x-y,z-x,x3-x,…,xn-x,t|s2|),N(x-y,z-x,x3-x,…,xn-x,t|1-s1-s2|)},N({t|s1|,t|s2|,t|1-s1-s2|}x-y+x-y+x-y,z-x,x3-x,…,xn-x,max{t,t|s1|,t|s2|})=N(1|3|max{t,t|s1|,t|s2|}x-y,z-x,x3-x,…,xn-x,1|3|max{t,t|s1|,t|s2|})=N(x-y,z-x,x3-x,…,xn-x,t)=min{{t|s1|,t|s2|,t|1-s1-s2|}N(x-y,z-x,x3-x,…,xn-x,t),N(x-y,z-x,x3-x,…,xn-x,t|s1|),N(x-y,z-x,x3-x,…,xn-x,t|s2|)}.By the strict convexity, we have |s1|=|s2|=|s3|=1. Then there exist integers k1, k2, and k3 such that s1=3k1, s2=3k2, and s3=3k3. Since 3k1+3k2+3k3=1, we know that k1<0, k2<0, k3<0. Without loss of generality, we let s1=3-n1, s2=3-n2, and s3=3-n3 with n1≥n2≥n3. Assume that the above one of the inequalities holds. Then 1=s1+s2+s3=3-n1(1+3n1-n2+3n1-n3). So 3n1=1+3n1-n2+3n1-n3. This is a contradiction. Thus, s1=s2=s3=1/3 which means u=v. This completes the proof.Theorem 16.
Let𝒳 and 𝒴 be non-Archimedean fuzzy n-normed spaces over a linear ordered non-Archimedean field 𝕂 with 𝒞={3n∣n∈ℤ} such that 𝒴 is strictly convex. If f:𝒳→𝒴 is an interior preserving fuzzy n-isometry, then f(x)-f(0) is additive.Proof.
Letg(x)=f(x)-f(0) for x∈𝒳. Then g is a fuzzy n-isometry and g(0)=0. For a,b,c∈𝒳, let Δabc be a triangle determined by the points a, b, c, and x, an interior point of Δabc. Since f is an interior preserving map, there exist si∈𝕂,si>0(i=1,2,3) with s1+s2+s3=1 such that f(x)=s1f(a)+s2f(b)+s3f(c). Then
(27)g(x)=s1f(a)+s2f(b)+s3f(c)-f(0)=s1(f(a)-f(0))+s2(f(b)-f(0))+s3(f(c)-f(0))=s1g(a)+s2g(b)+s3g(c),
and hence, g(x) is an interior point of Δg(a)g(b)g(c). Therefore, g is also an interior preserving mapping.
Now letx,y,z,x3,…,xn∈𝒳. Since g is a fuzzy n-isometry, we have
(28)N(g(x)-g(x+y+z3),g(y)-g(x+y+z3),g(x3)-g(x+y+z3),…,g(xn)-g(x+y+z3),t)=N(x-x+y+z3,y-x+y+z3,x3-x+y+z3,…,xn-x+y+z3,t)=N(x-y,x-x+y+z3,x3-x,…,xn-x,t)=N(x-y,2x-y-z,x3-x,…,xn-x,t)=N(x-y,x-z,x3-x,…,xn-x,t)=N(g(x)-g(y),g(x)-g(z),g(x3)-g(x),…,g(xn)-g(x),t),
and similarly, we can obtain(29)N(g(y)-g(x+y+z3),g(z)-g(x+y+z3),g(x3)-g(x+y+z3),…,g(xn)-g(x+y+z3),t)=N(g(x)-g(y),g(x)-g(z),g(x3)-g(x),…,g(xn)-g(x),t),(30)N(g(x)-g(x+y+z3),g(z)-g(x+y+z3),g(x3)-g(x+y+z3),…,g(xn)-g(x+y+z3),t)=N((x3)-g(x),…,g(xn)-g(x),tg(x)-g(y),g(x)-g(z),g(x3)-g(x),…,g(xn)-g(x),t).Since (x+y+z)/3 is an interior point of the triangle Δxyz and g is an interior preserving mapping, g((x+y+z)/3)∈{t1g(x)+t2g(y)+t3g(z)∣t1+t2+t3=1,ti∈𝕂,ti>0,i=1,2,3}. By Lemma 15,
(31)g(x+y+z3)=g(x)+g(y)+g(z)3.
Hence, g(x)=f(x)-f(0) is additive since g(0)=0. This completes the proof.



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*Source: 814067-2013-09-16.xml*