﻿ The Aleksandrov Problem in Non-Archimedean 2-Fuzzy 2-Normed Spaces

Journal of Applied Mathematics and Physics
Vol.07 No.08(2019), Article ID:94419,11 pages
10.4236/jamp.2019.78121

The Aleksandrov Problem in Non-Archimedean 2-Fuzzy 2-Normed Spaces

Meimei Song, Haixia Jin*

Science of College, Tianjin University of Technology, Tianjin, China    Received: July 19, 2019; Accepted: August 16, 2019; Published: August 19, 2019

ABSTRACT

We introduce the definition of non-Archimedean 2-fuzzy 2-normed spaces and the concept of isometry which is appropriate to represent the notion of area preserving mapping in the spaces above. And then we can get isometry when a mapping satisfies AOPP and (*) (in article) by applying the Benz’s theorem about the Aleksandrov problem in non-Archimedean 2-fuzzy 2-normed spaces.

Keywords:

Non-Archimedean 2-Fuzzy 2-Normed Space, Isometry, Benz’s Theorem 1. Introduction

Let $X,Y$ be two metric spaces. For a mapping $f:X\to Y$, for all ${x}_{1},{x}_{2}\in X$, if f satisfies,

${d}_{Y}\left(f\left({x}_{1}\right),f\left({x}_{2}\right)\right)={d}_{X}\left({x}_{1},{x}_{2}\right)$

where ${d}_{X}\left(\cdot ,\cdot \right),{d}_{Y}\left(\cdot ,\cdot \right)$ denote the metrics in the spaces $X,Y$, then f is called an isometry. It means that for some fixed number $p>0$, assume that f preserves distance p; i.e., for all ${x}_{1},{x}_{2}$ in X, if ${d}_{X}\left({x}_{1},{x}_{2}\right)=p$, we can get ${d}_{Y}\left(f\left({x}_{1}\right),f\left({x}_{2}\right)\right)=p$ . Then we say p is a conservative distance for the mapping f. Whether there exists a single conservative distance for some f such that f is an isometry from X to Y, is the basic issue of conservative distances. It is called the Aleksandrov problem.

Theorem 1.1. (  ) Let $X,Y$ be two real normed linear spaces (or NLS) with $\mathrm{dim}X>1$, $\mathrm{dim}Y>1$ and Y is strictly convex, assume that a fixed real number $p>0$ and that a fixed integer $N>1$ . Finally, if $f:X\to Y$ is a mapping satisfies

1) $‖{x}_{1}-{x}_{2}‖=p⇒‖f\left({x}_{1}\right)-f\left({x}_{2}\right)‖\le p$

2) $‖{x}_{1}-{x}_{2}‖=N\cdot p⇒‖f\left({x}_{1}\right)-f\left({x}_{2}\right)‖\ge N\cdot p$

for all ${x}_{1},{x}_{2}\in X$ . Then f is an affine isometry. we can call Benz’s theorem.

We can see some results about the Aleksandrov problem in different spaces in  -  . A natural question is that: Whether the Aleksandrov problem can be proved in non-Archimedean 2-fuzzy 2-normed spaces under some conditions. So in this article, we will give the definition of non-Archimedean 2-fuzzy 2-normed spaces according to     , then by applying the Benz’s theorem to fix the value of p and N to solve problems.

If a function from a field K to $\left[0,\infty \right)$ satisfies

(T1) $|a|\ge 0,|a|=0⇔a=0$ ;

(T2) $|ab|=|a||b|$ ;

(T3) $|a+b|\le \mathrm{max}\left\{|a|,|b|\right\}$ .

for all $a,b\in K$, then the field K is called a non-Archimedean field.

We can know $|-1|=|1|=1$, $|a|\le 1$ for all $a\in N$ from the above definition. An example of a non-Archimedean valuation (or NAV) is the function $|\text{ }\cdot \text{ }|$ taking $|0|=0$ and others into 1.

In 1897, Hence in  found that p-adic numbers play a vital role in the complex analysis, the norm derived from p-adic numbers is the non-Archimedean norm, the analysis of the non-Archimedean has important applications in physics.

