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

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.

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 $d_Y(f(x),f(y))=d_X(x,y)$ for every $x,y\in X$. Mazur and Ulam [1] 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 $f(0)=0$. Then f is linear.

It was natural to ask if the result holds without the onto assumption. Having asked this natural question, Baker [2] 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 [3–12].

Chu et al. [13] 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 [14], 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. [15] discussed characteristics of 2-isometries. In [16], Amyari and Sadeghi proved the Mazur-Ulam theorem in non-Archimedean 2-normed spaces under the condition of strict convexity. Recently, Choy et al. [17] 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 [20] investigated the stability of linear n-isometries in a linear n-normed Banach module. Chu et al. [21], 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 [22], 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, $f(x)-f(0)$ 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 $[0,\mathrm{\infty })$ such that for all $r,s\in \mathcal{K}$,

• (i) $|r|=0$ if and only if $r=0$;

• (ii) $|rs|=|r||s|$;

• (iii) $|r+s|\le max\{|r|,|s|\}$.

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 [23]).

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

• (i) $\parallel x\parallel =0$ if and only if $x=0$;

• (ii) $\parallel rx\parallel =|r|\parallel x\parallel$;

• (iii) the strong triangle inequality

  $$\parallel x+y\parallel \le max\{\parallel x\parallel ,\parallel y\parallel \}.$$

Then we say $(\mathcal{X},\parallel \cdot \parallel )$ 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 $\parallel \cdot ,\dots ,\cdot \parallel :\mathcal{X}\times \cdots \times \mathcal{X}\to [0,\mathrm{\infty })$ is said to be a non-Archimedean n-norm if

• (i) $\parallel x_1,\dots ,x_n\parallel =0\iff x_1,\dots ,x_n$ are linearly dependent;

• (ii) $\parallel x_1,\dots ,x_n\parallel =\parallel x_{j_1},\dots ,x_{j_n}\parallel$ for every permutation $(j_1,\dots ,j_n)$ of $(1,\dots ,n)$;

• (iii) $\parallel \alpha x_1,\dots ,x_n\parallel =|\alpha |\parallel x_1,\dots ,x_n\parallel$;

• (iv) $\parallel x+y,x_2,\dots ,x_n\parallel \le max\{\parallel x,x_2,\dots ,x_n\parallel ,\parallel y,x_2,\dots ,x_n\parallel \}$

for all $\alpha \in \mathcal{K}$ and all $x,y,x_1,\dots ,x_n\in \mathcal{X}$. Then $(\mathcal{X},\parallel \cdot ,\dots ,\cdot \parallel )$ 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

$$\parallel x_1-x_0,\dots ,x_n-x_0\parallel =\parallel f(x_1)-f(x_0),\dots ,f(x_n)-f(x_0)\parallel$$

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, $\{x_j-x_i\mid 0\le j\ne i\le n\}$ 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 $x_2-x_0=t(x_1-x_0)$ 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, $\mathcal{C}=\{\gamma \in \mathcal{K}:|\gamma |=1\}$.

Lemma 2.4 Let $x_i$ be an element of a non-Archimedean n-normed space $\mathcal{X}$ for every $i\in \{1,\dots ,n\}$ 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

$$\begin{array}{c}\parallel x_1,\dots ,x_i,\dots ,x_j+\gamma x_i,\dots ,x_n\parallel \hfill \\ \le max\{\parallel x_1,\dots ,x_i,\dots ,x_j,\dots ,x_n\parallel ,|\gamma |\parallel x_1,\dots ,x_i,\dots ,x_i,\dots ,x_n\parallel \}\hfill \\ =max\{\parallel x_1,\dots ,x_i,\dots ,x_j,\dots ,x_n\parallel ,0\}\hfill \\ =\parallel x_1,\dots ,x_i,\dots ,x_j,\dots ,x_n\parallel .\hfill \end{array}$$

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 [[22], 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 $\mathcal{C}=\{2^n|n\in \mathbb{Z}\}$ 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

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-u,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_1-u,x_1-x_2,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \end{array}$$

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

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-v,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_1-v,x_1-x_2,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \end{array}$$

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, $v=t{x}_0+(1-t)x_1$ for some $t\in \mathcal{K}$. Then we have

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-v,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-t{x}_0-(1-t)x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =|1-t|\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel ,\hfill \\ \parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_1-v,x_1-x_2,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \\ =\parallel x_1-t{x}_0-(1-t)x_1,x_1-x_2,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \\ =|t|\parallel x_0-x_1,x_1-x_2,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \\ =|t|\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel .\hfill \end{array}$$

