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

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 sn(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 sn(H).

Keywords

Share and Cite:

Song, M. and Zhao, S. (2018) Wigner’s Theorem in s* and sn(H) Spaces. American Journal of Computational Mathematics, 8, 209-221. doi: 10.4236/ajcm.2018.83017.

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 $V:X\to Y$ satisfying that

$\left\{‖V\left(x\right)-V\left(y\right)‖\right\}=\left\{‖x-y‖\right\}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x,y\in X\right).$

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 $\epsilon :X\to \left\{-1,1\right\}$ such that $J=\epsilon V$ is a linear isometry. we can say the mapping $V:X\to Y$ is phase equivalent to J.

If the two spaces are Hilbert spaces, Rätz proved that the phase isometries $V:X\to Y$ 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 $V:s\to s$ or ${s}_{n}\left(H\right)\to {s}_{n}\left(H\right)$ also have the properties, that are solutions of the functional equation

$\left\{‖V\left(x\right)-V\left(y\right)‖,‖V\left(x\right)+V\left(y\right)‖\right\}=\left\{‖x-y‖,‖x+y‖\right\}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(x,y\in X\right).$ (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 $x=\left\{{\xi }_{k}\right\}=\underset{k}{\sum }\text{ }{\xi }_{k}{e}_{k}$ , the F-norm of x is defined by $‖x‖=\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{|{\xi }_{k}|}{1+|{\xi }_{k}|}$ . 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 $‖x‖=\underset{k=1}{\overset{n}{\sum }}\frac{1}{{2}^{k}}\frac{|{\xi }_{k}|}{1+|{\xi }_{k}|}$ . where ${e}_{k}=\left\{{\xi }_{{k}^{\prime }}:{\xi }_{k}=1,{\xi }_{{k}^{\prime }}=0,{k}^{\prime }\ne k,\text{for}\text{\hspace{0.17em}}\text{all}\text{\hspace{0.17em}}{k}^{\prime }\in \Gamma \right\}$ . We denote the support of x by ${\Gamma }_{x}$ , i.e.,

$supp\left(x\right)={\Gamma }_{x}=\left\{\gamma \in \Gamma :{\xi }_{\gamma }\ne 0\right\}.$

For all $x,y\in s$ , 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 ${V}_{0}:{S}_{{r}_{0}}\left(s\right)\to {S}_{{r}_{0}}\left(s\right)$ is a mapping satisfying Equation (1). Then for any $x,y\in {S}_{{r}_{0}}\left(s\right)$ , we have

$x\perp y⇔{V}_{0}\left(x\right)\perp {V}_{0}\left(y\right)$

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

$\left\{‖{V}_{0}\left(x\right)-{V}_{0}\left(y\right)‖,‖{V}_{0}\left(x\right)+{V}_{0}\left(y\right)‖\right\}=\left\{‖x-y‖,‖x+y‖\right\}$ .

So

$‖{V}_{0}\left(x\right)-{V}_{0}\left(y\right)‖=‖x-y‖=‖x‖+‖y‖=2{r}_{0}=‖{V}_{0}\left(x\right)‖+‖{V}_{0}\left(y\right)‖$

or

$‖{V}_{0}\left(x\right)-{V}_{0}\left(y\right)‖=‖x+y‖=‖x‖+‖y‖=2{r}_{0}=‖{V}_{0}\left(x\right)‖+‖{V}_{0}\left(y\right)‖$

Thus

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

That means

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

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

For $‖{V}_{0}\left(x\right)+{V}_{0}\left(y\right)‖$ , similarty to the above $\left(|{{\xi }^{\prime }}_{n}+{{\eta }^{\prime }}_{n}|\le |{{\xi }^{\prime }}_{n}|+|{{\eta }^{\prime }}_{n}|\right)$ . 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

$‖x-y‖=‖{V}_{0}\left(x\right)-{V}_{0}\left(y\right)‖=‖{V}_{0}\left(x\right)‖+‖{V}_{0}\left(y\right)‖=2{r}_{0}$

so, it must have $‖x-y‖=‖x‖+‖y‖$ .

