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 AJCM  Vol.8 No.3 , September 2018
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).

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 ( 0 ) = 0 , then U is linear. Let X and Y be normed spaces, if the mapping V : X Y satisfying that

{ V ( x ) V ( y ) } = { x y } ( x , y X ) .

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 ε : X { 1,1 } such that J = ε V is a linear isometry. we can say the mapping V : X Y is phase equivalent to J.

If the two spaces are Hilbert spaces, Rätz proved that the phase isometries V : X 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 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 s or s n ( H ) s n ( H ) also have the properties, that are solutions of the functional equation

{ V ( x ) V ( y ) , V ( x ) + V ( y ) } = { x y , x + y } ( x , y X ) . (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 = { ξ k } = k ξ k e k , the F-norm of x is defined by x = n = 1 1 2 k | ξ k | 1 + | ξ k | . Let s ( n ) denote the set of all elements of the form x = { ξ 1 , , ξ n } with x = k = 1 n 1 2 k | ξ k | 1 + | ξ k | . where e k = { ξ k : ξ k = 1 , ξ k = 0 , k k , for all k Γ } . We denote the support of x by Γ x , i.e.,

s u p p ( x ) = Γ x = { γ Γ : ξ γ 0 } .

For all x , y s , if Γ x Γ y = , we say that x is orthogonal to y and write x y .

Lemma 2.1. Let S r 0 ( s ) be a sphere with radius r 0 and center 0 in s. Suppose that V 0 : S r 0 ( s ) S r 0 ( s ) is a mapping satisfying Equation (1). Then for any x , y S r 0 ( s ) , we have

x y V 0 ( x ) V 0 (y)

Proof: Necessity. Choosing x = { ξ n } , y = { η n } S r 0 ( s ) that satisfying x y . We can suppose V 0 ( x ) = { ξ n } , V 0 ( y ) = { η n } . And we also have

{ V 0 ( x ) V 0 ( y ) , V 0 ( x ) + V 0 ( y ) } = { x y , x + y } .

So

V 0 ( x ) V 0 ( y ) = x y = x + y = 2 r 0 = V 0 ( x ) + V 0 ( y )

or

V 0 ( x ) V 0 ( y ) = x + y = x + y = 2 r 0 = V 0 ( x ) + V 0 ( y )

Thus

n = 1 1 2 n | ξ n η n | 1 + | ξ n η n | = n = 1 1 2 n | ξ n | 1 + | ξ n | + n = 1 1 2 n | η n | 1 + | η n |

That means

n = 1 1 2 n [ | ξ n η n | 1 + | ξ n η n | | ξ n | 1 + | ξ n | | η n | 1 + | η n | ] = 0 (2)

It is easy to know f ( x ) = x 1 + x is strictly increasing. And | ξ n η n | | ξ n | + | η n | . We can get the result ξ n η n = 0 .

For V 0 ( x ) + V 0 ( y ) , similarty to the above ( | ξ n + η n | | ξ n | + | η n | ) . It is V 0 ( x ) V 0 ( y ) . Sufficiency. For V 0 ( x ) V 0 ( y ) , that is, ξ n η n = 0 , so (2) holds, and we have

x y = V 0 ( x ) V 0 ( y ) = V 0 ( x ) + V 0 ( y ) = 2 r 0

so, it must have x y = x + y .

or

x y = V 0 ( x ) + V 0 ( y ) = V 0 ( x ) + V 0 ( y ) = 2 r 0

as the same x y = x + y . It follows that

n = 1 1 2 n | ξ n η n | 1 + | ξ n η n | = n = 1 1 2 n | ξ n | 1 + | ξ n | + n = 1 1 2 n | η n | 1 + | η n | (3)

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

Lemma 2.2. Let S r 0 ( s ( n ) ) be a sphere with radius r 0 in the finite dimensional space s ( n ) , where r 0 < 1 2 n . Suppose that V 0 : S r 0 ( s ( n ) ) S r 0 ( s ( n ) ) is an phase isometry. Let λ k = 2 k r 0 1 2 k r 0 ( k , 1 k n ) , then there is a unique real θ with | θ | = 1 , such that V 0 ( λ k e k ) = θ λ k e k .

