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# A Mazur-Ulam problem in non-Archimedean n-normed spaces

Journal of Inequalities and Applications20132013:34

https://doi.org/10.1186/1029-242X-2013-34

• Received: 26 August 2012
• Accepted: 11 January 2013
• Published:

## Abstract

In this article, we study the notions of n-isometries in non-Archimedean n-normed spaces over linear ordered non-Archimedean fields and prove the Mazur-Ulam theorem in the spaces. Furthermore, we obtain some properties for n-isometries in non-Archimedean n-normed spaces.

MSC:46B20, 51M25, 46S10.

## Keywords

• Mazur-Ulam theorem
• n-isometry
• non-Archimedean n-normed space

## 1 Introduction

Let X and Y be metric spaces with metrics ${d}_{X}$ and ${d}_{Y}$, respectively. A map $f:X\to Y$ is called an isometry if ${d}_{Y}\left(f\left(x\right),f\left(y\right)\right)={d}_{X}\left(x,y\right)$ for every $x,y\in X$. Mazur and Ulam  treated the theory of isometry for the first time. They have proved the following theorem.

Mazur-Ulam theorem Let f be an isometric transformation from a real normed vector space X onto a real normed vector space Y with $f\left(0\right)=0$. Then f is linear.

It was natural to ask if the result holds without the onto assumption. Having asked this natural question, Baker  answered that every isometry of a real normed linear space into a strictly convex real normed linear space is affine. The Mazur-Ulam theorem has been widely studied by .

Chu et al.  have defined the notion of a 2-isometry which is suitable to represent the concept of an area-preserving mapping in linear 2-normed spaces. In , Chu proved that the Mazur-Ulam theorem holds in linear 2-normed spaces under the condition that a 2-isometry preserves collinearity. Chu et al.  discussed characteristics of 2-isometries. In , Amyari and Sadeghi proved the Mazur-Ulam theorem in non-Archimedean 2-normed spaces under the condition of strict convexity. Recently, Choy et al.  proved the theorem on non-Archimedean 2-normed spaces over linear ordered non-Archimedean fields without the strict convexity assumption.

Misiak [18, 19] defined the concept of an n-normed space and investigated the space. Park and Rassias  investigated the stability of linear n-isometries in a linear n-normed Banach module. Chu et al. , in linear n-normed spaces, defined the concept of an n-isometry that is suitable to represent the notion of a volume-preserving mapping. In , Chu et al. generalized the Mazur-Ulam theorem to n-normed spaces.

In this paper, without the condition of strict convexity, we prove the (additive) Mazur-Ulam theorem on non-Archimedean n-normed spaces. Firstly, we assert that an n-isometry f from a non-Archimedean space to a non-Archimedean space preserves the midpoint of a segment under some condition about the set of all elements of a valued field whose valuations are 1. Using the above result, we show the Mazur-Ulam theorem on non-Archimedean n-normed spaces over linear ordered non-Archimedean fields. In addition, we prove that the barycenter of a triangle in the non-Archimedean n-normed spaces is f-invariant under different conditions from those referred in previous statements. And then we also prove the (second type) Mazur-Ulam theorem in non-Archimedean n-normed spaces under some different conditions.

## 2 The Mazur-Ulam theorem I in non-Archimedean n-normed spaces

In this section, we introduce a non-Archimedean n-normed space which is a kind of generalization of a non-Archimedean 2-normed space, and we show the (additive) Mazur-Ulam theorem for an n-isometry f defined on a non-Archimedean n-normed space, that is, $f\left(x\right)-f\left(0\right)$ is additive. Firstly, we consider some definitions and lemmas which are needed to prove the theorem.

Recall that a non-Archimedean (or ultrametric) valuation is given by a map $|\cdot |$ from a field $\mathcal{K}$ into $\left[0,\mathrm{\infty }\right)$ such that for all $r,s\in \mathcal{K}$,
1. (i)

$|r|=0$ if and only if $r=0$;

2. (ii)

$|rs|=|r||s|$;

3. (iii)

$|r+s|\le max\left\{|r|,|s|\right\}$.