or

$‖x-y‖=‖{V}_{0}\left(x\right)+{V}_{0}\left(y\right)‖=‖{V}_{0}\left(x\right)‖+‖{V}_{0}\left(y\right)‖=2{r}_{0}$

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

$\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{n}}\frac{|{\xi }_{n}-{\eta }_{n}|}{1+|{\xi }_{n}-{\eta }_{n}|}=\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{n}}\frac{|{\xi }_{n}|}{1+|{\xi }_{n}|}+\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{n}}\frac{|{\eta }_{n}|}{1+|{\eta }_{n}|}$ (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 ${\lambda }_{k}=\frac{{2}^{k}{r}_{0}}{1-{2}^{k}{r}_{0}}\left(k\in ℕ,1\le k\le n\right)$ , then there is a unique real $\theta$ with $|\theta |=1$ , 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 $|\theta |=1$ 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 ${V}_{0}\left({\lambda }_{{k}_{0}}{e}_{{k}_{0}}\right)=\underset{k=1}{\overset{n}{\sum }}\text{ }{\eta }_{k}{e}_{k}$ and ${\eta }_{{k}_{1}}\ne 0,{\eta }_{{k}_{2}}\ne 0$ . In view of Lemma 1, we have

$\left[supp{V}_{0}\left({\lambda }_{{k}_{0}}{e}_{{k}_{0}}\right)\right]\cap \left[supp{V}_{0}\left({\lambda }_{k}{e}_{k}\right)\right]=\varnothing \text{ }\forall k\ne {k}_{0},1\le k\le n$ .

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 $|{\theta }_{1}|=|{\theta }_{2}|=1$ , then $l=p$ and ${\theta }_{2}=-{\theta }_{1}$ . Indeed, if $l\ne p$ , we have

$‖V\left({\lambda }_{k}{e}_{k}\right)-V\left(-{\lambda }_{k}{e}_{k}\right)‖=‖2{\lambda }_{k}{e}_{k}‖=\frac{1}{{2}^{k}}\frac{|2{\lambda }_{k}|}{1+|2{\lambda }_{k}|}\ne 2{r}_{0}$

or

$‖V\left({\lambda }_{k}{e}_{k}\right)-V\left(-{\lambda }_{k}{e}_{k}\right)‖=0$

and

$‖V\left({\lambda }_{k}{e}_{k}\right)-V\left(-{\lambda }_{k}{e}_{k}\right)‖=‖{\theta }_{1}{\lambda }_{l}{e}_{l}-{\theta }_{2}{\lambda }_{p}{e}_{p}‖=2{r}_{0}$ (4)

a contradiction which implies $l=p$ . From this ${\theta }_{1}=-{\theta }_{2}$ follows. Finally, there is a unique $\theta$ with $|\theta |=1$ 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

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

and

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

So, we get

$\frac{1}{{2}^{k}}\frac{|2{\lambda }_{k}|}{1+|2{\lambda }_{k}|}=\frac{1}{{2}^{l}}\frac{|2{\lambda }_{l}|}{1+|2{\lambda }_{l}|}$

and we also have

$\frac{1}{{2}^{k}}\frac{|{\lambda }_{k}|}{1+|{\lambda }_{k}|}=\frac{1}{{2}^{l}}\frac{|{\lambda }_{l}|}{1+|{\lambda }_{l}|},\left(={r}_{0}\right)$

through the two equalities of above

$\frac{\frac{1}{{2}^{k}}\frac{|2{\lambda }_{k}|}{1+|2{\lambda }_{k}|}}{\frac{1}{{2}^{k}}\frac{|{\lambda }_{k}|}{1+|{\lambda }_{k}|}}=\frac{\frac{1}{{2}^{l}}\frac{|2{\lambda }_{l}|}{1+|2{\lambda }_{l}|}}{\frac{1}{{2}^{l}}\frac{|{\lambda }_{l}|}{1+|{\lambda }_{l}|}}$