Definition 1.2. Let X be a vector space and dim $X\ge 2$ . A function $‖\cdot ,\cdot ‖:X\to \left[0,\infty \right)$ is called non-Archimedean 2-norm, if and only if it satisfies

(T1) $‖{x}_{1},{x}_{2}‖\ge 0$, $‖{x}_{1},{x}_{2}‖=0$ iff ${x}_{1},{x}_{2}$ are linearly dependent;

(T2) $‖{x}_{1},{x}_{2}‖=‖{x}_{2},{x}_{1}‖$ ;

(T3) $‖r{x}_{1},{x}_{2}‖=|r|‖{x}_{1},{x}_{2}‖$ ;

(T4) $‖{x}_{1}+{x}_{2},y‖\le \mathrm{max}\left\{‖{x}_{1},y‖,‖{x}_{2},y‖\right\}$

for all ${x}_{1},{x}_{2},y\in X,r\in K$ . Then $\left(X,‖\cdot ,\cdot ‖\right)$ is called non-Archimedean 2-normed space over the field K.

Definition 1.3. An NAV $|\text{ }\cdot \text{ }|$ in a linear space X over a field K. A function $F:X×ℝ\to \left[0,1\right]$ is said to be a non-Archimedean fuzzy norm on X, if and only if for all $x,{x}_{1},{x}_{2}\in X$ and $s,t\in ℝ$,

(F1) $F\left(x,s\right)=0$ with $s\le 0$,

(F2) $F\left(x,s\right)=1$ iff $x=0$ for all $s>0$,

(F3) $F\left(cx,s\right)=F\left(x,\frac{s}{|c|}\right)$, for $c\ne 0$ and $c\in K$,

(F4) $F\left({x}_{1}+{x}_{2},s+t\right)\ge \mathrm{min}\left\{F\left({x}_{1},s\right),F\left({x}_{2},t\right)\right\}$,

(F5) $F\left(x,\ast \right)$ is a nondecreasing function of $s\in R$ and ${\mathrm{lim}}_{s\to \infty }F\left(x,s\right)=1$ .

Then $\left(X,F\right)$ is known as a non-Archimedean fuzzy normed space (or F-NANS).

Theorem 1.4. Let $\left(X,F\right)$ be an F-NANS. Assume the condition that:

(F6) $F\left(x,s\right)>0$ for all $s>0$ $⇒x=0$ .

Define ${‖x‖}_{\alpha }=\mathrm{inf}\left\{s:F\left(x,s\right)\ge \alpha \right\},\alpha \in \left(0,1\right)$ . We call these α-norms on X or the fuzzy norm on X.

Proof: 1) Let ${‖x‖}_{\alpha }=0$, it implies that $\mathrm{inf}\left\{s:F\left(x,s\right)\ge \alpha \right\}=0$, then for all $s\in R$, $s>0$, $F\left(x,s\right)\ge \alpha >0$, so $x=0$ ;

Conversely, assume that $x=0$, by (F2), $F\left(x,s\right)=1$ for all $s>0$, then $\mathrm{inf}\left\{s:F\left(x,s\right)\ge \alpha \right\}=0$ for all $\alpha \in \left(0,1\right)$, so ${‖x‖}_{\alpha }=0$ .

2) By (F3), if $c\ne 0$, then

${‖cx‖}_{\alpha }=\mathrm{inf}\left\{s:F\left(cx,s\right)\ge \alpha \right\}=\mathrm{inf}\left\{s:F\left(x,\frac{s}{|c|}\right)\ge \alpha \right\}$

Let $t=\frac{s}{|c|}$, then

${‖cx‖}_{\alpha }=\mathrm{inf}\left\{|c|t:F\left(x,t\right)\ge \alpha \right\}=|c|\mathrm{inf}\left\{t:F\left(x,t\right)\ge \alpha \right\}=|c|\cdot {‖x‖}_{\alpha }$

If $c=0$, then

${‖cx‖}_{\alpha }=0=c{‖x‖}_{\alpha }$

3) We have

$\begin{array}{l}\mathrm{max}\left\{{‖x‖}_{\alpha },{‖y‖}_{\alpha }\right\}\\ =\mathrm{max}\left\{\mathrm{inf}\left\{s,F\left(x,s\right)\ge \alpha \right\},\mathrm{inf}\left\{t,F\left(x,t\right)\ge \alpha \right\}\right\}\\ =\mathrm{inf}\left\{\mathrm{max}\left\{s,t\right\},F\left(x,s\right)\ge \alpha ,F\left(x,t\right)\ge \alpha \right\}\\ \ge \mathrm{inf}\left\{s+t,F\left(x+y,s+t\right)\ge F\left(x+y,\mathrm{max}\left\{s,t\right\}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\ge \mathrm{min}\left\{F\left(x,s\right)\ge \alpha ,F\left(x,t\right)\ge \alpha \right\}\ge \alpha \right\}\\ \ge \mathrm{inf}\left\{r,F\left(x+y,r\right)\ge \alpha \right\}={‖x+y‖}_{\alpha }\end{array}$