If every element of $\mathcal{K}$ carries a valuation, then a field $\mathcal{K}$ is called a valued field; for convenience, we simply call it a field. It is obvious that $|1|=|-1|=1$ and $|n|\le 1$ for all $n\in \mathbb{N}$. A trivial example of a non-Archimedean valuation is the map $|\cdot |$ taking everything but 0 into 1 and $|0|=0$ (see ).

Let $\mathcal{X}$ be a vector space over a valued field $\mathcal{K}$. A non-Archimedean norm is a function $\parallel \cdot \parallel :\mathcal{X}\to \left[0,\mathrm{\infty }\right)$ such that for all $r\in \mathcal{K}$ and $x,y\in \mathcal{X}$,
1. (i)

$\parallel x\parallel =0$ if and only if $x=0$;

2. (ii)

$\parallel rx\parallel =|r|\parallel x\parallel$;

3. (iii)
the strong triangle inequality
$\parallel x+y\parallel \le max\left\{\parallel x\parallel ,\parallel y\parallel \right\}.$

Then we say $\left(\mathcal{X},\parallel \cdot \parallel \right)$ is a non-Archimedean space.

Definition 2.1 Let $\mathcal{X}$ be a vector space with the dimension greater than $n-1$ over a valued field $\mathcal{K}$ with a non-Archimedean valuation $|\cdot |$. A function $\parallel \cdot ,\dots ,\cdot \parallel :\mathcal{X}×\cdots ×\mathcal{X}\to \left[0,\mathrm{\infty }\right)$ is said to be a non-Archimedean n-norm if
1. (i)

$\parallel {x}_{1},\dots ,{x}_{n}\parallel =0⇔{x}_{1},\dots ,{x}_{n}$ are linearly dependent;

2. (ii)

$\parallel {x}_{1},\dots ,{x}_{n}\parallel =\parallel {x}_{{j}_{1}},\dots ,{x}_{{j}_{n}}\parallel$ for every permutation $\left({j}_{1},\dots ,{j}_{n}\right)$ of $\left(1,\dots ,n\right)$;

3. (iii)

$\parallel \alpha {x}_{1},\dots ,{x}_{n}\parallel =|\alpha |\parallel {x}_{1},\dots ,{x}_{n}\parallel$;

4. (iv)

$\parallel x+y,{x}_{2},\dots ,{x}_{n}\parallel \le max\left\{\parallel x,{x}_{2},\dots ,{x}_{n}\parallel ,\parallel y,{x}_{2},\dots ,{x}_{n}\parallel \right\}$

for all $\alpha \in \mathcal{K}$ and all $x,y,{x}_{1},\dots ,{x}_{n}\in \mathcal{X}$. Then $\left(\mathcal{X},\parallel \cdot ,\dots ,\cdot \parallel \right)$ is called a non-Archimedean n-normed space.

From now on, let $\mathcal{X}$ and $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$.

Definition 2.2 Let $\mathcal{X}$ and $\mathcal{Y}$ be non-Archimedean n-normed spaces and $f:\mathcal{X}\to \mathcal{Y}$ be a mapping. We call f an n-isometry if
$\parallel {x}_{1}-{x}_{0},\dots ,{x}_{n}-{x}_{0}\parallel =\parallel f\left({x}_{1}\right)-f\left({x}_{0}\right),\dots ,f\left({x}_{n}\right)-f\left({x}_{0}\right)\parallel$

for all ${x}_{0},{x}_{1},\dots ,{x}_{n}\in \mathcal{X}$.

Definition 2.3 The points ${x}_{0},{x}_{1},\dots ,{x}_{n}$ of a non-Archimedean n-normed space $\mathcal{X}$ are said to be n-collinear if for every i, $\left\{{x}_{j}-{x}_{i}\mid 0\le j\ne i\le n\right\}$ is linearly dependent.