In the end,

$|{\lambda }_{l}|=|{\lambda }_{k}|$ (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 ${V}_{0}:X\to Y$ is a surjective mapping satisfying Equation (1) and ${\lambda }_{k}$ as in Lemma 2.2. Then for any lement $x=\underset{k}{\sum }\text{ }{\xi }_{k}{e}_{k}\in X$ , we have ${V}_{0}\left(x\right)=\underset{k}{\sum }\text{ }{\eta }_{k}{e}_{k}$ , where $|{\xi }_{k}|=|{\eta }_{k}|$ 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 $x=\underset{k}{\sum }\text{ }{\xi }_{k}{e}_{k}$ , where $\underset{k}{\sum }\frac{1}{{2}^{k}}\frac{|{\xi }_{k}|}{1+|{\xi }_{k}|}={r}_{0}$ . we can write ${V}_{0}\left(x\right)=\underset{k}{\sum }\text{ }{\eta }_{k}{e}_{k}$ , where $\underset{k}{\sum }\frac{1}{{2}^{k}}\frac{|{\eta }_{k}|}{1+|{\eta }_{k}|}={r}_{0}$ . we have

$\begin{array}{l}‖{V}_{0}\left(x\right)+{V}_{0}\left({\lambda }_{{k}_{0}}{e}_{{k}_{0}}\right)‖+‖{V}_{0}\left(x\right)-{V}_{0}\left({\lambda }_{{k}_{0}}{e}_{{k}_{0}}\right)‖\\ =‖x+{\lambda }_{{k}_{0}}{e}_{{k}_{0}}‖+‖x-{\lambda }_{{k}_{0}}{e}_{{k}_{0}}‖\\ =‖\underset{k\ne {k}_{0}}{\sum }{\xi }_{k}{e}_{k}+\left({\xi }_{{k}_{0}}+{\lambda }_{{k}_{0}}\right){e}_{{k}_{0}}‖+‖\underset{k\ne {k}_{0}}{\sum }{\xi }_{k}{e}_{k}+\left({\xi }_{{k}_{0}}-{\lambda }_{{k}_{0}}\right){e}_{{k}_{0}}‖\\ ={r}_{0}+\frac{1}{{2}^{{k}_{0}}}\frac{|{\xi }_{{k}_{0}}+{\lambda }_{{k}_{0}}|}{1+{\xi }_{{k}_{0}}+{\lambda }_{{k}_{0}}}-\frac{1}{{2}^{{k}_{0}}}\frac{|{\xi }_{{k}_{0}}|}{1+|{\xi }_{{k}_{0}}|}+{r}_{0}+\frac{1}{{2}^{{k}_{0}}}\frac{|{\xi }_{{k}_{0}}-{\lambda }_{{k}_{0}}|}{1+{\xi }_{{k}_{0}}-{\lambda }_{{k}_{0}}}-\frac{1}{{2}^{{k}_{0}}}\frac{|{\xi }_{{k}_{0}}|}{1+|{\xi }_{{k}_{0}}|}.\end{array}$