Proof: We proof first that for any k ( 1 k n ) , there is a unique l ( 1 l n ) and a unique real θ with | θ | = 1 such that V 0 ( λ k e k ) = θ λ l e l (because the assumption of λ k implies λ k e k S r 0 ( s ( n ) ) ). To this end, suppose on the contrary that V 0 ( λ k 0 e k 0 ) = k = 1 n η k e k and η k 1 0, η k 2 0 . In view of Lemma 1, we have

[ s u p p V 0 ( λ k 0 e k 0 ) ] [ s u p p V 0 ( λ k e k ) ] = k k 0 ,1 k n .

Hence, by the “pigeon nest principle” (or Pigeonhole principle) there must exist k i 0 ( 1 k i 0 n ) such that V 0 ( λ k i 0 e k i 0 ) = θ , which leads to a contradiction.

Next, if V 0 ( λ k e k ) = θ 1 λ l e l , V 0 ( λ k e k ) = θ 2 λ p e p , where | θ 1 | = | θ 2 | = 1 , then l = p and θ 2 = θ 1 . Indeed, if l p , we have

V ( λ k e k ) V ( λ k e k ) = 2 λ k e k = 1 2 k | 2 λ k | 1 + | 2 λ k | 2 r 0

or

V ( λ k e k ) V ( λ k e k ) = 0

and

V ( λ k e k ) V ( λ k e k ) = θ 1 λ l e l θ 2 λ p e p = 2 r 0 (4)

a contradiction which implies l = p . From this θ 1 = θ 2 follows. Finally, there is a unique θ with | θ | = 1 such that V 0 ( λ k e k ) = θ λ k e k . Indeed, if V 0 ( λ k e k ) = θ λ l e l , by the result in the last step, we have V 0 ( λ k e k ) = θ λ l e l , thus

{ V ( λ k e k ) + V ( λ k e k ) , V ( λ k e k ) V ( λ k e k ) } = { 2 λ k e k , 0 } = { 1 2 k | 2 λ k | 1 + | 2 λ k | , 0 }

and

{ V ( λ k e k ) + V ( λ k e k ) , V ( λ k e k ) V ( λ k e k ) } = { 2 θ λ l e l , 0 } = { 1 2 l | 2 λ l | 1 + | 2 λ l | , 0 } (5)

So, we get

1 2 k | 2 λ k | 1 + | 2 λ k | = 1 2 l | 2 λ l | 1 + | 2 λ l |

and we also have

1 2 k | λ k | 1 + | λ k | = 1 2 l | λ l | 1 + | λ l | , ( = r 0 )

through the two equalities of above

1 2 k | 2 λ k | 1 + | 2 λ k | 1 2 k | λ k | 1 + | λ k | = 1 2 l | 2 λ l | 1 + | 2 λ l | 1 2 l | λ l | 1 + | λ l |

In the end,

| λ l | = | λ k | (6)

The proof is complete.

Lemma 2.3. Let X = S r 0 ( s ( n ) ) and Y = S r 0 ( s ( n ) ) . Suppose that V 0 : X Y is a surjective mapping satisfying Equation (1) and λ k as in Lemma 2.2. Then for any lement x = k ξ k e k X , we have V 0 ( x ) = k η k e k , where | ξ k | = | η k | for any 1 k 0 n .

Proof: Note that the defination of V 0 , we can easily get V 0 ( 0 ) = 0 . For any 0 x X , write x = k ξ k e k , where k 1 2 k | ξ k | 1 + | ξ k | = r 0 . we can write V 0 ( x ) = k η k e k , where k 1 2 k | η k | 1 + | η k | = r 0 . we have

V 0 ( x ) + V 0 ( λ k 0 e k 0 ) + V 0 ( x ) V 0 ( λ k 0 e k 0 ) = x + λ k 0 e k 0 + x λ k 0 e k 0 = k k 0 ξ k e k + ( ξ k 0 + λ k 0 ) e k 0 + k k 0 ξ k e k + ( ξ k 0 λ k 0 ) e k 0 = r 0 + 1 2 k 0 | ξ k 0 + λ k 0 | 1 + ξ k 0 + λ k 0 1 2 k 0 | ξ k 0 | 1 + | ξ k 0 | + r 0 + 1 2 k 0 | ξ k 0 λ k 0 | 1 + ξ k 0 λ k 0 1 2 k 0 | ξ k 0 | 1 + | ξ k 0 | .