The points ${x}_{0}$, ${x}_{1}$ and ${x}_{2}$ of a non-Archimedean n-normed space $\mathcal{X}$ are said to be 2-collinear if and only if ${x}_{2}-{x}_{0}=t\left({x}_{1}-{x}_{0}\right)$ for some element t of the non-Archimedean field $\mathcal{K}$. We denote the set of all elements of $\mathcal{K}$ whose valuations are 1 by $\mathcal{C}$, that is, $\mathcal{C}=\left\{\gamma \in \mathcal{K}:|\gamma |=1\right\}$.

Lemma 2.4 Let ${x}_{i}$ be an element of a non-Archimedean n-normed space $\mathcal{X}$ for every $i\in \left\{1,\dots ,n\right\}$ and $\gamma \in \mathcal{K}$. Then
$\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{j},\dots ,{x}_{n}\parallel =\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{j}+\gamma {x}_{i},\dots ,{x}_{n}\parallel$

for all $1\le i\ne j\le n$.

Proof

By Definition 2.1, we have
$\begin{array}{c}\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{j}+\gamma {x}_{i},\dots ,{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\le max\left\{\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{j},\dots ,{x}_{n}\parallel ,|\gamma |\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{i},\dots ,{x}_{n}\parallel \right\}\hfill \\ \phantom{\rule{1em}{0ex}}=max\left\{\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{j},\dots ,{x}_{n}\parallel ,0\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1},\dots ,{x}_{i},\dots ,{x}_{j},\dots ,{x}_{n}\parallel .\hfill \end{array}$

One can easily prove the converse using similar methods. This completes the proof. □

Remark 2.5 Let $\mathcal{X}$, $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$ and let $f:\mathcal{X}\to \mathcal{Y}$ be an n-isometry. One can show that the n-isometry f from $\mathcal{X}$ to $\mathcal{Y}$ preserves the 2-collinearity using a similar method to that in [, Lemma 3.2].

The midpoint of a segment with endpoints x and y in the non-Archimedean n-normed space $\mathcal{X}$ is defined by the point $\frac{x+y}{2}$.

Now, we prove the Mazur-Ulam theorem on non-Archimedean n-normed spaces. In the first step, we prove the following lemma. And then, using the lemma, we show that an n-isometry f from a non-Archimedean n-normed space $\mathcal{X}$ to a non-Archimedean n-normed space $\mathcal{Y}$ preserves the midpoint of a segment, i.e., the f-image of the midpoint of a segment in $\mathcal{X}$ is also the midpoint of a corresponding segment in $\mathcal{Y}$.

Lemma 2.6 Let $\mathcal{X}$ be a non-Archimedean n-normed space over a linear ordered non-Archimedean field $\mathcal{K}$ with $\mathcal{C}=\left\{{2}^{n}|n\in \mathbb{Z}\right\}$ and let ${x}_{0},{x}_{1}\in \mathcal{X}$ with ${x}_{0}\ne {x}_{1}$. Then $u:=\frac{{x}_{0}+{x}_{1}}{2}$ is the unique member of $\mathcal{X}$ satisfying
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-u,{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-u,{x}_{1}-{x}_{2},{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \end{array}$

for some ${x}_{2},\dots ,{x}_{n}\in \mathcal{X}$ with $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$ and u, ${x}_{0}$, ${x}_{1}$ are 2-collinear.

Proof Let $u:=\frac{{x}_{0}+{x}_{1}}{2}$. From the assumption for the dimension of $\mathcal{X}$, there exist $n-1$ elements ${x}_{2},\dots ,{x}_{n}$ in $\mathcal{X}$ such that $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$. One can easily prove that u satisfies the above equations and conditions. It suffices to show the uniqueness for u. Assume that there is another v satisfying
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-v,{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-v,{x}_{1}-{x}_{2},{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \end{array}$
for some elements ${x}_{2},\dots ,{x}_{n}$ of $\mathcal{X}$ with $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$ and v, ${x}_{0}$, ${x}_{1}$ are 2-collinear. Since v, ${x}_{0}$, ${x}_{1}$ are 2-collinear, $v=t{x}_{0}+\left(1-t\right){x}_{1}$ for some $t\in \mathcal{K}$. Then we have
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-v,{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-t{x}_{0}-\left(1-t\right){x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|1-t|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel ,\hfill \\ \parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-v,{x}_{1}-{x}_{2},{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-t{x}_{0}-\left(1-t\right){x}_{1},{x}_{1}-{x}_{2},{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|t|\parallel {x}_{0}-{x}_{1},{x}_{1}-{x}_{2},{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|t|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel .\hfill \end{array}$