On the other hand, we have

$\begin{array}{l}‖{V}_{0}\left(x\right)+{V}_{0}\left({\lambda }_{{k}_{0}}{e}_{{k}_{0}}\right)‖+‖{V}_{0}\left(x\right)-{V}_{0}\left({\lambda }_{{k}_{0}}{e}_{{k}_{0}}\right)‖\\ =‖\underset{k=1}{\overset{n}{\sum }}\text{ }{\eta }_{k}{e}_{k}+{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}{e}_{{k}_{0}}‖+‖\underset{k=1}{\overset{n}{\sum }}\text{ }{\eta }_{k}{e}_{k}-{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}{e}_{{k}_{0}}‖\\ =‖\underset{k\ne {k}_{0}}{\sum }{\eta }_{k}{e}_{k}+\left({\eta }_{{k}_{0}}+{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}\right){e}_{{k}_{0}}‖+‖\underset{k\ne {k}_{0}}{\sum }{\eta }_{k}{e}_{k}+\left({\eta }_{{k}_{0}}-{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}\right){e}_{{k}_{0}}‖\\ ={r}_{0}+\frac{1}{{2}^{{k}_{0}}}\frac{|{\eta }_{{k}_{0}}+{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}+{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}-\frac{1}{{2}^{{k}_{0}}}\frac{|{\eta }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}|}+{r}_{0}+\frac{1}{{2}^{{k}_{0}}}\frac{|{\eta }_{{k}_{0}}-{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}-{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}-\frac{1}{{2}^{{k}_{0}}}\frac{|{\eta }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}|}.\end{array}$

Combiniing the two equations, we obtain that

$\begin{array}{l}\frac{|{\xi }_{{k}_{0}}+{\lambda }_{{k}_{0}}|}{1+|{\xi }_{{k}_{0}}+{\lambda }_{{k}_{0}}|}-\frac{|2{\xi }_{{k}_{0}}|}{1+|{\xi }_{{k}_{0}}|}+\frac{|{\xi }_{{k}_{0}}-{\lambda }_{{k}_{0}}|}{1+|{\xi }_{{k}_{0}}-{\lambda }_{{k}_{0}}|}\\ =\frac{|{\eta }_{{k}_{0}}+{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}+{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}-\frac{|2{\eta }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}|}+\frac{|{\eta }_{{k}_{0}}-{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}{1+|{\eta }_{{k}_{0}}-{\theta }_{{k}_{0}}{\lambda }_{{k}_{0}}|}\end{array}$

As ${\lambda }_{{k}_{0}}\ge |{\xi }_{{k}_{0}}|$ and ${\lambda }_{{k}_{0}}\ge |{\eta }_{{k}_{0}}|$ , 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

$|{\xi }_{k}|=|{\eta }_{k}|$ (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 $V:X\to Y$ is a surjective mapping satisfying Equation (1). there exist two real numbers $\alpha$ and $\beta$ with absolute 1 such that

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

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 $x=\underset{k}{\sum }\text{ }{\xi }_{k}{e}_{k}$ , $y=\underset{k}{\sum }\text{ }{\eta }_{k}{e}_{k}$ .

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

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

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

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

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

Lemma 2.5. Let $X=s$ and $Y=s$ . Suppose that $V:X\to Y$ 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 $x,y\in X$ . Putting $y=x$ in the Equation (1), this yields

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

$V\left(x\right)=0$ 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 $x,y,z$ , we obtain

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

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

This yields $y,z\in \left\{x,-x\right\}$ . 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 $V:X\to Y$ 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

$V\left(z+\lambda {e}_{{\gamma }_{0}}\right)=\alpha \left(z,\lambda \right)V\left(z\right)+\beta \left(z,\lambda \right)V\left(\lambda {e}_{{\gamma }_{0}}\right),|\alpha \left(z,\lambda \right)|=|\beta \left(z,\lambda \right)|=1$

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

$J\left(z+\lambda {e}_{{\gamma }_{0}}\right)=\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(\lambda {e}_{{\gamma }_{0}}\right)$

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

$J\left({e}_{{\gamma }_{0}}\right)=V\left({e}_{{\gamma }_{0}}\right)$ , $J\left(-{e}_{{\gamma }_{0}}\right)=-V\left({e}_{{\gamma }_{0}}\right)$

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 $\forall 0\ne \lambda \in ℝ$ ,

