Wigner’s Theorem in s* and s_n(H) Spaces

Meimei Song, Shunxin Zhao^*

Department of Mathematics, Tianjin University of Technology, Tianjin, China.

DOI: 10.4236/ajcm.2018.83017                

Abstract

Wigner theorem is the cornerstone of the mathematical formula of quan-tum mechanics, it has promoted the research of basic theory of quantum mechanics. In this article, we give a certain pair of functional equations between two real spaces s or two real s_n(H), that we called “phase isometry”. It is obtained that all such solutions are phase equivalent to real linear isometries in the space s and the space s_n(H).

Keywords

s Space, Wigner’s Theorem, Phase Equivalent, Linear Isometry, s_n(H) Space

1. Introduction

Mazur and Ulam in [1] proved that every surjective isometry U between X and Y is a affine, also states that the mapping with $U\left(0\right)=0$ , then U is linear. Let X and Y be normed spaces, if the mapping $V\mathrm{<mo>\; :\; </mo>}X\to Y$ satisfying that

$\left\{\Vert V\left(x\right)-V\left(y\right)\Vert \right\}=\left\{\Vert x-y\Vert \right\}\left(x\mathrm{,}y\in X\right)\mathrm{.}$

It was called isometry. About it’s main properties in sequences spaces, Tingley, D, Ding Guanggui, Fu Xiaohong in [2] [3] [4] [5] [6] proved. So, we give a new definition that if there is a function $\epsilon \mathrm{<mo>\; :\; </mo>}X\to \left\{-\mathrm{1,1}\right\}$ such that $J=\epsilon V$ is a linear isometry. we can say the mapping $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is phase equivalent to J.

If the two spaces are Hilbert spaces, Rätz proved that the phase isometries $V\mathrm{<mo>\; :\; </mo>}X\to Y$ are precisely the solutions of functional equation in [7] . If the two spaces are not inner product spaces, Huang and Tan [8] gave a partial answer about the real atomic $L_p$ spaces with $p>0$ . Jia and Tan [9] get the conclusion about the $\mathcal{L}$ -type spaces. In [6] , xiaohong Fu proved the problem of isometry extension in the s space detailedly.

In this artical, we mainly discuss that all mappings $V\mathrm{<mo>\; :\; </mo>}s\to s$ or $s_n\left(H\right)\to s_n\left(H\right)$ also have the properties, that are solutions of the functional equation

$\left\{\Vert V\left(x\right)-V\left(y\right)\Vert \mathrm{,}\Vert V\left(x\right)+V\left(y\right)\Vert \right\}=\left\{\Vert x-y\Vert \mathrm{,}\Vert x+y\Vert \right\}\left(x\mathrm{,}y\in X\right)\mathrm{.}$ (1)

All metric spaces mentioned in this artical are assumed to be real.

2. Results about s

First, let us introduction some concepts. The s space in [10] , which consists of all scalar sequences and for each elements $x=\left\{{\xi }_k\right\}={\displaystyle \underset{k}{\sum }}{\xi }_k{e}_k$ , the F-norm of x is defined by $\Vert x\Vert ={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^k}\frac{\left|{\xi }_k\right|}{1+\left|{\xi }_k\right|}$ . Let $s_{\left(n\right)}$ denote the set of all elements of the form $x=\left\{{\xi }_1,\cdot \cdot \cdot ,{\xi }_n\right\}$ with $\Vert x\Vert ={\displaystyle \underset{k=1}{\overset{n}{\sum }}}\frac{1}{2^k}\frac{\left|{\xi }_k\right|}{1+\left|{\xi }_k\right|}$ . where $e_k=\left\{{\xi }_{k^{\prime }}:{\xi }_k=1,{\xi }_{k^{\prime }}=0,k^{\prime }\ne k,\text{for}\text{all}{k}^{\prime }\in \Gamma \right\}$ . We denote the support of x by ${\Gamma }_x$ , i.e.,

$supp\left(x\right)={\Gamma }_x=\left\{\gamma \in \Gamma \mathrm{<mo>\; :\; </mo>}{\xi }_{\gamma }\ne 0\right\}\mathrm{.}$

For all $x\mathrm{,}y\in s$ , if ${\Gamma }_x\cap {\Gamma }_y=\varnothing$ , we say that x is orthogonal to y and write $x\perp y$ .

Lemma 2.1. Let $S_{r_0}\left(s\right)$ be a sphere with radius $r_0$ and center 0 in s. Suppose that $V_0\mathrm{<mo>\; :\; </mo>}{S}_{r_0}\left(s\right)\to S_{r_0}\left(s\right)$ is a mapping satisfying Equation (1). Then for any $x\mathrm{,}y\in S_{r_0}\left(s\right)$ , we have

$x\perp y\iff V_0\left(x\right)\perp V_0(y)$

Proof: Necessity. Choosing $\forall x=\left\{{\xi }_n\right\}$ , $y=\left\{{\eta }_n\right\}\in S_{r_0}\left(s\right)$ that satisfying $x\perp y$ . We can suppose $V_0\left(x\right)=\left\{{{\xi }^{\prime }}_n\right\}$ , $V_0\left(y\right)=\left\{{{\eta }^{\prime }}_n\right\}$ . And we also have

$\left\{\Vert V_0\left(x\right)-V_0\left(y\right)\Vert ,\Vert V_0\left(x\right)+V_0\left(y\right)\Vert \right\}=\left\{\Vert x-y\Vert ,\Vert x+y\Vert \right\}$ .

So

$\Vert V_0\left(x\right)-V_0\left(y\right)\Vert =\Vert x-y\Vert =\Vert x\Vert +\Vert y\Vert =2r_0=\Vert V_0\left(x\right)\Vert +\Vert V_0\left(y\right)\Vert$

or

$\Vert V_0\left(x\right)-V_0\left(y\right)\Vert =\Vert x+y\Vert =\Vert x\Vert +\Vert y\Vert =2r_0=\Vert V_0\left(x\right)\Vert +\Vert V_0\left(y\right)\Vert$

Thus

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\frac{\left|{{\xi }^{\prime }}_n-{{\eta }^{\prime }}_n\right|}{1+\left|{{\xi }^{\prime }}_n-{{\eta }^{\prime }}_n\right|}={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\frac{\left|{{\xi }^{\prime }}_n\right|}{1+\left|{{\xi }^{\prime }}_n\right|}+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\frac{\left|{{\eta }^{\prime }}_n\right|}{1+\left|{{\eta }^{\prime }}_n\right|}$

That means

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\left[\frac{\left|{{\xi }^{\prime }}_n-{{\eta }^{\prime }}_n\right|}{1+\left|{{\xi }^{\prime }}_n-{{\eta }^{\prime }}_n\right|}-\frac{\left|{{\xi }^{\prime }}_n\right|}{1+\left|{{\xi }^{\prime }}_n\right|}-\frac{\left|{{\eta }^{\prime }}_n\right|}{1+\left|{{\eta }^{\prime }}_n\right|}\right]=0$ (2)

It is easy to know $f\left(x\right)=\frac{x}{1+x}$ is strictly increasing. And $\left|{{\xi }^{\prime }}_n-{{\eta }^{\prime }}_n\right|\le \left|{{\xi }^{\prime }}_n\right|+\left|{{\eta }^{\prime }}_n\right|$ . We can get the result ${{\xi }^{\prime }}_n\cdot {{\eta }^{\prime }}_n=0$ .