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 $1=2^{-n_1}+2^{-n_2}=2^{-n_1}(1+2^{n_1-n_2})$, 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 $\mathcal{C}=\{2^n|n\in \mathbb{Z}\}$ 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 $x_0\ne x_1$, $f(\frac{x_0+x_1}{2})$ is also the midpoint of a segment with endpoints $f(x_0)$ and $f(x_1)$ 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 $\mathcal{X}$, $f(x_0)$, $f(x_1)$, $f(\frac{x_0+x_1}{2})$ are also 2-collinear in $\mathcal{Y}$ by Remark 2.5. Since f is an n-isometry, we have the following:

$$\begin{array}{c}\parallel f(x_0)-f\left(\frac{x_0+x_1}{2}\right),f(x_0)-f(x_2),\dots ,f(x_0)-f(x_n)\parallel \hfill \\ =\parallel x_0-\frac{x_0+x_1}{2},x_0-x_2,\dots ,x_0-x_n\parallel \hfill \\ =\left|\frac{1}{2}\right|\parallel x_0-x_1,x_0-x_2,\dots ,x_0-x_n\parallel \hfill \\ =\parallel f(x_0)-f(x_1),f(x_0)-f(x_2),\dots ,f(x_0)-f(x_n)\parallel ,\hfill \\ \parallel f(x_1)-f\left(\frac{x_0+x_1}{2}\right),f(x_1)-f(x_2),\dots ,f(x_1)-f(x_n)\parallel \hfill \\ =\parallel x_1-\frac{x_0+x_1}{2},x_1-x_2,\dots ,x_1-x_n\parallel \hfill \\ =\left|\frac{1}{2}\right|\parallel x_1-x_0,x_1-x_2,\dots ,x_1-x_n\parallel \hfill \\ =\parallel f(x_0)-f(x_1),f(x_0)-f(x_2),\dots ,f(x_0)-f(x_n)\parallel .\hfill \end{array}$$

By Lemma 2.6, we obtain that $f(\frac{x_0+x_1}{2})=\frac{f(x_0)+f(x_1)}{2}$ 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.

• (i) The n-isometry f preserves the midpoint of a segment in $\mathcal{X}$, i.e., $f(\frac{x_0+x_1}{2})=\frac{f(x_0)+f(x_1)}{2}$ for all $x_0,x_1\in \mathcal{X}$ with $x_0\ne x_1$;

• (ii) The n-isometry f preserves the barycenter of a triangle in $\mathcal{X}$, i.e., $f(\frac{x_0+x_1+x_2}{3})=\frac{f(x_0)+f(x_1)+f(x_2)}{3}$ 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, $f(x_0)$, $f(\frac{x_0+x_1}{2})$, $f(x_1)$ are 2-collinear. So,

$$f\left(\frac{x_0+x_1}{2}\right)-f(x_0)=s(f(x_1)-f(x_0))$$

(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 $f(x_2)$, $f(\frac{x_0+x_1+x_2}{3})$, $f(\frac{x_0+x_1}{2})$ are also 2-collinear by Remark 2.5. So, we obtain that

$$f\left(\frac{x_0+x_1+x_2}{3}\right)-f(x_2)=t(f\left(\frac{x_0+x_1}{2}\right)-f(x_2))$$

(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

$$\frac{f(x_0)+f(x_1)+f(x_2)}{3}-f(x_2)=t(f(x_0)+sf(x_1)-sf(x_0)-f(x_2)).$$

Thus, we get

$$\frac{f(x_0)+f(x_1)-2f(x_2)}{3}=t(1-s)f(x_0)+tsf(x_1)-tf(x_2).$$

So, we have the following equation:

$$\frac{2}{3}(f(x_0)-f(x_2))-\frac{1}{3}(f(x_0)-f(x_1))=t(f(x_0)-f(x_2))-ts(f(x_0)-f(x_1)).$$

By a calculation, we obtain

$$(\frac{2}{3}-t)(f(x_0)-f(x_2))+(-\frac{1}{3}+ts)(f(x_0)-f(x_1))=0.$$

(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,

$$\begin{array}{c}\parallel f(x_0)-f(x_1),f(x_0)-f(x_2),f(x_0)-f(x_3),\dots ,f(x_0)-f(x_n)\parallel \hfill \\ =\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \ne 0.\hfill \end{array}$$

So, $f(x_0)-f(x_1)$ and $f(x_0)-f(x_2)$ 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

$$f\left(\frac{x_0+x_1}{2}\right)=\frac{f(x_0)+f(x_1)}{2}$$

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 $\mathcal{C}=\{2^n|n\in \mathbb{Z}\}$. If $f:\mathcal{X}\to \mathcal{Y}$ is an n-isometry, then $f(x)-f(0)$ is additive.