$\square$

Example 1.5. Let $\left(X,‖\text{ }\cdot \text{ }‖\right)$ be a non-Archimedean normed space. Define

$F\left(x,s\right)=\left(\begin{array}{ll}\frac{s}{s+‖x‖},\hfill & s>0,\hfill \\ 0,\hfill & s\le 0.\hfill \end{array}$

for all $x\in X$, Then $\left(X,F\right)$ is a F-NANS.

Definition 1.6. Let Z be any non-empty set and $\Im \left(Z\right)$ be the set of all fuzzy sets on Z. For ${Z}_{1},{Z}_{2}\in \Im \left(Z\right)$ and $\lambda \in K$, define

${Z}_{1}+{Z}_{2}=\left\{\left({z}_{1}+{z}_{2},{\mu }_{1}\wedge {\mu }_{2}\right)|\left({z}_{1},{\mu }_{1}\right)\in {Z}_{1},\left({z}_{2},{\mu }_{2}\right)\in {Z}_{2}\right\}$

and

$\lambda {Z}_{1}=\left\{\left(\lambda {z}_{1},{\mu }_{1}\right)|\left({z}_{1},{\mu }_{1}\right)\in {Z}_{1}\right\}$

Definition 1.7. A non-Archimedean fuzzy linear space $\stackrel{^}{X}=X×\left(0,1\right]$ over the field K, we define the addition and scalar multiplication operation of X as following: $\left({x}_{1},{\mu }_{1}\right)+\left({x}_{2},{\mu }_{2}\right)=\left({x}_{1}+{x}_{2},{\mu }_{1}\wedge {\mu }_{2}\right)$, $\lambda \left({x}_{1},{\mu }_{1}\right)=\left(\lambda {x}_{1},{\mu }_{1}\right)$, if for every $\left({x}_{1},{\mu }_{1}\right)\in X$, we have a related non-negative real numebr, $‖\left({x}_{1},{\mu }_{1}\right)‖$ is the fuzzy norm of $\left({x}_{1},{\mu }_{1}\right)$ in such that

(T1) $‖\left({x}_{1},{\mu }_{1}\right)‖=0⇔{x}_{1}=0,{\mu }_{1}\in \left(0,1\right]$ ;

(T2) $‖\lambda \left({x}_{1},{\mu }_{1}\right)‖=|\lambda |‖\left({x}_{1},{\mu }_{1}\right)‖$ ;

(T3) $‖\left({x}_{1},{\mu }_{1}\right)+\left({x}_{2},{\mu }_{2}\right)‖\le \mathrm{max}\left\{‖\left({x}_{1},{\mu }_{1}\wedge {\mu }_{2}\right),\left({x}_{2},{\mu }_{1}\wedge {\mu }_{2}\right)‖\right\}$ ;

(T4) $‖\left({x}_{1},{\vee }_{t}\text{ }{\mu }_{t}\right)‖={\wedge }_{t}‖\left({x}_{1},{\mu }_{t}\right)‖$ for all ${\mu }_{t}\in \left(0,1\right]$ .

for every $\left({x}_{1},{\mu }_{1}\right),\left({x}_{2},{\mu }_{2}\right)\in X,\lambda \in K$, then we say that X is an F-NANS.

Definition 1.8. Let X be a non-empty non-Archimedean field set, $\Im \left(X\right)$ be the set of all fuzzy sets on X. If ${f}_{1}\in \Im \left(X\right)$, then ${f}_{1}=\left\{\left({x}_{1},{\mu }_{1}\right):{x}_{1}\in X,{\mu }_{1}\in \left(0,1\right]\right\}$ . Clearly, $|{f}_{1}\left({x}_{1}\right)|\le 1$, so ${f}_{1}$ is a bounded function. Let $K\in ℚ$, then $\Im \left(X\right)$ is a non-Archimedean linear space over the field K and the addition, scalar multiplication are defined as follows

${f}_{1}+{f}_{2}=\left\{\left({x}_{1},{\mu }_{1}\right)+\left({x}_{2},{\mu }_{2}\right)\right\}=\left\{\left({x}_{1}+{x}_{2},{\mu }_{1}\wedge {\mu }_{2}\right)|\left({x}_{1},{\mu }_{1}\right)\in {f}_{1},\left({x}_{2},{\mu }_{2}\right)\in {f}_{2}\right\}$

and

$\lambda {f}_{1}=\left\{\left(\lambda {x}_{1},{\mu }_{1}\right)|\left({x}_{1},{\mu }_{1}\right)\in {f}_{1}\right\}$