For $\Vert V_0\left(x\right)+V_0\left(y\right)\Vert$ , similarty to the above $\left(\left|{{\xi }^{\prime }}_n+{{\eta }^{\prime }}_n\right|\le \left|{{\xi }^{\prime }}_n\right|+\left|{{\eta }^{\prime }}_n\right|\right)$ . It is $V_0\left(x\right)\perp V_0\left(y\right)$ . Sufficiency. For $V_0\left(x\right)\perp V_0\left(y\right)$ , that is, ${{\xi }^{\prime }}_n\cdot {{\eta }^{\prime }}_n=0$ , so (2) holds, and we have

$\Vert x-y\Vert =\Vert V_0\left(x\right)-V_0\left(y\right)\Vert =\Vert V_0\left(x\right)\Vert +\Vert V_0\left(y\right)\Vert =2r_0$

so, it must have $\Vert x-y\Vert =\Vert x\Vert +\Vert y\Vert$ .

or

$\Vert x-y\Vert =\Vert V_0\left(x\right)+V_0\left(y\right)\Vert =\Vert V_0\left(x\right)\Vert +\Vert V_0\left(y\right)\Vert =2r_0$

as the same $\Vert x-y\Vert =\Vert x\Vert +\Vert y\Vert$ . It follows that

${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\frac{\left|{\xi }_n-{\eta }_n\right|}{1+\left|{\xi }_n-{\eta }_n\right|}={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\frac{\left|{\xi }_n\right|}{1+\left|{\xi }_n\right|}+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{1}{2^n}\frac{\left|{\eta }_n\right|}{1+\left|{\eta }_n\right|}$ (3)

Similarly to the proof of necessity, we get $x\perp y$ .

Lemma 2.2. Let $S_{r_0}\left(s_{\left(n\right)}\right)$ be a sphere with radius $r_0$ in the finite dimensional space $s_{\left(n\right)}$ , where $r_0<\frac{1}{2^n}$ . Suppose that $V_0:S_{r_0}\left(s_{\left(n\right)}\right)\to S_{r_0}\left(s_{\left(n\right)}\right)$ is an phase isometry. Let ${\lambda }_k=\frac{2^k{r}_0}{1-2^k{r}_0}\left(k\in \mathbb{N},1\le k\le n\right)$ , then there is a unique real $\theta$ with $\left|\theta \right|=1$ , such that $V_0\left({\lambda }_k{e}_k\right)=\theta {\lambda }_k{e}_k$ .

Proof: We proof first that for any $k\left(1\le k\le n\right)$ , there is a unique $l\left(1\le l\le n\right)$ and a unique real $\theta$ with $\left|\theta \right|=1$ such that $V_0\left({\lambda }_k{e}_k\right)=\theta {\lambda }_l{e}_l$ (because the assumption of ${\lambda }_k$ implies ${\lambda }_k{e}_k\in S_{r_0}\left(s_{\left(n\right)}\right)$ ). To this end, suppose on the contrary that $V_0\left({\lambda }_{k_0}{e}_{k_0}\right)={\displaystyle \underset{k=1}{\overset{n}{\sum }}}{\eta }_k{e}_k$ and ${\eta }_{k_1}\ne \mathrm{0,}{\eta }_{k_2}\ne 0$ . In view of Lemma 1, we have

$\left[supp{V}_0\left({\lambda }_{k_0}{e}_{k_0}\right)\right]\cap \left[supp{V}_0\left({\lambda }_k{e}_k\right)\right]=\varnothing \forall k\ne k_0\mathrm{,1}\le k\le n$ .

Hence, by the “pigeon nest principle” (or Pigeonhole principle) there must exist $k_{i_0}\left(1\le k_{i_0}\le n\right)$ such that $V_0\left({\lambda }_{k_{i0}}{e}_{k_{i0}}\right)=\theta$ , which leads to a contradiction.

Next, if $V_0\left({\lambda }_k{e}_k\right)={\theta }_1{\lambda }_l{e}_l$ , $V_0\left(-{\lambda }_k{e}_k\right)={\theta }_2{\lambda }_p{e}_p$ , where $\left|{\theta }_1\right|=\left|{\theta }_2\right|=1$ , then $l=p$ and ${\theta }_2=-{\theta }_1$ . Indeed, if $l\ne p$ , we have

$\Vert V\left({\lambda }_k{e}_k\right)-V\left(-{\lambda }_k{e}_k\right)\Vert =\Vert 2{\lambda }_k{e}_k\Vert =\frac{1}{2^k}\frac{\left|2{\lambda }_k\right|}{1+\left|2{\lambda }_k\right|}\ne 2r_0$

or

$\Vert V\left({\lambda }_k{e}_k\right)-V\left(-{\lambda }_k{e}_k\right)\Vert =0$

and

$\Vert V\left({\lambda }_k{e}_k\right)-V\left(-{\lambda }_k{e}_k\right)\Vert =\Vert {\theta }_1{\lambda }_l{e}_l-{\theta }_2{\lambda }_p{e}_p\Vert =2r_0$ (4)

a contradiction which implies $l=p$ . From this ${\theta }_1=-{\theta }_2$ follows. Finally, there is a unique $\theta$ with $\left|\theta \right|=1$ such that $V_0\left({\lambda }_k{e}_k\right)=\theta {\lambda }_k{e}_k$ . Indeed, if $V_0\left({\lambda }_k{e}_k\right)=\theta {\lambda }_l{e}_l$ , by the result in the last step, we have $V_0\left(-{\lambda }_k{e}_k\right)=-\theta {\lambda }_l{e}_l$ , thus

$\begin{array}{l}\left\{\Vert V\left({\lambda }_k{e}_k\right)+V\left(-{\lambda }_k{e}_k\right)\Vert ,\Vert V\left({\lambda }_k{e}_k\right)-V\left(-{\lambda }_k{e}_k\right)\Vert \right\}\\ =\left\{\Vert 2{\lambda }_k{e}_k\Vert ,0\right\}=\left\{\frac{1}{2^k}\frac{\left|2{\lambda }_k\right|}{1+\left|2{\lambda }_k\right|},0\right\}\end{array}$

and

$\begin{array}{l}\left\{\Vert V\left({\lambda }_k{e}_k\right)+V\left(-{\lambda }_k{e}_k\right)\Vert ,\Vert V\left({\lambda }_k{e}_k\right)-V\left(-{\lambda }_k{e}_k\right)\Vert \right\}\\ =\left\{\Vert 2\theta {\lambda }_l{e}_l\Vert ,0\right\}=\left\{\frac{1}{2^l}\frac{\left|2{\lambda }_l\right|}{1+\left|2{\lambda }_l\right|},0\right\}\end{array}$ (5)