$\begin{array}{l}\left\{‖2z‖+\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1+\lambda |}{1+|1+\lambda |},\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1-\lambda |}{1+|1-\lambda |}\right\}\\ =\left\{‖J\left(z+{e}_{{\gamma }_{0}}\right)+J\left(z+\lambda {e}_{{\gamma }_{0}}\right)‖,‖J\left(z+{e}_{{\gamma }_{0}}\right)-J\left(z+\lambda {e}_{{\gamma }_{0}}\right)‖\right\}\\ =\left\{‖\alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left({e}_{{\gamma }_{0}}\right)+V\left(\lambda {e}_{{\gamma }_{0}}\right)‖,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}‖\alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left({e}_{{\gamma }_{0}}\right)-V\left(\lambda {e}_{{\gamma }_{0}}\right)‖\right\}\\ =\left\{‖|\alpha \left(z,1\right)\beta \left(z,1\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)|V\left(z\right)‖+\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1+\lambda |}{1+|1+\lambda |},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}‖|\alpha \left(z,1\right)\beta \left(z,1\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)|V\left(z\right)‖+\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1+\lambda |}{1+|1+\lambda |}\right\}\end{array}$

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

$\begin{array}{l}\left\{‖J\left(z\right)+J\left(-z\right)‖,‖J\left(z\right)-J\left(-z\right)+2V\left({e}_{{\gamma }_{0}}\right)‖\right\}\\ =\left\{‖J\left(z+{e}_{{\gamma }_{0}}\right)+J\left(-z-{e}_{{\gamma }_{0}}\right)‖,‖J\left(z+{e}_{{\gamma }_{0}}\right)-J\left(-z-{e}_{{\gamma }_{0}}\right)‖\right\}\\ =\left\{0,‖2\left(z+{e}_{{\gamma }_{0}}\right)‖\right\}\end{array}$

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

$\begin{array}{l}\left\{‖{z}_{1}+{z}_{2}‖+\frac{2}{3}\frac{1}{{2}^{{\gamma }_{0}}},‖{z}_{1}-{z}_{2}‖\right\}\\ =\left\{‖J\left({z}_{1}+{e}_{{\gamma }_{0}}\right)+J\left({z}_{2}+{e}_{{\gamma }_{0}}\right)‖,‖J\left({z}_{1}+{e}_{{\gamma }_{0}}\right)-J\left({z}_{2}+{e}_{{\gamma }_{0}}\right)‖\right\}\\ =\left\{‖J\left({z}_{1}\right)+J\left({z}_{2}\right)‖+\frac{2}{3}\frac{1}{{2}^{{\gamma }_{0}}},‖J\left({z}_{1}\right)-J\left({z}_{2}\right)‖\right\}\end{array}$

for $\forall {z}_{1},{z}_{2}\in Z$ , It follows that $‖J\left(x\right)-J\left(y\right)‖=‖x-y‖$ for all $x,y\in X$ , by assumed conditions, so J is a surjective isometry.,

Theorem 2.7. Let $X=s$ and $Y=s$ . Suppose that $V:X\to Y$ 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 ${s}_{n}\left(H\right)$

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 $‖x‖=\underset{k=1}{\overset{\infty }{\sum }}\frac{1}{{2}^{k}}\frac{‖x\left(k\right)‖}{1+‖x\left(k\right)‖}$ . 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 $‖x‖=\underset{k=1}{\overset{n}{\sum }}\frac{1}{{2}^{k}}\frac{‖x\left(k\right)‖}{1+‖x\left(k\right)‖}$ . where $x\left(i\right)\left(i=1,\cdot \cdot \cdot ,n\right)\in H$ .

Some notations used:

${e}_{x\left(k\right)}=\left(0,\cdot \cdot \cdot ,x\left(k\right),\cdot \cdot \cdot ,0\right)\in {s}_{n}\left(H\right)$ , where $‖x\left(k\right)‖=1$ .

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

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 $x,y\in {s}_{n}\left(H\right)$ , then

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

where $suppx=\left\{n:x\left(n\right)\ne 0,n\in ℕ\right\}$ .

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 ${V}_{0}:{S}_{{r}_{0}}\left({s}_{n}\left(H\right)\right)\to {S}_{{r}_{0}}\left({s}_{n}\left(H\right)\right)$ is an phase isometry, then we can get

$x\perp y⇔{V}_{0}\left(x\right)\perp {V}_{0}\left(y\right)$ .

