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 $X\mathrm{,}Y$ be two metric spaces. For a mapping $f\mathrm{<mo>\; :\; </mo>}X\to Y$, for all $x_1\mathrm{,}{x}_2\in X$, 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 $d_X\left(\cdot \mathrm{,}\cdot \right)\mathrm{,}{d}_Y\left(\cdot \mathrm{,}\cdot \right)$ denote the metrics in the spaces $X\mathrm{,}Y$, 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 $x_1\mathrm{,}{x}_2$ 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. ( [1] ) Let $X\mathrm{,}Y$ be two real normed linear spaces (or NLS) with $\dim X>1$, $\dim Y>1$ and Y is strictly convex, assume that a fixed real number $p>0$ and that a fixed integer $N>1$ . Finally, if $f\mathrm{<mo>\; :\; </mo>}X\to Y$ is a mapping satisfies

1) $\Vert x_1-x_2\Vert =p\Rightarrow \Vert f\left(x_1\right)-f\left(x_2\right)\Vert \le p$

2) $\Vert x_1-x_2\Vert =N\cdot p\Rightarrow \Vert f\left(x_1\right)-f\left(x_2\right)\Vert \ge N\cdot p$

for all $x_1\mathrm{,}{x}_2\in X$ . 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 [2] - [10] . 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 [11] [12] [13] [14] , 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 $\left[\mathrm{0,}\infty \right)$ satisfies

(T₁) $\left|a\right|\ge 0,\left|a\right|=0\iff a=0$ ;

(T₂) $\left|ab\right|=\left|a\right|\left|b\right|$ ;

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

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

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

In 1897, Hence in [15] 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 $\Vert \cdot \mathrm{,}\cdot \Vert \mathrm{<mo>\; :\; </mo>}X\to \left[\mathrm{0,}\infty \right)$ is called non-Archimedean 2-norm, if and only if it satisfies

(T₁) $\Vert x_1,x_2\Vert \ge 0$, $\Vert x_1,x_2\Vert =0$ iff $x_1\mathrm{,}{x}_2$ are linearly dependent;

(T₂) $\Vert x_1\mathrm{,}{x}_2\Vert =\Vert x_2\mathrm{,}{x}_1\Vert$ ;

(T₃) $\Vert r{x}_1,x_2\Vert =\left|r\right|\Vert x_1,x_2\Vert$ ;

(T₄) $\Vert x_1+x_2,y\Vert \le \max \left\{\Vert x_1,y\Vert ,\Vert x_2,y\Vert \right\}$

for all $x_1\mathrm{,}{x}_2\mathrm{,}y\in X\mathrm{,}r\in K$ . Then $\left(X\mathrm{,}\Vert \cdot \mathrm{,}\cdot \Vert \right)$ is called non-Archimedean 2-normed space over the field K.

Definition 1.3. An NAV $\left|\cdot \right|$ in a linear space X over a field K. A function $F\mathrm{<mo>\; :\; </mo>}X\times ℝ\to \left[\mathrm{0,1}\right]$ is said to be a non-Archimedean fuzzy norm on X, if and only if for all $x\mathrm{,}{x}_1\mathrm{,}{x}_2\in X$ and $s\mathrm{,}t\in ℝ$,

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

(F4) $F\left(x_1+x_2,s+t\right)\ge \min \left\{F\left(x_1,s\right),F\left(x_2,t\right)\right\}$,

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

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

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

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

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

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

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

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

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

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

If $c=0$, then

${\Vert cx\Vert }_{\alpha }=0=c{\Vert x\Vert }_{\alpha }$

3) We have

$\begin{array}{l}\max \left\{{\Vert x\Vert }_{\alpha },{\Vert y\Vert }_{\alpha }\right\}\\ =\max \left\{\inf \left\{s,F\left(x,s\right)\ge \alpha \right\},\inf \left\{t,F\left(x,t\right)\ge \alpha \right\}\right\}\\ =\inf \left\{\max \left\{s,t\right\},F\left(x,s\right)\ge \alpha ,F\left(x,t\right)\ge \alpha \right\}\\ \ge \inf \{s+t,F\left(x+y,s+t\right)\ge F\left(x+y,\max \left\{s,t\right\}\right)\\ \ge \min \left\{F\left(x,s\right)\ge \alpha ,F\left(x,t\right)\ge \alpha \right\}\ge \alpha \}\\ \ge \inf \left\{r,F\left(x+y,r\right)\ge \alpha \right\}={\Vert x+y\Vert }_{\alpha }\end{array}$