Since $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$, we have two equations $|1-t|=1$ and $|t|=1$. So, there are two integers ${k}_{1}$, ${k}_{2}$ such that $1-t={2}^{{k}_{1}}$, $t={2}^{{k}_{2}}$. Since ${2}^{{k}_{1}}+{2}^{{k}_{2}}=1$, ${k}_{i}<0$ for all $i=1,2$. Thus, we may assume that $1-t={2}^{-{n}_{1}}$, $t={2}^{-{n}_{2}}$ and ${n}_{1}\ge {n}_{2}\in \mathbb{N}$ without loss of generality. If ${n}_{1}⪈{n}_{2}$, then $1={2}^{-{n}_{1}}+{2}^{-{n}_{2}}={2}^{-{n}_{1}}\left(1+{2}^{{n}_{1}-{n}_{2}}\right)$, that is, ${2}^{{n}_{1}}=1+{2}^{{n}_{1}-{n}_{2}}$. This is a contradiction because the left-hand side of the equation is a multiple of 2 but the right-hand side of the equation is not. Thus, ${n}_{1}={n}_{2}=1$ and hence $v=\frac{1}{2}{x}_{0}+\frac{1}{2}{x}_{1}=u$. □

Theorem 2.7 Let $\mathcal{X}$, $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$ with $\mathcal{C}=\left\{{2}^{n}|n\in \mathbb{Z}\right\}$ and $f:\mathcal{X}\to \mathcal{Y}$ be an n-isometry. Then the midpoint of a segment is f-invariant, i.e., for every ${x}_{0},{x}_{1}\in \mathcal{X}$ with ${x}_{0}\ne {x}_{1}$, $f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)$ is also the midpoint of a segment with endpoints $f\left({x}_{0}\right)$ and $f\left({x}_{1}\right)$ in $\mathcal{Y}$.

Proof Let ${x}_{0},{x}_{1}\in \mathcal{X}$ with ${x}_{0}\ne {x}_{1}$. Since the dimension of $\mathcal{X}$ is greater than $n-1$, there exist $n-1$ elements ${x}_{2},\dots ,{x}_{n}$ of $\mathcal{X}$ satisfying $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$. Since ${x}_{0}$, ${x}_{1}$ and their midpoint $\frac{{x}_{0}+{x}_{1}}{2}$ are 2-collinear in $\mathcal{X}$, $f\left({x}_{0}\right)$, $f\left({x}_{1}\right)$, $f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)$ are also 2-collinear in $\mathcal{Y}$ by Remark 2.5. Since f is an n-isometry, we have the following:
$\begin{array}{c}\parallel f\left({x}_{0}\right)-f\left(\frac{{x}_{0}+{x}_{1}}{2}\right),f\left({x}_{0}\right)-f\left({x}_{2}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-\frac{{x}_{0}+{x}_{1}}{2},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{1}{2}|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel f\left({x}_{0}\right)-f\left({x}_{1}\right),f\left({x}_{0}\right)-f\left({x}_{2}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel ,\hfill \\ \parallel f\left({x}_{1}\right)-f\left(\frac{{x}_{0}+{x}_{1}}{2}\right),f\left({x}_{1}\right)-f\left({x}_{2}\right),\dots ,f\left({x}_{1}\right)-f\left({x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-\frac{{x}_{0}+{x}_{1}}{2},{x}_{1}-{x}_{2},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{1}{2}|\parallel {x}_{1}-{x}_{0},{x}_{1}-{x}_{2},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel f\left({x}_{0}\right)-f\left({x}_{1}\right),f\left({x}_{0}\right)-f\left({x}_{2}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel .\hfill \end{array}$