On the other hand, we have

V 0 ( x ) + V 0 ( λ k 0 e k 0 ) + V 0 ( x ) V 0 ( λ k 0 e k 0 ) = k = 1 n η k e k + θ k 0 λ k 0 e k 0 + k = 1 n η k e k θ k 0 λ k 0 e k 0 = k k 0 η k e k + ( η k 0 + θ k 0 λ k 0 ) e k 0 + k k 0 η k e k + ( η k 0 θ k 0 λ k 0 ) e k 0 = r 0 + 1 2 k 0 | η k 0 + θ k 0 λ k 0 | 1 + | η k 0 + θ k 0 λ k 0 | 1 2 k 0 | η k 0 | 1 + | η k 0 | + r 0 + 1 2 k 0 | η k 0 θ k 0 λ k 0 | 1 + | η k 0 θ k 0 λ k 0 | 1 2 k 0 | η k 0 | 1 + | η k 0 | .

Combiniing the two equations, we obtain that

| ξ k 0 + λ k 0 | 1 + | ξ k 0 + λ k 0 | | 2 ξ k 0 | 1 + | ξ k 0 | + | ξ k 0 λ k 0 | 1 + | ξ k 0 λ k 0 | = | η k 0 + θ k 0 λ k 0 | 1 + | η k 0 + θ k 0 λ k 0 | | 2 η k 0 | 1 + | η k 0 | + | η k 0 θ k 0 λ k 0 | 1 + | η k 0 θ k 0 λ k 0 |

As λ k 0 | ξ k 0 | and λ k 0 | η k 0 | , we have

ξ k 0 + λ k 0 1 + ξ k 0 + λ k 0 2 ξ k 0 1 + ξ k 0 + λ k 0 ξ k 0 1 + λ k 0 ξ k 0 = λ k 0 + θ k 0 η k 0 1 + λ k 0 + θ k 0 η k 0 2 η k 0 1 + η k 0 + λ k 0 θ k 0 η k 0 1 + λ k 0 θ k 0 η k 0

Therefore,

λ k 0 + λ k 0 2 ξ k 0 2 ( 1 + λ k 0 ) 2 ξ k 0 2 + η k 0 1 + η k 0 ξ k 0 1 + ξ k 0 = λ k 0 + λ k 0 2 η k 0 2 ( 1 + λ k 0 ) 2 η k 0 2

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

| ξ k | = | η 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 ( n ) and Y = s ( n ) . Suppose that V : X Y is a surjective mapping satisfying Equation (1). there exist two real numbers α and β with absolute 1 such that

V ( x + y ) = α V ( x ) + β 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 = k ξ k e k , y = k η k e k .

V ( x ) = k ξ k e k , V ( y ) = k η k e k

V ( x + y ) = k ξ k e k + k η k e k ,

where | ξ k | = | ξ k | = | ξ k | and | η k | = | η k | = | η k | . We infer from Equation (1) that

{ 2 x + y , y } = { V ( x + y ) + V ( x ) , V ( x + y ) V ( x ) } = { k ξ k e k + k η k e k + k ξ k e k , k ξ k e k + k η k e k + k η k e k } = { 1 2 k | ξ k + ξ k | 1 + | ξ k + ξ k | + y , 1 2 k | ξ k ξ k | 1 + | ξ k ξ k | + y }

Through the above equation we can get ξ k + ξ k = 0 or ξ k ξ k = 0 . This implies that k ξ k e k = ± V ( x ) , and similarly k η k e k = ± V ( y ) . The proof is complete.,

Lemma 2.5. Let X = s and Y = s . Suppose that V : X Y is a surjective mapping satisfying Equation (1). Then V is injective and V ( x ) = V ( x ) for all x X .