If for every $f\in \Im \left(X\right)$, there is a related non-negative real number $‖f‖$ called the norm of f in such that for all ${f}_{1}=\left({x}_{1},{\mu }_{1}\right),{f}_{2}=\left({x}_{2},{\mu }_{2}\right)\in \Im \left(X\right)$

(T1) $‖f‖=0$ iff $f=0$ . For

$‖f‖=\left\{‖\left({x}_{1},{\mu }_{1}\right)‖\right\}=0$

$⇔{x}_{1}=0,{\mu }_{1}\in \left(0,1\right]$

$⇔f=0.$

(T2) $‖\lambda f‖=|\lambda |‖f‖,\lambda \in K$ . For

$‖\lambda f‖=\left\{‖\lambda \left({x}_{1},{\mu }_{1}\right)‖\right\}=\left\{|\lambda |‖\left({x}_{1},{\mu }_{1}\right)‖\right\}=|\lambda |‖f‖$

(T3) $‖{f}_{1}+{f}_{2}‖\le \mathrm{max}\left\{‖{f}_{1}‖,‖{f}_{2}‖\right\}$ . For

$\begin{array}{c}‖{f}_{1}+{f}_{2}‖=\left\{‖\left({x}_{1},{\mu }_{1}\right)+\left({x}_{2},{\mu }_{2}\right)‖\right\}\\ =\left\{‖\left({x}_{1}+{x}_{2}\right),\left({\mu }_{1}\wedge {\mu }_{2}\right)‖\right\}\\ \le \mathrm{max}\left\{‖\left({x}_{1},{\mu }_{1}\wedge {\mu }_{2}\right)‖,‖\left({x}_{2},{\mu }_{1}\wedge {\mu }_{2}\right)‖\right\}\\ \le \mathrm{max}\left\{‖{f}_{1}‖,‖{f}_{2}‖\right\}\end{array}$

Then the linear space $\Im \left(X\right)$ is a non-Archimedean normed space.

Definition 1.9. (  ) A 2-fuzzy set on X is a fuzzy set on $\Im \left(X\right)$ .

Definition 1.10. A NAV $|\cdot ,\cdot |$ in a linear space $\Im \left(X\right)$ over a field K. If a function $F:\Im {\left(X\right)}^{2}×ℝ\to \left[0,1\right]$ is a non-Archimedean 2-fuzzy 2-norm on X (or a fuzzy 2-norm on $\Im \left(X\right)$ ), iff for all ${f}_{1},{f}_{2},{f}_{3}\in \Im \left(X\right)$, $s,t\in ℝ$,

(F1) $F\left({f}_{1},{f}_{2},s\right)=0$ for $s\le 0$ ;

(F2) $F\left({f}_{1},{f}_{2},s\right)=1$ iff ${f}_{1},{f}_{2}$ are linearly dependent for all $s>0$ ;

(F3) $F\left({f}_{1},{f}_{2},s\right)=N\left({f}_{2},{f}_{1},s\right)$ ;

(F4) $F\left(c{f}_{1},{f}_{2},s\right)=N\left({f}_{1},{f}_{2},\frac{s}{|c|}\right)$, for $c\ne 0$ and $c\in K$ ;

(F5) $F\left({f}_{1},{f}_{2}+{f}_{3},s+t\right)\ge \mathrm{min}\left\{F\left({f}_{1},{f}_{2},s\right),F\left({f}_{1},{f}_{3},t\right)\right\}$ ;

(F6) $F\left({f}_{1},{f}_{2},\ast \right)$ is a nondecreasing function of R and ${\mathrm{lim}}_{s\to \infty }F\left({f}_{1},{f}_{2},s\right)=1$ ;

Then $\left(\Im \left(X\right),F\right)$ is called a non-Archimedean fuzzy 2-normed space (or FNA-2) or $\left(X,F\right)$ is a non-Archimedean 2-fuzzy 2-normed space.

Theorem 1.11. Let $\left(\Im \left(X\right),F\right)$ be an FNA-2. Suppose the condition that:

(F7) $N\left({f}_{1},{f}_{2},s\right)>0$ for all $s>0$ $⇒{f}_{1}$ and ${f}_{2}$ are linearly dependent.