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

$2{r}_{0}=‖x‖+‖y‖=‖x-y‖=‖{V}_{0}\left(x\right)-{V}_{0}\left(y\right)‖=\underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{x}^{\prime }\left(i\right)-{y}^{\prime }\left(i\right)‖}{1+‖{x}^{\prime }\left(i\right)-{y}^{\prime }\left(i\right)‖}$

or

$2{r}_{0}=‖x‖+‖y‖=‖x-y‖=‖{V}_{0}\left(x\right)+{V}_{0}\left(y\right)‖=\underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{x}^{\prime }\left(i\right)-{y}^{\prime }\left(i\right)‖}{1+‖{x}^{\prime }\left(i\right)-{y}^{\prime }\left(i\right)‖}$ (8)

at the same time, we have

$\underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{x}^{\prime }\left(i\right)-{y}^{\prime }\left(i\right)‖}{1+‖{x}^{\prime }\left(i\right)-{y}^{\prime }\left(i\right)‖}\le \underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{x}^{\prime }\left(i\right)‖}{1+‖{x}^{\prime }\left(i\right)‖}+\underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{y}^{\prime }\left(i\right)‖}{1+‖{y}^{\prime }\left(i\right)‖}=2{r}_{0}$

$\underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{x}^{\prime }\left(i\right)+{y}^{\prime }\left(i\right)‖}{1+‖{x}^{\prime }\left(i\right)+{y}^{\prime }\left(i\right)‖}\le \underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{x}^{\prime }\left(i\right)‖}{1+‖{x}^{\prime }\left(i\right)‖}+\underset{i=1}{\overset{n}{\sum }}\frac{1}{{2}^{i}}\frac{‖{y}^{\prime }\left(i\right)‖}{1+‖{y}^{\prime }\left(i\right)‖}=2{r}_{0}$ (9)

That means $‖{V}_{0}\left(x\right)-{V}_{0}\left(y\right)‖=‖{V}_{0}\left(x\right)+{V}_{0}\left(y\right)‖=‖{V}_{0}\left(x\right)‖+‖+{V}_{0}\left(y\right)‖$ , 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, ${\lambda }_{k}=\frac{{2}^{k}{r}_{0}}{1-{2}^{k}{r}_{0}}\left(k\in ℕ\right),\left(1\le k\le n\right)$ , and ${e}_{x\left(k\right)}={s}_{n}\left(H\right)$ . $\left(‖x\left(k\right)‖=1\right)$ . Then there exists ${x}^{\prime }\left(k\right)\in H\left(‖{x}^{\prime }\left(k\right)‖=1\right)$ , such that ${V}_{0}\left(±{\lambda }_{k}{e}_{x\left(k\right)}\right)=±{\lambda }_{k}{e}_{{x}^{\prime }\left(k\right)}$ .

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 ${x}^{\prime }\left(l\right)\left(‖{x}^{\prime }\left(l\right)‖=1\right)$ 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 ${V}_{0}\left(±{\lambda }_{k}{e}_{x\left(k\right)}\right)=±{\lambda }_{k}{e}_{{x}^{\prime }\left(k\right)}$ . 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 ${V}_{0}\left(-{\lambda }_{k}{e}_{x\left(k\right)}\right)={\lambda }_{l}{e}_{{x}^{″}\left(l\right)}$ ,

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

Therefore,

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

or

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

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

Lemma 3.4. Let $X={s}_{n}\left(H\right)$ and $Y={s}_{n}\left(H\right)$ . Suppose that $V:X\to Y$ is a surjective mapping satisfying Equation (1). there exist two real numbers $\alpha$ and $\beta$ with absolute 1 such that

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

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.