By Lemma 2.6, we obtain that $f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)=\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)}{2}$ for all ${x}_{0},{x}_{1}\in \mathcal{X}$ with ${x}_{0}\ne {x}_{1}$. This completes the proof. □

Lemma 2.8 Let $\mathcal{X}$ and $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$ and $f:\mathcal{X}\to \mathcal{Y}$ be an n-isometry. Then the following conditions are equivalent.
1. (i)

The n-isometry f preserves the midpoint of a segment in $\mathcal{X}$, i.e., $f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)=\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)}{2}$ for all ${x}_{0},{x}_{1}\in \mathcal{X}$ with ${x}_{0}\ne {x}_{1}$;

2. (ii)

The n-isometry f preserves the barycenter of a triangle in $\mathcal{X}$, i.e., $f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right)=\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)+f\left({x}_{2}\right)}{3}$ for all ${x}_{0},{x}_{1},{x}_{2}\in \mathcal{X}$ satisfying that ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ are not 2-collinear.

Proof Assume that the n-isometry f preserves the barycenter of a triangle in $\mathcal{X}$. Let ${x}_{0}$, ${x}_{1}$ be in $\mathcal{X}$ with ${x}_{0}\ne {x}_{1}$. Since the n-isometry f preserves the 2-collinearity, $f\left({x}_{0}\right)$, $f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)$, $f\left({x}_{1}\right)$ are 2-collinear. So,
$f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)-f\left({x}_{0}\right)=s\left(f\left({x}_{1}\right)-f\left({x}_{0}\right)\right)$
(2.1)
for some element s of $\mathcal{K}$. By the hypothesis for the dimension of $\mathcal{X}$, we can choose the element ${x}_{2}$ of $\mathcal{X}$ satisfying that ${x}_{0}$, ${x}_{1}$ and ${x}_{2}$ are not 2-collinear. Since ${x}_{2}$, $\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}$, $\frac{{x}_{0}+{x}_{1}}{2}$ are 2-collinear, we have that $f\left({x}_{2}\right)$, $f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right)$, $f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)$ are also 2-collinear by Remark 2.5. So, we obtain that
$f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right)-f\left({x}_{2}\right)=t\left(f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)-f\left({x}_{2}\right)\right)$
(2.2)
for some element t of the non-Archimedean field $\mathcal{K}$. By the equations (2.1), (2.2) and the barycenter-preserving property for the n-isometry f, we have
$\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)+f\left({x}_{2}\right)}{3}-f\left({x}_{2}\right)=t\left(f\left({x}_{0}\right)+sf\left({x}_{1}\right)-sf\left({x}_{0}\right)-f\left({x}_{2}\right)\right).$
Thus, we get
$\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)-2f\left({x}_{2}\right)}{3}=t\left(1-s\right)f\left({x}_{0}\right)+tsf\left({x}_{1}\right)-tf\left({x}_{2}\right).$
So, we have the following equation:
$\frac{2}{3}\left(f\left({x}_{0}\right)-f\left({x}_{2}\right)\right)-\frac{1}{3}\left(f\left({x}_{0}\right)-f\left({x}_{1}\right)\right)=t\left(f\left({x}_{0}\right)-f\left({x}_{2}\right)\right)-ts\left(f\left({x}_{0}\right)-f\left({x}_{1}\right)\right).$
By a calculation, we obtain
$\left(\frac{2}{3}-t\right)\left(f\left({x}_{0}\right)-f\left({x}_{2}\right)\right)+\left(-\frac{1}{3}+ts\right)\left(f\left({x}_{0}\right)-f\left({x}_{1}\right)\right)=0.$
(2.3)
Since ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ are not 2-collinear, ${x}_{0}-{x}_{1}$, ${x}_{0}-{x}_{2}$ are linearly independent. Since $dim\mathcal{X}\ge n$, there are ${x}_{3},\dots ,{x}_{n}\in \mathcal{X}$ such that $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$. Since f is an n-isometry,
$\begin{array}{c}\parallel f\left({x}_{0}\right)-f\left({x}_{1}\right),f\left({x}_{0}\right)-f\left({x}_{2}\right),f\left({x}_{0}\right)-f\left({x}_{3}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0.\hfill \end{array}$
So, $f\left({x}_{0}\right)-f\left({x}_{1}\right)$ and $f\left({x}_{0}\right)-f\left({x}_{2}\right)$ are linearly independent. Hence, from equation (2.3), we have $\frac{2}{3}-t=0$ and $-\frac{1}{3}+ts=0$, i.e., we obtain $t=\frac{2}{3}$, $s=\frac{1}{2}$, which imply the equation
$f\left(\frac{{x}_{0}+{x}_{1}}{2}\right)=\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)}{2}$