So, we get

$\frac{1}{2^k}\frac{\left|2{\lambda }_k\right|}{1+\left|2{\lambda }_k\right|}=\frac{1}{2^l}\frac{\left|2{\lambda }_l\right|}{1+\left|2{\lambda }_l\right|}$

and we also have

$\frac{1}{2^k}\frac{\left|{\lambda }_k\right|}{1+\left|{\lambda }_k\right|}=\frac{1}{2^l}\frac{\left|{\lambda }_l\right|}{1+\left|{\lambda }_l\right|},\left(=r_0\right)$

through the two equalities of above

$\frac{\frac{1}{2^k}\frac{\left|2{\lambda }_k\right|}{1+\left|2{\lambda }_k\right|}}{\frac{1}{2^k}\frac{\left|{\lambda }_k\right|}{1+\left|{\lambda }_k\right|}}=\frac{\frac{1}{2^l}\frac{\left|2{\lambda }_l\right|}{1+\left|2{\lambda }_l\right|}}{\frac{1}{2^l}\frac{\left|{\lambda }_l\right|}{1+\left|{\lambda }_l\right|}}$

In the end,

$\left|{\lambda }_l\right|=\left|{\lambda }_k\right|$ (6)

The proof is complete.

Lemma 2.3. Let $X=S_{r_0}\left(s_{\left(n\right)}\right)$ and $Y=S_{r_0}\left(s_{\left(n\right)}\right)$ . Suppose that $V_0\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1) and ${\lambda }_k$ as in Lemma 2.2. Then for any lement $x={\displaystyle \underset{k}{\sum }}{\xi }_k{e}_k\in X$ , we have $V_0\left(x\right)={\displaystyle \underset{k}{\sum }}{\eta }_k{e}_k$ , where $\left|{\xi }_k\right|=\left|{\eta }_k\right|$ for any $1\le k_0\le n$ .

Proof: Note that the defination of $V_0$ , we can easily get $V_0\left(0\right)=0$ . For any $0\ne x\in X$ , write $x={\displaystyle \underset{k}{\sum }}{\xi }_k{e}_k$ , where ${\displaystyle \underset{k}{\sum }}\frac{1}{2^k}\frac{\left|{\xi }_k\right|}{1+\left|{\xi }_k\right|}=r_0$ . we can write $V_0\left(x\right)={\displaystyle \underset{k}{\sum }}{\eta }_k{e}_k$ , where ${\displaystyle \underset{k}{\sum }}\frac{1}{2^k}\frac{\left|{\eta }_k\right|}{1+\left|{\eta }_k\right|}=r_0$ . we have

$\begin{array}{l}\Vert V_0\left(x\right)+V_0\left({\lambda }_{k_0}{e}_{k_0}\right)\Vert +\Vert V_0\left(x\right)-V_0\left({\lambda }_{k_0}{e}_{k_0}\right)\Vert \\ =\Vert x+{\lambda }_{k_0}{e}_{k_0}\Vert +\Vert x-{\lambda }_{k_0}{e}_{k_0}\Vert \\ =\Vert {\displaystyle \underset{k\ne k_0}{\sum }}{\xi }_k{e}_k+\left({\xi }_{k_0}+{\lambda }_{k_0}\right)e_{k_0}\Vert +\Vert {\displaystyle \underset{k\ne k_0}{\sum }}{\xi }_k{e}_k+\left({\xi }_{k_0}-{\lambda }_{k_0}\right)e_{k_0}\Vert \\ =r_0+\frac{1}{2^{k_0}}\frac{\left|{\xi }_{k_0}+{\lambda }_{k_0}\right|}{1+{\xi }_{k_0}+{\lambda }_{k_0}}-\frac{1}{2^{k_0}}\frac{\left|{\xi }_{k_0}\right|}{1+\left|{\xi }_{k_0}\right|}+r_0+\frac{1}{2^{k_0}}\frac{\left|{\xi }_{k_0}-{\lambda }_{k_0}\right|}{1+{\xi }_{k_0}-{\lambda }_{k_0}}-\frac{1}{2^{k_0}}\frac{\left|{\xi }_{k_0}\right|}{1+\left|{\xi }_{k_0}\right|}.\end{array}$

On the other hand, we have

$\begin{array}{l}\Vert V_0\left(x\right)+V_0\left({\lambda }_{k_0}{e}_{k_0}\right)\Vert +\Vert V_0\left(x\right)-V_0\left({\lambda }_{k_0}{e}_{k_0}\right)\Vert \\ =\Vert {\displaystyle \underset{k=1}{\overset{n}{\sum }}}{\eta }_k{e}_k+{\theta }_{k_0}{\lambda }_{k_0}{e}_{k_0}\Vert +\Vert {\displaystyle \underset{k=1}{\overset{n}{\sum }}}{\eta }_k{e}_k-{\theta }_{k_0}{\lambda }_{k_0}{e}_{k_0}\Vert \\ =\Vert {\displaystyle \underset{k\ne k_0}{\sum }}{\eta }_k{e}_k+\left({\eta }_{k_0}+{\theta }_{k_0}{\lambda }_{k_0}\right)e_{k_0}\Vert +\Vert {\displaystyle \underset{k\ne k_0}{\sum }}{\eta }_k{e}_k+\left({\eta }_{k_0}-{\theta }_{k_0}{\lambda }_{k_0}\right)e_{k_0}\Vert \\ =r_0+\frac{1}{2^{k_0}}\frac{\left|{\eta }_{k_0}+{\theta }_{k_0}{\lambda }_{k_0}\right|}{1+\left|{\eta }_{k_0}+{\theta }_{k_0}{\lambda }_{k_0}\right|}-\frac{1}{2^{k_0}}\frac{\left|{\eta }_{k_0}\right|}{1+\left|{\eta }_{k_0}\right|}+r_0+\frac{1}{2^{k_0}}\frac{\left|{\eta }_{k_0}-{\theta }_{k_0}{\lambda }_{k_0}\right|}{1+\left|{\eta }_{k_0}-{\theta }_{k_0}{\lambda }_{k_0}\right|}-\frac{1}{2^{k_0}}\frac{\left|{\eta }_{k_0}\right|}{1+\left|{\eta }_{k_0}\right|}.\end{array}$