$\square$

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

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

for all $x\in X$, Then $\left(X\mathrm{,}F\right)$ 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 $Z_1\mathrm{,}{Z}_2\in \Im \left(Z\right)$ 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

$\lambda Z_1=\left\{\left(\lambda z_1\mathrm{,}{\mu }_1\right)\mathrm{<mo>\; |\; </mo>}\left(z_1\mathrm{,}{\mu }_1\right)\in Z_1\right\}$

Definition 1.7. A non-Archimedean fuzzy linear space $\hat{X}=X\times \left(0,1\right]$ 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 $\left(x_1\mathrm{,}{\mu }_1\right)\in X$, we have a related non-negative real numebr, $\Vert \left(x_1,{\mu }_1\right)\Vert$ is the fuzzy norm of $\left(x_1\mathrm{,}{\mu }_1\right)$ in such that

(T₁) $\Vert \left(x_1,{\mu }_1\right)\Vert =0\iff x_1=0,{\mu }_1\in \left(0,1\right]$ ;

(T₂) $\Vert \lambda \left(x_1,{\mu }_1\right)\Vert =\left|\lambda \right|\Vert \left(x_1,{\mu }_1\right)\Vert$ ;

(T₃) $\Vert \left(x_1,{\mu }_1\right)+\left(x_2,{\mu }_2\right)\Vert \le \max \left\{\Vert \left(x_1,{\mu }_1\wedge {\mu }_2\right),\left(x_2,{\mu }_1\wedge {\mu }_2\right)\Vert \right\}$ ;

(T₄) $\Vert \left(x_1\mathrm{,}{{\displaystyle \vee }}_t{\mu }_t\right)\Vert ={{\displaystyle \wedge }}_t\Vert \left(x_1\mathrm{,}{\mu }_t\right)\Vert$ for all ${\mu }_t\in \left(\mathrm{0,1}\right]$ .

for every $\left(x_1\mathrm{,}{\mu }_1\right)\mathrm{,}\left(x_2\mathrm{,}{\mu }_2\right)\in X\mathrm{,}\lambda \in K$, 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, $\left|f_1\left(x_1\right)\right|\le 1$, so $f_1$ is a bounded function. Let $K\in \mathbb{Q}$, 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

$\lambda f_1=\left\{\left(\lambda x_1\mathrm{,}{\mu }_1\right)\mathrm{<mo>\; |\; </mo>}\left(x_1\mathrm{,}{\mu }_1\right)\in f_1\right\}$

If for every $f\in \Im \left(X\right)$, there is a related non-negative real number $\Vert f\Vert$ called the norm of f in such that for all $f_1=\left(x_1,{\mu }_1\right),f_2=\left(x_2,{\mu }_2\right)\in \Im (X)$

(T₁) $\Vert f\Vert =0$ iff $f=0$ . For

$\Vert f\Vert =\left\{\Vert \left(x_1,{\mu }_1\right)\Vert \right\}=0$

$\iff x_1=0,{\mu }_1\in \left(0,1\right]$

$\iff f=0.$

(T₂) $\Vert \lambda f\Vert =\left|\lambda \right|\Vert f\Vert ,\lambda \in K$ . For

$\Vert \lambda f\Vert =\left\{\Vert \lambda \left(x_1,{\mu }_1\right)\Vert \right\}=\left\{\left|\lambda \right|\Vert \left(x_1,{\mu }_1\right)\Vert \right\}=\left|\lambda \right|\Vert f\Vert$

(T₃) $\Vert f_1+f_2\Vert \le \max \left\{\Vert f_1\Vert ,\Vert f_2\Vert \right\}$ . For