for all ${x}_{0},{x}_{1}\in \mathcal{X}$ with ${x}_{0}\ne {x}_{1}$.

Conversely, (i) trivially implies (ii). This completes the proof of this lemma. □

Remark 2.9 One can prove that the above lemma also holds in the case of linear n-normed spaces.

Theorem 2.10 Let $\mathcal{X}$ and $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$ with $\mathcal{C}=\left\{{2}^{n}|n\in \mathbb{Z}\right\}$. If $f:\mathcal{X}\to \mathcal{Y}$ is an n-isometry, then $f\left(x\right)-f\left(0\right)$ is additive.

Proof Let $g\left(x\right):=f\left(x\right)-f\left(0\right)$. Then it is clear that $g\left(0\right)=0$ and g is also an n-isometry.

From Theorem 2.7, for ${x}_{0},{x}_{1}\in \mathcal{X}$ (${x}_{0}\ne {x}_{1}$), we have
$g\left(\frac{{x}_{0}+{x}_{1}}{2}\right)=\frac{g\left({x}_{0}\right)+g\left({x}_{1}\right)}{2}.$

Since $g\left(0\right)=0$, we obtain that g is additive which completes the proof. □

## 3 The Mazur-Ulam theorem II in non-Archimedean n-normed spaces

In this section, under different conditions from those previously referred in Theorem 2.10, we also prove the (second type) Mazur-Ulam theorem on a non-Archimedean n-normed space. Firstly, we show that an n-isometry f from a non-Archimedean n-normed space $\mathcal{X}$ to a non-Archimedean n-normed space $\mathcal{Y}$ preserves the barycenter of a triangle, i.e., the f-image of the barycenter of a triangle is also the barycenter of a corresponding triangle. Then, using Lemma 2.8, we also prove the Mazur-Ulam theorem (a non-Archimedean n-normed space version) under some different conditions.

Lemma 3.1 Let $\mathcal{X}$ be a non-Archimedean n-normed space over a linear ordered non-Archimedean field $\mathcal{K}$ with $\mathcal{C}=\left\{{3}^{n}|n\in \mathbb{Z}\right\}$ and let ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ be elements of $\mathcal{X}$ such that ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ are not 2-collinear. Then $u:=\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}$ is the unique member of $\mathcal{X}$ satisfying
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-u,{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-{x}_{2},{x}_{1}-u,{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{2}-{x}_{0},{x}_{2}-u,{x}_{2}-{x}_{3},\dots ,{x}_{2}-{x}_{n}\parallel \hfill \end{array}$

for some ${x}_{3},\dots ,{x}_{n}\in \mathcal{X}$ with $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$ and u is an interior point of ${\mathrm{△}}_{{x}_{0}{x}_{1}{x}_{2}}$.