Define ${‖{f}_{1},{f}_{2}‖}_{\alpha }=\mathrm{inf}\left\{t:N\left({f}_{1},{f}_{2},s\right)\ge \alpha ,\alpha \in \left(0,1\right)\right\}$ . We call these α-2-norms on $\Im \left(X\right)$ or the 2-fuzzy 2-norm on X.

Proof: It is similar to the proof of Theorem 1.4. $\square$

2. Main Result

From now on, if we have no other explanation, let $\mathrm{dim}\Im \left(X\right)\ge 2$, $\mathrm{dim}\Im \left(Y\right)\ge 2$ . $▲\text{\hspace{0.17em}}={‖f-h,g-h‖}_{\alpha }$, $▼\text{\hspace{0.17em}}={‖\psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }$

Definition 2.1. Let $\Im \left(X\right),\Im \left(Y\right)$ be two FNA-2 and a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ . If for all $f,g,h\in \Im \left(X\right)$ and $\alpha ,\beta \in \left(0,1\right)$, we have

${‖\psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }={‖f-h,g-h‖}_{\alpha }$ $\left( \nabla \right)$

then $\psi$ is called 2-isometry.

Definition 2.2. For a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ and $f,g,h\in \Im \left(X\right)$

1) If $▲\text{\hspace{0.17em}}=1$, then $▼\text{\hspace{0.17em}}=1$, we say $\psi$ satisfies the area one preserving property (AOPP).

2) If $▲\text{\hspace{0.17em}}=n$, then $▼\text{\hspace{0.17em}}=n$, we say $\psi$ satisfies the area n for each n (AnPP).

Definition 2.3. We say a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ preserves collinear, if $f,g,h$ mutually disjoint elements of $\Im \left(X\right)$, then exist some real number t we have

$\psi \left(g\right)-\psi \left(h\right)=t\left(\psi \left(f\right)-\psi \left(h\right)\right)$

Next, we denote ${‖\psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }\le {‖f-h,g-h‖}_{\alpha }$ $\left(\ast \right)$ .

Lemma 2.4. Let $\Im \left(X\right)$ and $\Im \left(Y\right)$ be two FNA-2. If $▲\text{\hspace{0.17em}}\le 1$, a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ satisfies $\left(\ast \right)$ and AOPP, then we can get $\left(\nabla \right)$ where $▲\text{\hspace{0.17em}}\le 1$ .

Proof: 1) Firstly, we prove that f preserves collinear. We assume that $▲\text{\hspace{0.17em}}=0$, according to $\left(\ast \right)$, we get

${‖\psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }=0$

then $\psi \left(f\right)-\psi \left(h\right)$ and $\psi \left(g\right)-\psi \left(h\right)$ are linearly dependent. So we obtain that $\psi$ preserves collinear.

2) Secondly, we prove that when $▲\text{\hspace{0.17em}}\le 1$, we can get $\left(\nabla \right)$ .

If

$▼\text{\hspace{0.17em}}<\text{\hspace{0.17em}}▲$

Let $\omega =h+\frac{f-h}{{‖f-h,g-h‖}_{\alpha }}$, then ${‖\omega -h,g-h‖}_{\alpha }=1$, so

${‖\psi \left(\omega \right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }=1$ $\left(\Delta \right)$

Since

${‖\omega -f,g-h‖}_{\alpha }=‖\frac{f-h}{‖f-h,g-h‖}-\left(f-h\right),g-h‖=1-\text{\hspace{0.17em}}▲$

according to $\left(\ast \right)$, we have

${‖\psi \left(\omega \right)-\psi \left(f\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }\le {‖\omega -f,g-h‖}_{\alpha }=1-\text{\hspace{0.17em}}▲$

Since f preserves collinear, so there exists a real number s such that

$\psi \left(\omega \right)-\psi \left(h\right)=s\left(\psi \left(f\right)-\psi \left(h\right)\right)$

and

$\psi \left(\omega \right)-\psi \left(f\right)=\left(s-1\right)\left(\psi \left(f\right)-\psi \left(h\right)\right)$

So, we get

$\begin{array}{l}{‖\psi \left(\omega \right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }\\ =|s|▼\\ \le |s-1|▼\text{\hspace{0.17em}}+\text{\hspace{0.17em}}▼\\ ={‖\psi \left(\omega \right)-\psi \left(f\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }+\text{\hspace{0.17em}}▼\\ <1-\text{\hspace{0.17em}}▲\text{\hspace{0.17em}}+\text{\hspace{0.17em}}▲\text{\hspace{0.17em}}=1\end{array}$