Proof Let $g(x):=f(x)-f(0)$. Then it is clear that $g(0)=0$ 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

$$g\left(\frac{x_0+x_1}{2}\right)=\frac{g(x_0)+g(x_1)}{2}.$$

Since $g(0)=0$, 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 $\mathcal{C}=\{3^n|n\in \mathbb{Z}\}$ 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

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-u,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_1-x_2,x_1-u,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \\ =\parallel x_2-x_0,x_2-u,x_2-x_3,\dots ,x_2-x_n\parallel \hfill \end{array}$$

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

$$\begin{array}{c}\parallel x_0-x_1,x_0-u,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-\frac{x_0+x_1+x_2}{3},x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\left|\frac{1}{3}\right|\parallel x_0-x_1,x_0-x_1+x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel .\hfill \end{array}$$

And we can also obtain that

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_1-x_2,x_1-u,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \\ =\parallel x_2-x_0,x_2-u,x_2-x_3,\dots ,x_2-x_n\parallel .\hfill \end{array}$$

For the proof of uniqueness, let v be another interior point of ${\mathrm{△}}_{x_0{x}_1{x}_2}$ satisfying

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-v,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_1-x_2,x_1-v,x_1-x_3,\dots ,x_1-x_n\parallel \hfill \\ =\parallel x_2-x_0,x_2-v,x_2-x_3,\dots ,x_2-x_n\parallel \hfill \end{array}$$

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 $\{t_0{x}_0+t_1{x}_1+t_2{x}_2|t_0+t_1+t_2=1,t_i\in \mathcal{K},t_i>0\text{ for all }i\}$, 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

$$\begin{array}{c}\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-v,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-s_0{x}_0-s_1{x}_1-s_2{x}_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,(s_0-1)x_0+s_1{x}_1+(1-s_0-s_1)x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,(s_0+s_1-1)x_0+(1-s_0-s_1)x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =|s_0+s_1-1|\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =|s_2|\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \end{array}$$

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 $1=s_0+s_1+s_2=3^{-n_0}(1+3^{n_0-n_1}+3^{n_0-n_2})$, 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 $\mathcal{C}=\{3^n|n\in \mathbb{Z}\}$ 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, $f(\frac{x_0+x_1+x_2}{3})$ is also the interior point of a triangle ${\mathrm{△}}_{f(x_0)f(x_1)f(x_2)}$. 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

$$\begin{array}{c}\parallel f(x_0)-f(x_1),f(x_0)-f\left(\frac{x_0+x_1+x_2}{3}\right),f(x_0)-f(x_3),\dots ,f(x_0)-f(x_n)\parallel \hfill \\ =\parallel x_0-x_1,x_0-\frac{x_0+x_1+x_2}{3},x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\left|\frac{1}{3}\right|\parallel x_0-x_1,x_0-x_1+x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel x_0-x_1,x_0-x_2,x_0-x_3,\dots ,x_0-x_n\parallel \hfill \\ =\parallel f(x_0)-f(x_1),f(x_0)-f(x_2),f(x_0)-f(x_3),\dots ,f(x_0)-f(x_n)\parallel .\hfill \end{array}$$

Similarly, we obtain

$$\begin{array}{c}\parallel f(x_1)-f(x_2),f(x_1)-f\left(\frac{x_0+x_1+x_2}{3}\right),f(x_1)-f(x_3),\dots ,f(x_1)-f(x_n)\parallel \hfill \\ =\parallel f(x_2)-f(x_1),f(x_2)-f\left(\frac{x_0+x_1+x_2}{3}\right),f(x_2)-f(x_3),\dots ,f(x_2)-f(x_n)\parallel \hfill \\ =\parallel f(x_0)-f(x_1),f(x_0)-f(x_2),f(x_0)-f(x_3),\dots ,f(x_0)-f(x_n)\parallel .\hfill \end{array}$$

From Lemma 3.1, we get

$$f\left(\frac{x_0+x_1+x_2}{3}\right)=\frac{f(x_0)+f(x_1)+f(x_2)}{3}$$

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 $\mathcal{C}=\{3^n|n\in \mathbb{Z}\}$. If $f:\mathcal{X}\to \mathcal{Y}$ is an interior-preserving n-isometry, then $f(x)-f(0)$ is additive.

Proof Let $g(x):=f(x)-f(0)$. One can easily check that $g(0)=0$ and g is also an n-isometry. Using a similar method in [[17], 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,

$$g\left(\frac{x_0+x_1+x_2}{3}\right)=\frac{g(x_0)+g(x_1)+g(x_2)}{3}$$

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 $g(0)=0$, we obtain that the interior-preserving n-isometry g is additive, which completes the proof. □