Proof: Suppose that V is surjective and V ( x ) = V ( y ) for some x , y X . Putting y = x in the Equation (1), this yields

{ 2 V ( x ) ,0 } = { 2 x ,0 }

V ( x ) = 0 if and only if x = 0 . Assume that V ( x ) = V ( y ) 0 choose z X such that V ( z ) = V ( x ) , using the Equation (1) for x , y , z , we obtain

{ x y , x + y } = { V ( x ) + V ( y ) , V ( x ) V ( y ) } = { 2 V ( x ) ,0 }

{ x z , x + z } = { V ( x ) + V ( z ) , V ( x ) V ( z ) } = { 2 V ( x ) ,0 }

This yields y , z { x , x } . If z = x , then V ( x ) = V ( x ) = 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 ( x ) = V ( y ) = V ( z ) = V ( x ) = 0

This lead to the contradiction that V ( x ) 0 .

Theorem 2.6. Let X = s ( n ) and Y = s ( n ) . Suppose that V : X Y is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.

Proof: Fix γ 0 Γ , and let Z = { z X : z e γ 0 } . By Lemma 2.4 we can write

V ( z + λ e γ 0 ) = α ( z , λ ) V ( z ) + β ( z , λ ) V ( λ e γ 0 ) , | α ( z , λ ) | = | β ( z , λ ) | = 1

for any z Z . Then, we can define a mapping J : s ( n ) s ( n ) as follows:

J ( z + λ e γ 0 ) = α ( z , λ ) β ( z , λ ) V ( z ) + V ( λ e γ 0 )

J ( λ z ) = α ( z , λ ) β ( z , λ ) V (λz)

J ( e γ 0 ) = V ( e γ 0 ) , J ( e γ 0 ) = V ( e γ 0 )

for 0 λ . The J is phase equivalent to V. So it is easily to know that J satisfies functional Equation (1). For any z Z , and 0 λ ,

{ 2 z + 1 2 γ 0 | 1 + λ | 1 + | 1 + λ | , 1 2 γ 0 | 1 λ | 1 + | 1 λ | } = { J ( z + e γ 0 ) + J ( z + λ e γ 0 ) , J ( z + e γ 0 ) J ( z + λ e γ 0 ) } = { α ( z , 1 ) β ( z , 1 ) V ( z ) + α ( z , λ ) β ( z , λ ) V ( z ) + V ( e γ 0 ) + V ( λ e γ 0 ) , α ( z , 1 ) β ( z , 1 ) V ( z ) α ( z , λ ) β ( z , λ ) V ( z ) + V ( e γ 0 ) V ( λ e γ 0 ) } = { | α ( z , 1 ) β ( z , 1 ) + α ( z , λ ) β ( z , λ ) | V ( z ) + 1 2 γ 0 | 1 + λ | 1 + | 1 + λ | , | α ( z , 1 ) β ( z , 1 ) α ( z , λ ) β ( z , λ ) | V ( z ) + 1 2 γ 0 | 1 + λ | 1 + | 1 + λ | }

That means α ( z , 1 ) β ( z , 1 ) = α ( z , λ ) β ( z , λ ) ,

J ( z + λ e γ 0 ) = J ( z ) + V ( λ e γ 0 ) for any z Z , and 0 λ .