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

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

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

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

$\begin{array}{l}\left\{‖2x‖+‖y‖,‖y‖\right\}\\ =\left\{‖V\left\{x\left(i\right)+y\left(i\right)\right\}+V\left\{x\left(i\right)\right\}‖,‖V\left\{x\left(i\right)+y\left(i\right)\right\}+V\left\{y\left(i\right)\right\}‖\right\}\\ =\left\{\underset{i=1}{\overset{n}{\sum }}\frac{‖{x}^{\prime }\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{{x}^{\prime }\left(i\right)}{‖{x}^{\prime }\left(i\right)‖}}\right)+\underset{i=1}{\overset{n}{\sum }}\frac{‖{y}^{\prime }\left(i\right)‖}{{\mu }_{i}}V\left({\mu }_{i}{e}_{\frac{{y}^{\prime }\left(i\right)}{‖{y}^{\prime }\left(i\right)‖}}\right)+\underset{i=1}{\overset{n}{\sum }}\frac{‖x\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{x\left(i\right)}{‖x\left(i\right)‖}}\right),\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{i=1}{\overset{n}{\sum }}\frac{‖{x}^{\prime }\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{{x}^{\prime }\left(i\right)}{‖{x}^{\prime }\left(i\right)‖}}\right)+\underset{i=1}{\overset{n}{\sum }}\frac{‖{y}^{\prime }\left(i\right)‖}{{\mu }_{i}}V\left({\mu }_{i}{e}_{\frac{{y}^{\prime }\left(i\right)}{‖{y}^{\prime }\left(i\right)‖}}\right)+\underset{i=1}{\overset{n}{\sum }}\frac{‖y\left(i\right)‖}{{\mu }_{i}}V\left({\mu }_{i}{e}_{\frac{y\left(i\right)}{‖y\left(i\right)‖}}\right)\right\}\\ =\left\{\underset{i=1}{\overset{n}{\sum }}\frac{‖{x}^{\prime }\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{{x}^{\prime }\left(i\right)}{‖{x}^{\prime }\left(i\right)‖}}\right)+\underset{i=1}{\overset{n}{\sum }}\frac{‖x\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{x\left(i\right)}{‖x\left(i\right)‖}}\right)+‖\left\{y\left(i\right)\right\}‖,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\underset{i=1}{\overset{n}{\sum }}\frac{‖{x}^{\prime }\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{{x}^{\prime }\left(i\right)}{‖{x}^{\prime }\left(i\right)‖}}\right)-\underset{i=1}{\overset{n}{\sum }}\frac{‖x\left(i\right)‖}{{\lambda }_{i}}V\left({\lambda }_{i}{e}_{\frac{x\left(i\right)}{‖x\left(i\right)‖}}\right)+‖\left\{y\left(i\right)\right\}‖\right\}\end{array}$

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

Lemma 3.5. Let $X={s}_{n}\left(H\right)$ and $Y={s}_{n}\left(H\right)$ . Suppose that $V:X\to Y$ 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 $x,y\in X$ . Putting $y=x$ in the Equation (1), this yields

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

$V\left(x\right)=0$ 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 $x,y,z$ , we obtain

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

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

This yields $y,z\in \left\{x,-x\right\}$ . 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 $V:X\to Y$ 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 $Z=\left\{z\in X:z\perp {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right\}$ . By Lemma 3.4 we can write

$V\left(z+\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)=\alpha \left(z,\lambda \right)V\left(z\right)+\beta \left(z,\lambda \right)V\left(\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right),|\alpha \left(z,\lambda \right)|=|\beta \left(z,\lambda \right)|=1$

for any $z\in Z$ . Then, we can define a mapping $J:{s}_{n}\left(H\right)\to {s}_{n}\left(H\right)$ as follows:

$J\left(z+\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)=\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left(\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)$

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

$J\left({e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)=V\left({e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)$ , $J\left(-{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)=-V\left({e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)$

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 $\forall 0\ne \lambda \in ℝ$ ,