Combiniing the two equations, we obtain that

$\begin{array}{l}\frac{\left|{\xi }_{k_0}+{\lambda }_{k_0}\right|}{1+\left|{\xi }_{k_0}+{\lambda }_{k_0}\right|}-\frac{\left|2{\xi }_{k_0}\right|}{1+\left|{\xi }_{k_0}\right|}+\frac{\left|{\xi }_{k_0}-{\lambda }_{k_0}\right|}{1+\left|{\xi }_{k_0}-{\lambda }_{k_0}\right|}\\ =\frac{\left|{\eta }_{k_0}+{\theta }_{k_0}{\lambda }_{k_0}\right|}{1+\left|{\eta }_{k_0}+{\theta }_{k_0}{\lambda }_{k_0}\right|}-\frac{\left|2{\eta }_{k_0}\right|}{1+\left|{\eta }_{k_0}\right|}+\frac{\left|{\eta }_{k_0}-{\theta }_{k_0}{\lambda }_{k_0}\right|}{1+\left|{\eta }_{k_0}-{\theta }_{k_0}{\lambda }_{k_0}\right|}\end{array}$

As ${\lambda }_{k_0}\ge \left|{\xi }_{k_0}\right|$ and ${\lambda }_{k_0}\ge \left|{\eta }_{k_0}\right|$ , we have

$\begin{array}{l}\frac{{\xi }_{k_0}+{\lambda }_{k_0}}{1+{\xi }_{k_0}+{\lambda }_{k_0}}-\frac{2{\xi }_{k_0}}{1+{\xi }_{k_0}}+\frac{{\lambda }_{k_0}-{\xi }_{k_0}}{1+{\lambda }_{k_0}-{\xi }_{k_0}}\\ =\frac{{\lambda }_{k_0}+{\theta }_{k_0}{\eta }_{k_0}}{1+{\lambda }_{k_0}+{\theta }_{k_0}{\eta }_{k_0}}-\frac{2{\eta }_{k_0}}{1+{\eta }_{k_0}}+\frac{{\lambda }_{k_0}-{\theta }_{k_0}{\eta }_{k_0}}{1+{\lambda }_{k_0}-{\theta }_{k_0}{\eta }_{k_0}}\end{array}$

Therefore,

$\frac{{\lambda }_{k_0}+{\lambda }_{k_0}^2-{\xi }_{k_0}^2}{{\left(1+{\lambda }_{k_0}\right)}^2-{\xi }_{k_0}^2}+\frac{{\eta }_{k_0}}{1+{\eta }_{k_0}}-\frac{{\xi }_{k_0}}{1+{\xi }_{k_0}}=\frac{{\lambda }_{k_0}+{\lambda }_{k_0}^2-{\eta }_{k_0}^2}{{\left(1+{\lambda }_{k_0}\right)}^2-{\eta }_{k_0}^2}$

Analysis of the equation, according to the monotony of the function, that is

$\left|{\xi }_k\right|=\left|{\eta }_k\right|$ (7)

The proof is complete.,

The next result shows that a mapping satisfying functional Equation (1) has a property close to linearity.

Lemma 2.4. Let $X=s_{\left(n\right)}$ and $Y=s_{\left(n\right)}$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). there exist two real numbers $\alpha$ and $\beta$ with absolute 1 such that

$V\left(x+y\right)=\alpha V\left(x\right)+\beta V(y)$

for all nonzero vectors x and y in X, x and y are orthogonal.

Proof: Let x and y be nonzero orthogonal vectors in X, we write $x={\displaystyle \underset{k}{\sum }}{\xi }_k{e}_k$ , $y={\displaystyle \underset{k}{\sum }}{\eta }_k{e}_k$ .

$V\left(x\right)={\displaystyle \underset{k}{\sum }}{{\xi }^{\prime }}_k{e}_k$ , $V\left(y\right)={\displaystyle \underset{k}{\sum }}{{\eta }^{\prime }}_k{e}_k$

$V\left(x+y\right)={\displaystyle \underset{k}{\sum }}{{\xi }^{″}}_k{e}_k+{\displaystyle \underset{k}{\sum }}{{\eta }^{″}}_k{e}_k$ ,

where $\left|{{\xi }^{\prime }}_k\right|=\left|{{\xi }^{″}}_k\right|=\left|{\xi }_k\right|$ and $\left|{{\eta }^{\prime }}_k\right|=\left|{{\eta }^{″}}_k\right|=\left|{\eta }_k\right|$ . We infer from Equation (1) that

$\begin{array}{l}\left\{\Vert 2x\Vert +\Vert y\Vert ,\Vert y\Vert \right\}\\ =\left\{\Vert V\left(x+y\right)+V\left(x\right)\Vert ,\Vert V\left(x+y\right)-V\left(x\right)\Vert \right\}\\ =\left\{\Vert {\displaystyle \underset{k}{\sum }}{{\xi }^{″}}_k{e}_k+{\displaystyle \underset{k}{\sum }}{{\eta }^{″}}_k{e}_k+{\displaystyle \underset{k}{\sum }}{{\xi }^{\prime }}_k{e}_k\Vert ,\Vert {\displaystyle \underset{k}{\sum }}{{\xi }^{″}}_k{e}_k+{\displaystyle \underset{k}{\sum }}{{\eta }^{″}}_k{e}_k+{\displaystyle \underset{k}{\sum }}{{\eta }^{\prime }}_k{e}_k\Vert \right\}\\ =\left\{\frac{1}{2^k}\frac{\left|{{\xi }^{″}}_k+{{\xi }^{\prime }}_k\right|}{1+\left|{{\xi }^{″}}_k+{{\xi }^{\prime }}_k\right|}+\Vert y\Vert ,\frac{1}{2^k}\frac{\left|{{\xi }^{″}}_k-{{\xi }^{\prime }}_k\right|}{1+\left|{{\xi }^{″}}_k-{{\xi }^{\prime }}_k\right|}+\Vert y\Vert \right\}\end{array}$

Through the above equation we can get ${{\xi }^{″}}_k+{{\xi }^{\prime }}_k=0$ or ${{\xi }^{″}}_k-{{\xi }^{\prime }}_k=0$ . This implies that ${\displaystyle \underset{k}{\sum }}{{\xi }^{″}}_k{e}_k=\pm V\left(x\right)$ , and similarly ${\displaystyle \underset{k}{\sum }}{{\eta }^{″}}_k{e}_k=\pm V\left(y\right)$ . The proof is complete.,

Lemma 2.5. Let $X=s$ and $Y=s$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). Then V is injective and $V\left(-x\right)=-V\left(x\right)$ for all $x\in X$ .