This contradicts with $\Delta$ . $\square$

Lemma 2.5. Let $\Im \left(X\right)$ and $\Im \left(Y\right)$ be two FNA-2. If a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ satisfies AOPP and preserves collinear, then

1) $\psi$ is an injective;

2) if $\varphi \left(f\right)=\psi \left(f\right)-\psi \left(0\right)$, then $\varphi \left(f+g\right)=\varphi \left(f\right)+\varphi \left(g\right)$ and $\varphi \left(\lambda f\right)=\lambda \varphi \left(f\right)$ with $0<\lambda <1$ .

Proof: 1) We prove $\psi$ is injective. Let $f,g\in \Im \left(X\right)$, since dim $\Im \left(X\right)\ge 2$, there exists an element $h\in \Im \left(X\right)$ such that $f-h,g-h$ are linearly independent. Hence $▲\text{\hspace{0.17em}}\ne 0$ .

Let $\gamma =h+\frac{g-h}{{‖f-h,g-h‖}_{\alpha }}$, then ${‖f-h,\gamma -h‖}_{\alpha }=1$, and $\psi$ satisfies AOPP, so

${‖\psi \left(f\right)-\psi \left(h\right),\psi \left(\gamma \right)-\psi \left(h\right)‖}_{\beta }=1$

we can see $\psi \left(h\right)\ne \psi \left(f\right)$ . So the mapping $\psi$ is injective.

2) Let $f,g,h$ mutually disjoint elements of $\Im \left(X\right)$ and $f=\frac{g+h}{2}$, so $f-h=g-f$ $\left(\star \right)$ . Since $\psi$ is injective and preserves collinear, there exist $s\ne 0$ such that

$\psi \left(g\right)-\psi \left(f\right)=s\left(\psi \left(h\right)-\psi \left(f\right)\right)$

Since dim $\Im \left(X\right)\ge 2$, there exist an element ${f}_{1}\in \Im \left(X\right)$ such that ${‖g-f,{f}_{1}-f‖}_{\alpha }\ne 0$ . Let $\eta =f+\frac{{f}_{1}-f}{{‖g-f,{f}_{1}-f‖}_{\alpha }}$, then ${‖g-f,\eta -f‖}_{\alpha }=1$ and

${‖\psi \left(g\right)-\psi \left(f\right),\psi \left(\eta \right)-\psi \left(f\right)‖}_{\beta }=1.$

So,

${‖\psi \left(h\right)-\psi \left(f\right),\psi \left(\eta \right)-\psi \left(f\right)‖}_{\beta }=|\frac{1}{s}|.$

Since $\left(\star \right)$, we get ${‖h-f,\eta -f‖}_{\alpha }=1$ and

${‖\psi \left(h\right)-\psi \left(f\right),\psi \left(\eta \right)-\psi \left(f\right)‖}_{\beta }=1.$

According to the mapping $\psi$ is injective, so $s=-1$, and

$\psi \left(\frac{g+h}{2}\right)=\frac{\psi \left(g\right)+\psi \left(h\right)}{2}$

Let $\varphi \left(f\right)=\psi \left(f\right)-\psi \left(0\right)$, so we have

$\varphi \left(\frac{g+h}{2}\right)=\frac{\varphi \left(g\right)+\varphi \left(h\right)}{2}$

Therefore

$\varphi \left(\frac{f}{2}\right)=\varphi \left(\frac{f+0}{2}\right)=\frac{\varphi \left(f\right)}{2}$

and

$\varphi \left(f+g\right)=\varphi \left(\frac{2f+2g}{2}\right)=\frac{\varphi \left(2f\right)}{2}+\frac{\varphi \left(2g\right)}{2}=\varphi \left(f\right)+\varphi \left(g\right)$

So $\varphi$ is additive.

From the lemma 2.4, we know that if $▲\text{\hspace{0.17em}}\le 1$, then $\varphi$ satisfies 2-isometry.

$0={‖\lambda f,f‖}_{\alpha }={‖\psi \left(\lambda f\right)-\psi \left(0\right),\psi \left(f\right)-\psi \left(0\right)‖}_{\beta }={‖\varphi \left(\lambda f\right),\varphi \left(f\right)‖}_{\beta }$

so $\varphi \left(\lambda f\right)$ and $\varphi \left(f\right)$ is linearly dependent i.e. $\varphi \left(\lambda f\right)=s\varphi \left(f\right)$ .