Proof Let $u:=\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}$. Thus u is an interior point of ${\mathrm{△}}_{{x}_{0}{x}_{1}{x}_{2}}$. Since $dim\mathcal{X}>n-1$, there are $n-2$ elements ${x}_{3},\dots ,{x}_{n}$ of $\mathcal{X}$ such that $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$. Applying Lemma 2.4, we have that
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-u,{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{1}{3}|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{1}+{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel .\hfill \end{array}$
And we can also obtain that
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-{x}_{2},{x}_{1}-u,{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{2}-{x}_{0},{x}_{2}-u,{x}_{2}-{x}_{3},\dots ,{x}_{2}-{x}_{n}\parallel .\hfill \end{array}$
For the proof of uniqueness, let v be another interior point of ${\mathrm{△}}_{{x}_{0}{x}_{1}{x}_{2}}$ satisfying
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-v,{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{1}-{x}_{2},{x}_{1}-v,{x}_{1}-{x}_{3},\dots ,{x}_{1}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{2}-{x}_{0},{x}_{2}-v,{x}_{2}-{x}_{3},\dots ,{x}_{2}-{x}_{n}\parallel \hfill \end{array}$
with $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$. Since v is an element of the set , there are elements ${s}_{0}$, ${s}_{1}$, ${s}_{2}$ of $\mathcal{K}$ with ${s}_{0}+{s}_{1}+{s}_{2}=1$, ${s}_{i}>0$ such that $v={s}_{0}{x}_{0}+{s}_{1}{x}_{1}+{s}_{2}{x}_{2}$. Then we have
$\begin{array}{c}\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-v,{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-{s}_{0}{x}_{0}-{s}_{1}{x}_{1}-{s}_{2}{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},\left({s}_{0}-1\right){x}_{0}+{s}_{1}{x}_{1}+\left(1-{s}_{0}-{s}_{1}\right){x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},\left({s}_{0}+{s}_{1}-1\right){x}_{0}+\left(1-{s}_{0}-{s}_{1}\right){x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|{s}_{0}+{s}_{1}-1|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|{s}_{2}|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \end{array}$

and hence $|{s}_{2}|=1$ since $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \ne 0$. Similarly, we obtain $|{s}_{0}|=|{s}_{1}|=1$. By the hypothesis of $\mathcal{C}$, there are integers ${k}_{0}$, ${k}_{1}$, ${k}_{2}$ such that ${s}_{0}={3}^{{k}_{1}}$, ${s}_{1}={3}^{{k}_{2}}$, ${s}_{2}={3}^{{k}_{2}}$. Since ${s}_{0}+{s}_{1}+{s}_{2}=1$, every ${k}_{i}$ is less than 0. So, one may let ${s}_{0}={3}^{-{n}_{0}}$, ${s}_{1}={3}^{-{n}_{1}}$, ${s}_{2}={3}^{-{n}_{2}}$ and ${n}_{0}\ge {n}_{1}\ge {n}_{2}\in \mathbb{N}$. Assume that one of the above inequalities holds. Then $1={s}_{0}+{s}_{1}+{s}_{2}={3}^{-{n}_{0}}\left(1+{3}^{{n}_{0}-{n}_{1}}+{3}^{{n}_{0}-{n}_{2}}\right)$, i.e., ${3}^{{n}_{0}}=1+{3}^{{n}_{0}-{n}_{1}}+{3}^{{n}_{0}-{n}_{2}}$. This is a contradiction, because the left-hand side is a multiple of 3 whereas the right-hand side is not. Thus, ${n}_{0}={n}_{1}={n}_{2}$. Consequently, ${s}_{0}={s}_{1}={s}_{2}=\frac{1}{3}$. This means that u is unique. □

Theorem 3.2 Let $\mathcal{X}$, $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$ with $\mathcal{C}=\left\{{3}^{n}|n\in \mathbb{Z}\right\}$ and $f:\mathcal{X}\to \mathcal{Y}$ be an interior-preserving n-isometry. Then the barycenter of a triangle is f-invariant.