Proof: Suppose that V is surjective and $V\left(x\right)=V\left(y\right)$ for some $x\mathrm{,}y\in X$ . Putting $y=x$ in the Equation (1), this yields

$\left\{\Vert 2V\left(x\right)\Vert \mathrm{,0}\right\}=\left\{\Vert 2x\Vert \mathrm{,0}\right\}$

$V\left(x\right)=0$ if and only if $x=0$ . Assume that $V\left(x\right)=V\left(y\right)\ne 0$ choose $z\in X$ such that $V\left(z\right)=-V\left(x\right)$ , using the Equation (1) for $x\mathrm{,}y\mathrm{,}z$ , we obtain

$\left\{\Vert x-y\Vert ,\Vert x+y\Vert \right\}=\left\{\Vert V\left(x\right)+V\left(y\right)\Vert ,\Vert V\left(x\right)-V\left(y\right)\Vert \right\}=\left\{\Vert 2V\left(x\right)\Vert \mathrm{,0}\right\}$

$\left\{\Vert x-z\Vert ,\Vert x+z\Vert \right\}=\left\{\Vert V\left(x\right)+V\left(z\right)\Vert ,\Vert V\left(x\right)-V\left(z\right)\Vert \right\}=\left\{\Vert 2V\left(x\right)\Vert \mathrm{,0}\right\}$

This yields $y\mathrm{,}z\in \left\{x\mathrm{,}-x\right\}$ . If $z=x$ , then $V\left(x\right)=-V\left(x\right)=0$ , which is a contradiction. So we obtain $z=-x$ , and we must have $y=x$ . For otherwise we get $y=z=-x$ and

$V\left(x\right)=V\left(y\right)=V\left(z\right)=-V\left(x\right)=0$

This lead to the contradiction that $V\left(x\right)\ne 0$ .

Theorem 2.6. Let $X=s_{\left(n\right)}$ and $Y=s_{\left(n\right)}$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.

Proof: Fix ${\gamma }_0\in \Gamma$ , and let $Z=\left\{z\in X:z\perp e_{{\gamma }_0}\right\}$ . By Lemma 2.4 we can write

$V\left(z+\lambda e_{{\gamma }_0}\right)=\alpha \left(z,\lambda \right)V\left(z\right)+\beta \left(z,\lambda \right)V\left(\lambda e_{{\gamma }_0}\right),\left|\alpha \left(z,\lambda \right)\right|=\left|\beta \left(z,\lambda \right)\right|=1$

for any $z\in Z$ . Then, we can define a mapping $J\mathrm{<mo>\; :\; </mo>}{s}_{\left(n\right)}\to s_{\left(n\right)}$ as follows:

$J\left(z+\lambda e_{{\gamma }_0}\right)=\alpha \left(z\mathrm{,}\lambda \right)\beta \left(z\mathrm{,}\lambda \right)V\left(z\right)+V\left(\lambda e_{{\gamma }_0}\right)$

$J\left(\lambda z\right)=\alpha \left(z\mathrm{,}\lambda \right)\beta \left(z\mathrm{,}\lambda \right)V(\lambda z)$

$J\left(e_{{\gamma }_0}\right)=V\left(e_{{\gamma }_0}\right)$ , $J\left(-e_{{\gamma }_0}\right)=-V\left(e_{{\gamma }_0}\right)$

for $\forall 0\ne \lambda \in ℝ$ . The J is phase equivalent to V. So it is easily to know that J satisfies functional Equation (1). For any $z\in Z$ , and $\forall 0\ne \lambda \in ℝ$ ,

$\begin{array}{l}\left\{\Vert 2z\Vert +\frac{1}{2^{{\gamma }_0}}\frac{\left|1+\lambda \right|}{1+\left|1+\lambda \right|},\frac{1}{2^{{\gamma }_0}}\frac{\left|1-\lambda \right|}{1+\left|1-\lambda \right|}\right\}\\ =\left\{\Vert J\left(z+e_{{\gamma }_0}\right)+J\left(z+\lambda e_{{\gamma }_0}\right)\Vert ,\Vert J\left(z+e_{{\gamma }_0}\right)-J\left(z+\lambda e_{{\gamma }_0}\right)\Vert \right\}\\ =\{\Vert \alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(e_{{\gamma }_0}\right)+V\left(\lambda e_{{\gamma }_0}\right)\Vert ,\\ \Vert \alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(e_{{\gamma }_0}\right)-V\left(\lambda e_{{\gamma }_0}\right)\Vert \}\\ =\{\Vert \left|\alpha \left(z,1\right)\beta \left(z,1\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)\right|V\left(z\right)\Vert +\frac{1}{2^{{\gamma }_0}}\frac{\left|1+\lambda \right|}{1+\left|1+\lambda \right|},\\ \Vert \left|\alpha \left(z,1\right)\beta \left(z,1\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)\right|V\left(z\right)\Vert +\frac{1}{2^{{\gamma }_0}}\frac{\left|1+\lambda \right|}{1+\left|1+\lambda \right|}\}\end{array}$

That means $\alpha \left(z,1\right)\beta \left(z,1\right)=\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)$ ,

$J\left(z+\lambda e_{{\gamma }_0}\right)=J\left(z\right)+V\left(\lambda e_{{\gamma }_0}\right)$ for any $z\in Z$ , and $\forall 0\ne \lambda \in ℝ$ .

That yields

$\begin{array}{l}\left\{\Vert J\left(z\right)+J\left(-z\right)\Vert ,\Vert J\left(z\right)-J\left(-z\right)+2V\left(e_{{\gamma }_0}\right)\Vert \right\}\\ =\left\{\Vert J\left(z+e_{{\gamma }_0}\right)+J\left(-z-e_{{\gamma }_0}\right)\Vert ,\Vert J\left(z+e_{{\gamma }_0}\right)-J\left(-z-e_{{\gamma }_0}\right)\Vert \right\}\\ =\left\{0,\Vert 2\left(z+e_{{\gamma }_0}\right)\Vert \right\}\end{array}$

That means $J\left(-z\right)=-J\left(z\right)$ . On the other hand,

$\begin{array}{l}\left\{\Vert z_1+z_2\Vert +\frac{2}{3}\frac{1}{2^{{\gamma }_0}},\Vert z_1-z_2\Vert \right\}\\ =\left\{\Vert J\left(z_1+e_{{\gamma }_0}\right)+J\left(z_2+e_{{\gamma }_0}\right)\Vert ,\Vert J\left(z_1+e_{{\gamma }_0}\right)-J\left(z_2+e_{{\gamma }_0}\right)\Vert \right\}\\ =\left\{\Vert J\left(z_1\right)+J\left(z_2\right)\Vert +\frac{2}{3}\frac{1}{2^{{\gamma }_0}},\Vert J\left(z_1\right)-J\left(z_2\right)\Vert \right\}\end{array}$

for $\forall z_1\mathrm{,}{z}_2\in Z$ , It follows that $\Vert J\left(x\right)-J\left(y\right)\Vert =\Vert x-y\Vert$ for all $x\mathrm{,}y\in X$ , by assumed conditions, so J is a surjective isometry.,

Theorem 2.7. Let $X=s$ and $Y=s$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.