Next we assume ${‖f,g‖}_{\alpha }=\lambda$,

$\frac{1}{\lambda }{‖f,g‖}_{\alpha }={‖\frac{f}{\lambda }-0,g-0‖}_{\alpha }=1$

and

$\begin{array}{c}1={‖\varphi \left(\frac{f}{\lambda }\right)-\varphi \left(0\right),\varphi \left(g\right)-\varphi \left(0\right)‖}_{\beta }\\ ={‖\varphi \left(\frac{f}{\lambda }\right),\varphi \left(g\right)‖}_{\beta }\\ =\frac{1}{|s|}{‖\varphi \left(f\right),\varphi \left(g\right)‖}_{\beta }\\ =\frac{1}{|s|}{‖f,g‖}_{\alpha }\end{array}$

Thus $\lambda =|s|$, if $s=-\lambda$, then $\varphi \left(\lambda f\right)=-\lambda \varphi \left(f\right)$, but

$\begin{array}{c}|\lambda -1|{‖f,g‖}_{\alpha }={‖\lambda f-f,g-0‖}_{\alpha }\\ ={‖\varphi \left(\lambda f\right)-\varphi \left(f\right),\varphi \left(g\right)-\varphi \left(0\right)‖}_{\beta }\\ ={‖-\lambda \varphi \left(f\right)-\varphi \left(f\right),\varphi \left(g\right)-\varphi \left(0\right)‖}_{\beta }\\ =\left(\lambda +1\right){‖\varphi \left(f\right),\varphi \left(g\right)‖}_{\beta }\\ =\left(\lambda +1\right){‖f,g‖}_{\alpha }\end{array}$

so $|\lambda -1|=\left(\lambda +1\right)$ . It contradicts with $0<\lambda <1$ . Thus $\varphi \left(\lambda f\right)=\lambda \varphi \left(f\right)$ . $\square$

Lemma 2.6. Let $\Im \left(X\right)$ and $\Im \left(Y\right)$ be FNA-2. If $▲\text{\hspace{0.17em}}\le 1$, a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ satisfies $\left(\ast \right)$ and AOPP, then we can get for all $f,g,h\in \Im \left(X\right)$, we can get $\left(\nabla \right)$ .

Proof: From lemma 2.4, we know $\psi$ preserves collinear.

For any $f,g,h\in \Im \left(X\right)$, there exist two numbers $m,n\in {ℕ}^{\ast }$ such that $▲\text{\hspace{0.17em}}\le \frac{m}{n}$ .

So,

${‖\psi \left(\frac{f}{m}\right)-\psi \left(\frac{h}{m}\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }\le {‖\frac{f-h}{m},g-h‖}_{\alpha }\le \frac{1}{n}$

and

${‖\left(\frac{f}{m}\right)-\varphi \left(\frac{h}{m}\right),\varphi \left(g\right)-\varphi \left(h\right)‖}_{\beta }\le \frac{1}{n}$

By lemma 2.5, we have

${‖\frac{1}{m}\left(\varphi \left(f\right)-\varphi \left(h\right)\right),\varphi \left(g\right)-\varphi \left(h\right)‖}_{\beta }\le \frac{1}{n}$

${‖\varphi \left(f\right)-\varphi \left(h\right),\varphi \left(g\right)-\varphi \left(h\right)‖}_{\beta }\le \frac{m}{n}$

Thus

${‖\psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }\le \frac{m}{n}$

$\square$

Lemma 2.7. Let $\Im \left(X\right)$ and $\Im \left(Y\right)$ be two FNA-2. If a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ satisfies AOPP and $\left(\ast \right)$ for all $f,g,h\in \Im \left(X\right)$ with $▲\text{\hspace{0.17em}}\le 1$, then $\psi$ satisfies AnPP.

Proof: Let $f,g,h\in \Im \left(X\right)$ and $n\in N$ . Let

$▲\text{\hspace{0.17em}}=n,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{g}_{i}=h+\frac{i}{n}\left(g-h\right)$

and

${‖f-h,{g}_{i+1}-{g}_{i}‖}_{\alpha }=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=0,1,\cdots ,n-1.$

So,

${‖\psi \left(f\right)-\psi \left(h\right),\psi \left({g}_{i+1}\right)-\psi \left({g}_{i}\right)‖}_{\beta }=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=0,1,\cdots ,n-1.$

We know $\psi$ preserves collinear. So there exist a number $t\in ℝ$ such that

$\psi \left({g}_{2}\right)-\psi \left({g}_{1}\right)=t\left(\psi \left({g}_{1}\right)-\psi \left({g}_{0}\right)\right)$

Therefore

Then we have $t=±1$ . By lemma 2.5, $t=1$, so

$\psi \left({g}_{2}\right)-\psi \left({g}_{1}\right)=\psi \left({g}_{1}\right)-\psi \left({g}_{0}\right)$ .