$\begin{array}{c}\Vert f_1+f_2\Vert =\left\{\Vert \left(x_1,{\mu }_1\right)+\left(x_2,{\mu }_2\right)\Vert \right\}\\ =\left\{\Vert \left(x_1+x_2\right),\left({\mu }_1\wedge {\mu }_2\right)\Vert \right\}\\ \le \max \left\{\Vert \left(x_1,{\mu }_1\wedge {\mu }_2\right)\Vert ,\Vert \left(x_2,{\mu }_1\wedge {\mu }_2\right)\Vert \right\}\\ \le \max \left\{\Vert f_1\Vert ,\Vert f_2\Vert \right\}\end{array}$

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

Definition 1.9. ( [4] ) A 2-fuzzy set on X is a fuzzy set on $\Im \left(X\right)$ .

Definition 1.10. A NAV $\left|\cdot \mathrm{,}\cdot \right|$ in a linear space $\Im \left(X\right)$ over a field K. If a function $F\mathrm{<mo>\; :\; </mo>}\Im {\left(X\right)}^2\times ℝ\to \left[\mathrm{0,1}\right]$ is a non-Archimedean 2-fuzzy 2-norm on X (or a fuzzy 2-norm on $\Im \left(X\right)$ ), iff for all $f_1\mathrm{,}{f}_2\mathrm{,}{f}_3\in \Im \left(X\right)$, $s\mathrm{,}t\in ℝ$,

(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 $f_1\mathrm{,}{f}_2$ 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) $F\left(c{f}_1,f_2,s\right)=N\left(f_1,f_2,\frac{s}{\left|c\right|}\right)$, for $c\ne 0$ and $c\in K$ ;

(F5) $F\left(f_1,f_2+f_3,s+t\right)\ge \min \left\{F\left(f_1,f_2,s\right),F\left(f_1,f_3,t\right)\right\}$ ;

(F6) $F\left(f_1\mathrm{,}{f}_2\mathrm{,}\ast \right)$ is a nondecreasing function of R and ${\lim }_{s\to \infty }F\left(f_1,f_2,s\right)=1$ ;

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

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

(F7) $N\left(f_1,f_2,s\right)>0$ for all $s>0$ $\Rightarrow f_1$ and $f_2$ are linearly dependent.

Define ${\Vert f_1,f_2\Vert }_{\alpha }=\inf \left\{t:N\left(f_1,f_2,s\right)\ge \alpha ,\alpha \in \left(0,1\right)\right\}$ . 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 $\dim \Im \left(X\right)\ge 2$, $\dim \Im \left(Y\right)\ge 2$ . $▲={\Vert f-h,g-h\Vert }_{\alpha }$, $▼={\Vert \psi \left(f\right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)\Vert }_{\beta }$

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

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

then $\psi$ is called 2-isometry.

Definition 2.2. For a mapping $\psi \mathrm{<mo>\; :\; </mo>}\Im \left(X\right)\to \Im \left(Y\right)$ and $f\mathrm{,}g\mathrm{,}h\in \Im (X)$

1) If $▲=1$, then $▼=1$, we say $\psi$ satisfies the area one preserving property (AOPP).

2) If $▲=n$, then $▼=n$, we say $\psi$ satisfies the area n for each n (AnPP).

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

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

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

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

Proof: 1) Firstly, we prove that f preserves collinear. We assume that $▲=0$, according to $(\ast )$, we get

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

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 $▲\le 1$, we can get $(\nabla )$ .

If

$▼<▲$

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

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

Since

${\Vert \omega -f,g-h\Vert }_{\alpha }=\Vert \frac{f-h}{\Vert f-h,g-h\Vert }-\left(f-h\right),g-h\Vert =1-▲$

according to $(\ast )$, we have

${\Vert \psi \left(\omega \right)-\psi \left(f\right),\psi \left(g\right)-\psi \left(h\right)\Vert }_{\beta }\le {\Vert \omega -f,g-h\Vert }_{\alpha }=1-▲$

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

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

and

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

So, we get

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

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 $\psi \mathrm{<mo>\; :\; </mo>}\Im \left(X\right)\to \Im \left(Y\right)$ 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 $f\mathrm{,}g\in \Im \left(X\right)$, since dim $\Im \left(X\right)\ge 2$, there exists an element $h\in \Im \left(X\right)$ such that $f-h\mathrm{,}g-h$ are linearly independent. Hence $▲\ne 0$ .