Proof: According to [10] Theorem 1, Theorem 2 the author presents some results of extension from some spheres in the finite dimensional spaces $s_{\left(n\right)}$ . And also we have the above Theorem 2.6, so we can get the result easily.

3. Results about $s_n(H)$

In this part, we mainly introduce the space $s_n\left(H\right)$ , where H is a Hilbert space. In [11] mainly discussed the isometric extension in the space $s_n\left(H\right)$ . For each element $x=\left\{x\left(k\right)\right\}$ , the F-norm of x is defined by $\Vert x\Vert ={\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}\frac{1}{2^k}\frac{\Vert x\left(k\right)\Vert }{1+\Vert x\left(k\right)\Vert }$ . Let $s_n\left(H\right)$ denote the set of all elements of the form $x=\left(x\left(1\right),\cdot \cdot \cdot ,x\left(n\right)\right)$ with $\Vert x\Vert ={\displaystyle \underset{k=1}{\overset{n}{\sum }}}\frac{1}{2^k}\frac{\Vert x\left(k\right)\Vert }{1+\Vert x\left(k\right)\Vert }$ . where $x\left(i\right)\left(i=1,\cdot \cdot \cdot ,n\right)\in H$ .

Some notations used:

$e_{x\left(k\right)}=\left(\mathrm{0,}\cdot \cdot \cdot \mathrm{,}x\left(k\right)\mathrm{,}\cdot \cdot \cdot \mathrm{,0}\right)\in s_n\left(H\right)$ , where $\Vert x\left(k\right)\Vert =1$ .

Specially, when $\Vert x\left(k\right)\Vert =0$ , we have $e_{\frac{x\left(k\right)}{\Vert x\left(k\right)\Vert }}=\left(0,\cdot \cdot \cdot ,0\right)$ .

Next, we study the phase isometry between the space $s_n\left(H\right)$ to $s_n\left(H\right)$ , that if V is a surjective phase isometry, then V is phase equivalent to a linear isometry J.

Lemma 3.1. If $x\mathrm{,}y\in s_n\left(H\right)$ , then

$\Vert x-y\Vert =\Vert x\Vert +\Vert y\Vert$ if and only if $suppx\cap suppy=\varnothing$

where $suppx=\left\{n\mathrm{<mo>\; :\; </mo>}x\left(n\right)\ne \mathrm{0,}n\in \mathbb{N}\right\}$ .

Proof: It has a detailed proof process in [11] .

Lemma 3.2. Let $S_{r_0}\left(s_n\left(H\right)\right)$ be a sphere with radius $r_0$ in the finite dimensional space $s_n\left(H\right)$ , where $r_0<\frac{1}{2^n}$ . Defined $V_0\mathrm{<mo>\; :\; </mo>}{S}_{r_0}\left(s_n\left(H\right)\right)\to S_{r_0}\left(s_n\left(H\right)\right)$ is an phase isometry, then we can get

$x\perp y\iff V_0\left(x\right)\perp V_0\left(y\right)$ .

Proof: “Þ” Take any two elements $x=\left\{x\left(i\right)\right\}$ , $y=\left\{y\left(i\right)\right\}$ , let $V_0\left(x\right)=\left\{x^{\prime }\left(i\right)\right\}$ , $V_0\left(y\right)=\left\{y^{\prime }\left(i\right)\right\}$ . Then we have

$2r_0=\Vert x\Vert +\Vert y\Vert =\Vert x-y\Vert =\Vert V_0\left(x\right)-V_0\left(y\right)\Vert ={\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert x^{\prime }\left(i\right)-y^{\prime }\left(i\right)\Vert }{1+\Vert x^{\prime }\left(i\right)-y^{\prime }\left(i\right)\Vert }$

or

$2r_0=\Vert x\Vert +\Vert y\Vert =\Vert x-y\Vert =\Vert V_0\left(x\right)+V_0\left(y\right)\Vert ={\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert x^{\prime }\left(i\right)-y^{\prime }\left(i\right)\Vert }{1+\Vert x^{\prime }\left(i\right)-y^{\prime }\left(i\right)\Vert }$ (8)

at the same time, we have

${\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert x^{\prime }\left(i\right)-y^{\prime }\left(i\right)\Vert }{1+\Vert x^{\prime }\left(i\right)-y^{\prime }\left(i\right)\Vert }\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert x^{\prime }\left(i\right)\Vert }{1+\Vert x^{\prime }\left(i\right)\Vert }+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert y^{\prime }\left(i\right)\Vert }{1+\Vert y^{\prime }\left(i\right)\Vert }=2r_0$

${\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert x^{\prime }\left(i\right)+y^{\prime }\left(i\right)\Vert }{1+\Vert x^{\prime }\left(i\right)+y^{\prime }\left(i\right)\Vert }\le {\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert x^{\prime }\left(i\right)\Vert }{1+\Vert x^{\prime }\left(i\right)\Vert }+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{1}{2^i}\frac{\Vert y^{\prime }\left(i\right)\Vert }{1+\Vert y^{\prime }\left(i\right)\Vert }=2r_0$ (9)

That means $\Vert V_0\left(x\right)-V_0\left(y\right)\Vert =\Vert V_0\left(x\right)+V_0\left(y\right)\Vert =\Vert V_0\left(x\right)\Vert +\Vert +V_0\left(y\right)\Vert$ , it is $V_0\left(x\right)\perp V_0\left(y\right)$ . “Ü” The proof of sufficiency is similar to the Lemma 2.1.

Lemma 3.3. Let $V_0$ be as in Lemma 3.2, ${\lambda }_k=\frac{2^k{r}_0}{1-2^k{r}_0}\left(k\in \mathbb{N}\right),\left(1\le k\le n\right)$ , and $e_{x\left(k\right)}=s_n\left(H\right)$ . $\left(\Vert x\left(k\right)\Vert =1\right)$ . Then there exists $x^{\prime }\left(k\right)\in H\left(\Vert x^{\prime }\left(k\right)\Vert =1\right)$ , such that $V_0\left(\pm {\lambda }_k{e}_{x\left(k\right)}\right)=\pm {\lambda }_k{e}_{x^{\prime }\left(k\right)}$ .

Proof: We prove first that, for any $k\left(1\le k\le n\right)$ , there exist $l\left(1\le l\le n\right)$ and $x^{\prime }\left(l\right)\left(\Vert x^{\prime }\left(l\right)\Vert =1\right)$ such that $V_0\left({\lambda }_k{e}_{x\left(k\right)}\right)={\lambda }_l{e}_{x^{\prime }\left(k\right)}$ . And then prove $l=p$ . It is the same an Lemma 2.2.