$\begin{array}{l}\left\{‖2z‖+\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1+\lambda |}{1+|1+\lambda |},\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1-\lambda |}{1+|1-\lambda |}\right\}\\ =\left\{‖J\left(z+{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)+J\left(z+\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)‖,‖J\left(z+{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)-J\left(z+\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)‖\right\}\\ =\left\{‖\alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left({e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)+V\left(\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)‖,\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}‖\alpha \left(z,1\right)\beta \left(z,1\right)V\left(z\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)V\left(z\right)+V\left({e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)-V\left(\lambda {e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)‖\right\}\\ =\left\{‖|\alpha \left(z,1\right)\beta \left(z,1\right)+\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)|V\left(z\right)‖+\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1+\lambda |}{1+|1+\lambda |},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}‖|\alpha \left(z,1\right)\beta \left(z,1\right)-\alpha \left(z,\lambda \right)\beta \left(z,\lambda \right)|V\left(z\right)‖+\frac{1}{{2}^{{\gamma }_{0}}}\frac{|1+\lambda |}{1+|1+\lambda |}\right\}\end{array}$

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

That yields

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

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

$\begin{array}{l}\left\{‖{z}_{1}+{z}_{2}‖+\frac{2}{3}\frac{1}{{2}^{{\gamma }_{0}}},‖{z}_{1}-{z}_{2}‖\right\}\\ =\left\{‖J\left({z}_{1}+{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)+J\left({z}_{2}+{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)‖,‖J\left({z}_{1}+{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)-J\left({z}_{2}+{e}_{\frac{x\left({\gamma }_{0}\right)}{‖{\gamma }_{0}‖}}\right)‖\right\}\\ =\left\{‖J\left({z}_{1}\right)+J\left({z}_{2}\right)‖+\frac{2}{3}\frac{1}{{2}^{{\gamma }_{0}}},‖J\left({z}_{1}\right)-J\left({z}_{2}\right)‖\right\}\end{array}$

for $\forall {z}_{1},{z}_{2}\in Z$ , It follows that $‖J\left(x\right)-J\left(y\right)‖=‖x-y‖$ for all $x,y\in X$ , 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.

 [1] Mazur, S. and Ulam, S.M. (1932) Sur les transformations isométriques d’espaces vectoriels normés. Comptes Rendus de l'Académie des Sciences Paris, 194, 946-948. [2] Tingley, D. (1987) Isometries of the Unit Sphere. Geometriae Dedicata, 22, 371-378. https://doi.org/10.1007/BF00147942 [3] Ding, G.G. (2003) On the Extension of Isometries between Unit Spheres of E and . Acta Mathematica Scientia, New Series to Appear. [4] Ding, G.G. (2002) The 1-Lipschitz Mapping between Unit Spheres of Two Hilbert Spaces Can Extended to a Real Linear Isometry of the Whole Space. Science in China Series A, 45, 479-483. https://doi.org/10.1007/BF02872336 [5] Fu, X.H. (2002) A Note on the Isometric Extension of Unit Spheres in Hilbert Space. Acta Mathematica Scientia, No. 6, 1147-1148 (In Chinese). [6] Fu, X.H. (2006) Isometries on the Space s*. Acta Mathematica Scientia, 26B, 502-508. https://doi.org/10.1016/S0252-9602(06)60075-1 [7] Maksa, G. and Pales, Z. (2012) Wigner’s Theorem Revisited. Mathematicae Debrecen, 81, 243-249. https://doi.org/10.5486/PMD.2012.5359 [8] Tan, D.N. (2017) Wigner’s Theorem in Atomic -Spaces ( ). Publicationes Mathematicae Debrecen. [9] Jia, W.K. and Tan, D.N. (2017) Wigner’s Theorem in -Type Spaces. Bulletin of the Australian Mathematical Society, 97, 279-284. [10] Rudin, W. (1985) Functional Analysis. 8th Edition, McGraw-Hill Inc, New York. [11] Fu, X.H. (2014) On Isometrie Extension in the Space . Journal of Function Spaces, 2014, 4.