In the same way, we can get

$\psi \left({g}_{i+1}\right)-\psi \left({g}_{i}\right)=\psi \left({g}_{i}\right)-\psi \left({g}_{i-1}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}i=0,1,\cdots ,n-1.$

Hence

$\begin{array}{c}\psi \left(g\right)-\psi \left(h\right)=\psi \left({g}_{n}\right)-\psi \left({g}_{0}\right)\\ =\psi \left({g}_{n}\right)-\psi \left({g}_{n-1}\right)+\psi \left({g}_{n-1}\right)-\psi \left({g}_{n-2}\right)+\cdots +\psi \left({g}_{1}\right)-\psi \left({g}_{0}\right)\\ =n\left(\psi \left({g}_{1}\right)-\psi \left({g}_{0}\right)\right)\end{array}$

Therefore

$\begin{array}{c}▼\text{\hspace{0.17em}}={‖\psi \left(f\right)-\psi \left(h\right),n\left(\psi \left({g}_{1}\right)-\psi \left({g}_{0}\right)\right)‖}_{\beta }\\ =n{‖\psi \left(f\right)-\psi \left(h\right),\psi \left({g}_{1}\right)-\psi \left({g}_{0}\right)‖}_{\beta }=n\end{array}$

$\square$

Theorem 2.8. Let $\Im \left(X\right)$ and $\Im \left(Y\right)$ be two FNA-2. If a mapping $\psi :\Im \left(X\right)\to \Im \left(Y\right)$ satisfies AOPP and $\left(\ast \right)$ for all $f,g,h\in \Im \left(X\right)$ with $▲\text{\hspace{0.17em}}\le 1$, then $\psi$ is 2-isometry.

Proof: Since lemma 2.4, we just need to prove that $\left(\nabla \right)$ with $▲\text{\hspace{0.17em}}>1$ .

We can assume that when $▲\text{\hspace{0.17em}}>1$, for all $f,g,h\in \Im \left(X\right)$, we have $▼\text{\hspace{0.17em}}<{n}_{0}+1$ . and there exist a number ${n}_{0}\in {ℕ}^{\ast }$ such that

Let $\tau =f+\frac{{n}_{0}+1}{{‖f-h,g-h‖}_{\alpha }}\left(f-h\right)$, then

${‖\tau -f,g-h‖}_{\alpha }={n}_{0}+1$

and

${‖\tau -h,g-h‖}_{\alpha }={n}_{0}+1-\text{\hspace{0.17em}}▲$

Since $\psi$ preserves collinear, there exist a number $c\in ℝ$ such that

$\psi \left(\tau \right)-\psi \left(f\right)=c\left(\psi \left(h\right)-\psi \left(f\right)\right)$

Since 2),

$\begin{array}{c}{n}_{0}+1={‖\psi \left(\tau \right)-\psi \left(f\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }\\ =|c|▼\\ \le |c-1|▼\text{\hspace{0.17em}}+\text{\hspace{0.17em}}▼\\ ={‖\psi \left(\tau \right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }+\text{\hspace{0.17em}}▼\\ <{n}_{0}+1-\text{\hspace{0.17em}}▲\text{\hspace{0.17em}}+\text{\hspace{0.17em}}▲\text{\hspace{0.17em}}={n}_{0}+1\end{array}$

$▼\text{\hspace{0.17em}}\ge {n}_{0}+1$

Therefore, we get $\left(\nabla \right)$ with $▲\text{\hspace{0.17em}}>1$ . Hence

${‖\psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)‖}_{\beta }={‖f-h,g-h‖}_{\alpha }$

for all $f,g,h\in \Im \left(X\right)$ . $\square$

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

Cite this paper

Song, M.M. and Jin, H.X. (2019) The Aleksandrov Problem in Non-Archimedean 2-Fuzzy 2-Normed Spaces. Journal of Applied Mathematics and Physics, 7, 1775-1785. https://doi.org/10.4236/jamp.2019.78121

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