Finally, we assert that, there exists $x^{\prime }\left(k\right)$ such that $V_0\left(\pm {\lambda }_k{e}_{x\left(k\right)}\right)=\pm {\lambda }_k{e}_{x^{\prime }\left(k\right)}$ . Indeed, if $V_0\left({\lambda }_k{e}_{x\left(k\right)}\right)={\lambda }_l{e}_{x^{\prime }\left(l\right)}$ , by the result in the last step, we have $V_0\left(-{\lambda }_k{e}_{x\left(k\right)}\right)={\lambda }_l{e}_{x^{″}\left(l\right)}$ ,

$\begin{array}{l}\left\{0,\frac{1}{2^k}\frac{2{\lambda }_k}{1+2{\lambda }_k}\right\}\\ =\left\{\Vert V_0\left({\lambda }_k{e}_{x\left(k\right)}\right)-V_0\left(-{\lambda }_k{e}_{x\left(k\right)}\right)\Vert ,\Vert V_0\left({\lambda }_k{e}_{x\left(k\right)}\right)+V_0\left(-{\lambda }_k{e}_{x\left(k\right)}\right)\Vert \right\}\\ =\left\{\Vert {\lambda }_l{e}_{x^{\prime }\left(l\right)}-{\lambda }_l{e}_{x^{″}\left(l\right)}\Vert ,\Vert {\lambda }_l{e}_{x^{\prime }\left(l\right)}-{\lambda }_l{e}_{x^{″}\left(l\right)}\Vert \right\}\\ =\left\{\frac{1}{2^l}\frac{{\lambda }_l\Vert x^{\prime }\left(l\right)-x^{″}\left(l\right)\Vert }{1+{\lambda }_l\Vert x^{\prime }\left(l\right)-x^{″}\left(l\right)\Vert },\frac{1}{2^l}\frac{{\lambda }_l\Vert x^{\prime }\left(l\right)+x^{″}\left(l\right)\Vert }{1+{\lambda }_l\Vert x^{\prime }\left(l\right)+x^{″}\left(l\right)\Vert }\right\}\end{array}$

Therefore,

$\frac{1}{2^k}\frac{2{\lambda }_k}{1+2{\lambda }_k}=\frac{1}{2^l}\frac{{\lambda }_l\Vert x^{\prime }\left(l\right)-x^{″}\left(l\right)\Vert }{1+{\lambda }_l\Vert x^{\prime }\left(l\right)-x^{″}\left(l\right)\Vert }\le \frac{1}{2^l}\frac{2{\lambda }_l}{1+2{\lambda }_l}$

or

$\frac{1}{2^k}\frac{2{\lambda }_k}{1+2{\lambda }_k}=\frac{1}{2^l}\frac{{\lambda }_l\Vert x^{\prime }\left(l\right)+x^{″}\left(l\right)\Vert }{1+{\lambda }_l\Vert x^{\prime }\left(l\right)+x^{″}\left(l\right)\Vert }\le \frac{1}{2^l}\frac{2{\lambda }_l}{1+2{\lambda }_l}$ (10)

So, we can get $k=l$ . And $\Vert x^{\prime }\left(l\right)-x^{″}\left(l\right)\Vert =\Vert x^{\prime }\left(l\right)+x^{″}\left(l\right)\Vert =2$ , that means $x^{\prime }\left(l\right)=\pm x^{″}\left(l\right)$ .

Lemma 3.4. Let $X=s_n\left(H\right)$ and $Y=s_n\left(H\right)$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). there exist two real numbers $\alpha$ and $\beta$ with absolute 1 such that

$V\left(x+y\right)=\alpha V\left(x\right)+\beta V(y)$

for all nonzero vectors x and y in X, x and y are orthogonal. Proof: Let $x=\left\{x\left(i\right)\right\}$ and $y=\left\{y\left(i\right)\right\}$ be nonzero orthogonal vectors in X.

$V\left\{x\left(i\right)\right\}={\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x\left(i\right)}{\Vert x\left(i\right)\Vert }}\right)$ ,

$V\left\{y\left(i\right)\right\}={\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert y\left(i\right)\Vert }{{\mu }_i}V\left({\mu }_i{e}_{\frac{y\left(i\right)}{\Vert y\left(i\right)\Vert }}\right)$

$V\left\{x\left(i\right)+y\left(i\right)\right\}={\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x^{\prime }\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x^{\prime }\left(i\right)}{\Vert x^{\prime }\left(i\right)\Vert }}\right)+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert y^{\prime }\left(i\right)\Vert }{{\mu }_i}V\left({\mu }_i{e}_{\frac{y^{\prime }\left(i\right)}{\Vert y^{\prime }\left(i\right)\Vert }}\right)$ ,

where $\Vert x^{\prime }\left(i\right)\Vert =\Vert x\left(i\right)\Vert$ and $\Vert y^{\prime }\left(i\right)\Vert =\Vert y\left(i\right)\Vert$ . We infer from Equation (1) that

$\begin{array}{l}\left\{\Vert 2x\Vert +\Vert y\Vert ,\Vert y\Vert \right\}\\ =\left\{\Vert V\left\{x\left(i\right)+y\left(i\right)\right\}+V\left\{x\left(i\right)\right\}\Vert ,\Vert V\left\{x\left(i\right)+y\left(i\right)\right\}+V\left\{y\left(i\right)\right\}\Vert \right\}\\ =\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x^{\prime }\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x^{\prime }\left(i\right)}{\Vert x^{\prime }\left(i\right)\Vert }}\right)+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert y^{\prime }\left(i\right)\Vert }{{\mu }_i}V\left({\mu }_i{e}_{\frac{y^{\prime }\left(i\right)}{\Vert y^{\prime }\left(i\right)\Vert }}\right)+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x\left(i\right)}{\Vert x\left(i\right)\Vert }}\right),\\ {\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x^{\prime }\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x^{\prime }\left(i\right)}{\Vert x^{\prime }\left(i\right)\Vert }}\right)+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert y^{\prime }\left(i\right)\Vert }{{\mu }_i}V\left({\mu }_i{e}_{\frac{y^{\prime }\left(i\right)}{\Vert y^{\prime }\left(i\right)\Vert }}\right)+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert y\left(i\right)\Vert }{{\mu }_i}V\left({\mu }_i{e}_{\frac{y\left(i\right)}{\Vert y\left(i\right)\Vert }}\right)\}\\ =\{{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x^{\prime }\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x^{\prime }\left(i\right)}{\Vert x^{\prime }\left(i\right)\Vert }}\right)+{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x\left(i\right)}{\Vert x\left(i\right)\Vert }}\right)+\Vert \left\{y\left(i\right)\right\}\Vert ,\\ {\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x^{\prime }\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x^{\prime }\left(i\right)}{\Vert x^{\prime }\left(i\right)\Vert }}\right)-{\displaystyle \underset{i=1}{\overset{n}{\sum }}}\frac{\Vert x\left(i\right)\Vert }{{\lambda }_i}V\left({\lambda }_i{e}_{\frac{x\left(i\right)}{\Vert x\left(i\right)\Vert }}\right)+\Vert \left\{y\left(i\right)\right\}\Vert \}\end{array}$

Through the above equation we can get $\Vert x^{\prime }\left(i\right)\Vert =\Vert x\left(i\right)\Vert$ or $\Vert x^{\prime }\left(i\right)\Vert =-\Vert x\left(i\right)\Vert$ . The proof is complete.,

Lemma 3.5. Let $X=s_n\left(H\right)$ and $Y=s_n\left(H\right)$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). Then V is injective and $V\left(-x\right)=-V\left(x\right)$ for all $x\in X$ .