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

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

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

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

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

Since dim $\Im \left(X\right)\ge 2$, there exist an element $f_1\in \Im \left(X\right)$ such that ${\Vert g-f\mathrm{,}{f}_1-f\Vert }_{\alpha }\ne 0$ . Let $\eta =f+\frac{f_1-f}{{\Vert g-f,f_1-f\Vert }_{\alpha }}$, then ${\Vert g-f,\eta -f\Vert }_{\alpha }=1$ and

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

So,

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

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

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

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

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

So $\varphi$ is additive.

From the lemma 2.4, we know that if $▲\le 1$, then $\varphi$ satisfies 2-isometry.

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

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 ${\Vert f,g\Vert }_{\alpha }=\lambda$,

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

and

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

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

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

so $\left|\lambda -1\right|=\left(\lambda +1\right)$ . 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 $▲\le 1$, a mapping $\psi \mathrm{<mo>\; :\; </mo>}\Im \left(X\right)\to \Im \left(Y\right)$ satisfies $(\ast )$ and AOPP, then we can get for all $f\mathrm{,}g\mathrm{,}h\in \Im \left(X\right)$, we can get $(\nabla )$ .

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

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

So,

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

and

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

By lemma 2.5, we have

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

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

Thus

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

$\square$

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

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

$▲=n,g_i=h+\frac{i}{n}\left(g-h\right)$

and

${\Vert f-h,g_{i+1}-g_i\Vert }_{\alpha }=1,i=0,1,\cdots ,n-1.$

So,

${\Vert \psi \left(f\right)-\psi \left(h\right),\psi \left(g_{i+1}\right)-\psi \left(g_i\right)\Vert }_{\beta }=1,i=0,1,\cdots ,n-1.$

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 $t=\pm 1$ . 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

$\psi \left(g_{i+1}\right)-\psi \left(g_i\right)=\psi \left(g_i\right)-\psi \left(g_{i-1}\right),i=0,1,\cdots ,n-1.$

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

$\begin{array}{c}▼={\Vert \psi \left(f\right)-\psi \left(h\right),n\left(\psi \left(g_1\right)-\psi \left(g_0\right)\right)\Vert }_{\beta }\\ =n{\Vert \psi \left(f\right)-\psi \left(h\right),\psi \left(g_1\right)-\psi \left(g_0\right)\Vert }_{\beta }=n\end{array}$

$\square$

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

Proof: Since lemma 2.4, we just need to prove that $(\nabla )$ with $▲>1$ .

We can assume that when $▲>1$, for all $f\mathrm{,}g\mathrm{,}h\in \Im \left(X\right)$, we have $▼<n_0+1$ . and there exist a number $n_0\in {\mathbb{N}}^{\ast }$ such that

Let $\tau =f+\frac{n_0+1}{{\Vert f-h,g-h\Vert }_{\alpha }}\left(f-h\right)$, then

${\Vert \tau -f,g-h\Vert }_{\alpha }=n_0+1$

and

${\Vert \tau -h,g-h\Vert }_{\alpha }=n_0+1-▲$

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

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

Since 2),

$\begin{array}{c}{n}_0+1={\Vert \psi \left(\tau \right)-\psi \left(f\right),\psi \left(g\right)-\psi \left(h\right)\Vert }_{\beta }\\ =\left|c\right|▼\\ \le \left|c-1\right|▼+▼\\ ={\Vert \psi \left(\tau \right)-\psi \left(h\right),\psi \left(g\right)-\psi \left(h\right)\Vert }_{\beta }+▼\\ <n_0+1-▲+▲=n_0+1\end{array}$

which is contradiction, so

$▼\ge n_0+1$

Therefore, we get $(\nabla )$ with $▲>1$ . Hence

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

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

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

References