Proof Let ${x}_{0}$, ${x}_{1}$ and ${x}_{2}$ be elements of $\mathcal{X}$ satisfying that ${x}_{0}$, ${x}_{1}$ and ${x}_{2}$ are not 2-collinear. It is obvious that the barycenter $\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}$ of a triangle ${\mathrm{△}}_{{x}_{0}{x}_{1}{x}_{2}}$ is an interior point of the triangle. By the assumption, $f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right)$ is also the interior point of a triangle ${\mathrm{△}}_{f\left({x}_{0}\right)f\left({x}_{1}\right)f\left({x}_{2}\right)}$. Since $dim\mathcal{X}>n-1$, there exist $n-2$ elements ${x}_{3},\dots ,{x}_{n}$ in $\mathcal{X}$ such that $\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel$ is not zero. Since f is an n-isometry, we have
$\begin{array}{c}\parallel f\left({x}_{0}\right)-f\left({x}_{1}\right),f\left({x}_{0}\right)-f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right),f\left({x}_{0}\right)-f\left({x}_{3}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=|\frac{1}{3}|\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{1}+{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel {x}_{0}-{x}_{1},{x}_{0}-{x}_{2},{x}_{0}-{x}_{3},\dots ,{x}_{0}-{x}_{n}\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel f\left({x}_{0}\right)-f\left({x}_{1}\right),f\left({x}_{0}\right)-f\left({x}_{2}\right),f\left({x}_{0}\right)-f\left({x}_{3}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel .\hfill \end{array}$
Similarly, we obtain
$\begin{array}{c}\parallel f\left({x}_{1}\right)-f\left({x}_{2}\right),f\left({x}_{1}\right)-f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right),f\left({x}_{1}\right)-f\left({x}_{3}\right),\dots ,f\left({x}_{1}\right)-f\left({x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel f\left({x}_{2}\right)-f\left({x}_{1}\right),f\left({x}_{2}\right)-f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right),f\left({x}_{2}\right)-f\left({x}_{3}\right),\dots ,f\left({x}_{2}\right)-f\left({x}_{n}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\parallel f\left({x}_{0}\right)-f\left({x}_{1}\right),f\left({x}_{0}\right)-f\left({x}_{2}\right),f\left({x}_{0}\right)-f\left({x}_{3}\right),\dots ,f\left({x}_{0}\right)-f\left({x}_{n}\right)\parallel .\hfill \end{array}$
From Lemma 3.1, we get
$f\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right)=\frac{f\left({x}_{0}\right)+f\left({x}_{1}\right)+f\left({x}_{2}\right)}{3}$

for all ${x}_{0},{x}_{1},{x}_{2}\in \mathcal{X}$ satisfying that ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ are not 2-collinear. □

The next theorem is the Mazur-Ulam theorem II in non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$. The assumptions of this theorem are different from that of Theorem 2.10. In particular, in the proof of the theorem, we use the f-preserving property for the barycenter of a triangle.

Theorem 3.3 Let $\mathcal{X}$ and $\mathcal{Y}$ be non-Archimedean n-normed spaces over a linear ordered non-Archimedean field $\mathcal{K}$ with $\mathcal{C}=\left\{{3}^{n}|n\in \mathbb{Z}\right\}$. If $f:\mathcal{X}\to \mathcal{Y}$ is an interior-preserving n-isometry, then $f\left(x\right)-f\left(0\right)$ is additive.

Proof Let $g\left(x\right):=f\left(x\right)-f\left(0\right)$. One can easily check that $g\left(0\right)=0$ and g is also an n-isometry. Using a similar method in [, Theorem 2.4], we can easily prove that g is also an interior-preserving mapping.

Now, let ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ be elements of $\mathcal{X}$ satisfying that ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ are not 2-collinear. Since g is an interior-preserving n-isometry, by Theorem 3.2,
$g\left(\frac{{x}_{0}+{x}_{1}+{x}_{2}}{3}\right)=\frac{g\left({x}_{0}\right)+g\left({x}_{1}\right)+g\left({x}_{2}\right)}{3}$

for any ${x}_{0},{x}_{1},{x}_{2}\in \mathcal{X}$ satisfying that ${x}_{0}$, ${x}_{1}$, ${x}_{2}$ are not 2-collinear. Using Lemma 2.8 and the property $g\left(0\right)=0$, we obtain that the interior-preserving n-isometry g is additive, which completes the proof. □

## Declarations

### Acknowledgements

The authors are deeply grateful to the referees whose helped to improve our manuscript. This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2010-0013784).

## Authors’ Affiliations

(1)
Department of Mathematics, Chungnam National University, 79, Daehangno, Yuseong-Gu, Daejeon, 305-764, Republic of Korea

## References 