Proof: Suppose that V is surjective and $V\left(x\right)=V\left(y\right)$ for some $x\mathrm{,}y\in X$ . Putting $y=x$ in the Equation (1), this yields

$\left\{\Vert 2V\left(x\right)\Vert ,0\right\}=\left\{\Vert 2x\Vert ,0\right\}$

$V\left(x\right)=0$ if and only if $x=0$ . Assume that $V\left(x\right)=V\left(y\right)\ne 0$ choose $z\in X$ such that $V\left(z\right)=-V\left(x\right)$ , using the Equation (1) for $x\mathrm{,}y\mathrm{,}z$ , we obtain

$\left\{\Vert x+y\Vert ,\Vert x-y\Vert \right\}=\left\{\Vert V\left(x\right)+V\left(y\right)\Vert ,\Vert V\left(x\right)-V\left(y\right)\Vert \right\}=\left\{\Vert 2V\left(x\right)\Vert ,0\right\}$

$\left\{\Vert x+z\Vert ,\Vert x-z\Vert \right\}=\left\{\Vert V\left(x\right)+V\left(z\right)\Vert ,\Vert V\left(x\right)-V\left(z\right)\Vert \right\}=\left\{\Vert 2V\left(x\right)\Vert ,0\right\}$

This yields $y\mathrm{,}z\in \left\{x\mathrm{,}-x\right\}$ . If $z=x$ , then $V\left(x\right)=-V\left(x\right)=0$ , which is a contradiction. So we obtain $z=-x$ , and we must have $y=x$ . For otherwise we get $y=z=-x$ and

$V\left(x\right)=V\left(y\right)=V\left(z\right)=-V\left(x\right)=0$

This lead to the contradiction that $V\left(x\right)\ne 0$ .

Theorem 3.6. Let $X=s_n\left(H\right)$ and $Y=s_n\left(H\right)$ . Suppose that $V\mathrm{<mo>\; :\; </mo>}X\to Y$ is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.

Proof: Fix ${\gamma }_0\in \Gamma$ , and let $Z=\left\{z\in X:z\perp e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right\}$ . By Lemma 3.4 we can write

$V\left(z+\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)=\alpha \left(z,\lambda \right)V(z)+\beta \left(z,\lambda \right)V\left(\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right),\left|\alpha \left(z,\lambda \right)\right|=\left|\beta \left(z,\lambda \right)\right|=1$

for any $z\in Z$ . Then, we can define a mapping $J\mathrm{<mo>\; :\; </mo>}{s}_n\left(H\right)\to s_n\left(H\right)$ as follows:

$J\left(z+\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)=\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)$

$J\left(\lambda z\right)=\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V(\lambda z)$

$J\left(e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)=V\left(e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)$ , $J\left(-e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)=-V\left(e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)$

for $\forall 0\ne \lambda \in ℝ$ . The J is phase equivalent to V. So it is easily to know that J satisfies functional Equation (1). For any $z\in Z$ , and $\forall 0\ne \lambda \in ℝ$ ,

$\begin{array}{l}\left\{\Vert 2z\Vert +\frac{1}{2^{{\gamma }_0}}\frac{\left|1+\lambda \right|}{1+\left|1+\lambda \right|},\frac{1}{2^{{\gamma }_0}}\frac{\left|1-\lambda \right|}{1+\left|1-\lambda \right|}\right\}\\ =\left\{\Vert J\left(z+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)+J\left(z+\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert ,\Vert J\left(z+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)-J\left(z+\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert \right\}\\ =\{\Vert \alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)+V\left(\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert ,\\ \Vert \alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)-V\left(\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert \}\\ =\{\Vert \left|\alpha \left(z,1\right)\beta \left(z,1\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)\right|V\left(z\right)\Vert +\frac{1}{2^{{\gamma }_0}}\frac{\left|1+\lambda \right|}{1+\left|1+\lambda \right|},\\ \Vert \left|\alpha \left(z,1\right)\beta \left(z,1\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)\right|V\left(z\right)\Vert +\frac{1}{2^{{\gamma }_0}}\frac{\left|1+\lambda \right|}{1+\left|1+\lambda \right|}\}\end{array}$

That means $\alpha \left(z,1\right)\beta \left(z,1\right)=\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)$ , $J\left(z+\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)=J\left(z\right)+V\left(\lambda e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)$ for any $z\in Z$ , and $\forall 0\ne \lambda \in ℝ$ .

That yields

$\begin{array}{l}\left\{\Vert J\left(z\right)+J\left(-z\right)\Vert ,\Vert J\left(z\right)-J\left(-z\right)+2V\left(e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert \right\}\\ =\left\{\Vert J\left(z+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)+J\left(-z-e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert ,\Vert J\left(z+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)-J\left(-z-e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert \right\}\\ =\left\{0,\Vert 2\left(z+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert \right\}\end{array}$

That means $J\left(-z\right)=-J\left(z\right)$ . On the other hand,

$\begin{array}{l}\left\{\Vert z_1+z_2\Vert +\frac{2}{3}\frac{1}{2^{{\gamma }_0}},\Vert z_1-z_2\Vert \right\}\\ =\left\{\Vert J\left(z_1+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)+J\left(z_2+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert ,\Vert J\left(z_1+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)-J\left(z_2+e_{\frac{x\left({\gamma }_0\right)}{\Vert {\gamma }_0\Vert }}\right)\Vert \right\}\\ =\left\{\Vert J\left(z_1\right)+J\left(z_2\right)\Vert +\frac{2}{3}\frac{1}{2^{{\gamma }_0}},\Vert J\left(z_1\right)-J\left(z_2\right)\Vert \right\}\end{array}$

for $\forall z_1\mathrm{,}{z}_2\in Z$ , It follows that $\Vert J\left(x\right)-J\left(y\right)\Vert =\Vert x-y\Vert$ for all $x\mathrm{,}y\in X$ , by assumed conditions, so J is a surjective isometry.,

4. Conclusion

Through the analysis of this article, we can get the conclusion that if a surjective mapping satisfying phase-isometry, then it can phase equivalent to a linear isometry in the space s and the space $s\left(H\right)$ .

Acknowledgements

The author wish to express his appreciation to Professor Meimei Song for several valuable comments.

Conflicts of Interest

The authors declare no conflicts of interest.

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