That yields

{ J ( z ) + J ( z ) , J ( z ) J ( z ) + 2 V ( e γ 0 ) } = { J ( z + e γ 0 ) + J ( z e γ 0 ) , J ( z + e γ 0 ) J ( z e γ 0 ) } = { 0 , 2 ( z + e γ 0 ) }

That means J ( z ) = J ( z ) . On the other hand,

{ z 1 + z 2 + 2 3 1 2 γ 0 , z 1 z 2 } = { J ( z 1 + e γ 0 ) + J ( z 2 + e γ 0 ) , J ( z 1 + e γ 0 ) J ( z 2 + e γ 0 ) } = { J ( z 1 ) + J ( z 2 ) + 2 3 1 2 γ 0 , J ( z 1 ) J ( z 2 ) }

for z 1 , z 2 Z , It follows that J ( x ) J ( y ) = x y for all x , y X , by assumed conditions, so J is a surjective isometry.,

Theorem 2.7. Let X = s and Y = s . Suppose that V : X 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 ( n ) . 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 ( H ) , where H is a Hilbert space. In [11] mainly discussed the isometric extension in the space s n ( H ) . For each element x = { x ( k ) } , the F-norm of x is defined by x = k = 1 1 2 k x ( k ) 1 + x ( k ) . Let s n ( H ) denote the set of all elements of the form x = ( x ( 1 ) , , x ( n ) ) with x = k = 1 n 1 2 k x ( k ) 1 + x ( k ) . where x ( i ) ( i = 1 , , n ) H .

Some notations used:

e x ( k ) = ( 0, , x ( k ) , ,0 ) s n ( H ) , where x ( k ) = 1 .

Specially, when x ( k ) = 0 , we have e x ( k ) x ( k ) = ( 0 , , 0 ) .

Next, we study the phase isometry between the space s n ( H ) to s n ( H ) , that if V is a surjective phase isometry, then V is phase equivalent to a linear isometry J.

Lemma 3.1. If x , y s n ( H ) , then

x y = x + y if and only if s u p p x s u p p y =

where s u p p x = { n : x ( n ) 0, n } .

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

Lemma 3.2. Let S r 0 ( s n ( H ) ) be a sphere with radius r 0 in the finite dimensional space s n ( H ) , where r 0 < 1 2 n . Defined V 0 : S r 0 ( s n ( H ) ) S r 0 ( s n ( H ) ) is an phase isometry, then we can get

x y V 0 ( x ) V 0 ( y ) .

Proof: “Þ” Take any two elements x = { x ( i ) } , y = { y ( i ) } , let V 0 ( x ) = { x ( i ) } , V 0 ( y ) = { y ( i ) } . Then we have

2 r 0 = x + y = x y = V 0 ( x ) V 0 ( y ) = i = 1 n 1 2 i x ( i ) y ( i ) 1 + x ( i ) y ( i )

or

2 r 0 = x + y = x y = V 0 ( x ) + V 0 ( y ) = i = 1 n 1 2 i x ( i ) y ( i ) 1 + x ( i ) y ( i ) (8)

at the same time, we have

i = 1 n 1 2 i x ( i ) y ( i ) 1 + x ( i ) y ( i ) i = 1 n 1 2 i x ( i ) 1 + x ( i ) + i = 1 n 1 2 i y ( i ) 1 + y ( i ) = 2 r 0

i = 1 n 1 2 i x ( i ) + y ( i ) 1 + x ( i ) + y ( i ) i = 1 n 1 2 i x ( i ) 1 + x ( i ) + i = 1 n 1 2 i y ( i ) 1 + y ( i ) = 2 r 0 (9)

That means V 0 ( x ) V 0 ( y ) = V 0 ( x ) + V 0 ( y ) = V 0 ( x ) + + V 0 ( y ) , it is V 0 ( x ) V 0 ( y ) . “Ü” The proof of sufficiency is similar to the Lemma 2.1.

Lemma 3.3. Let V 0 be as in Lemma 3.2, λ k = 2 k r 0 1 2 k r 0 ( k ) , ( 1 k n ) , and e x ( k ) = s n ( H ) . ( x ( k ) = 1 ) . Then there exists x ( k ) H ( x ( k ) = 1 ) , such that V 0 ( ± λ k e x ( k ) ) = ± λ k e x ( k ) .

Proof: We prove first that, for any k ( 1 k n ) , there exist l ( 1 l n ) and x ( l ) ( x ( l ) = 1 ) such that V 0 ( λ k e x ( k ) ) = λ l e x ( k ) . And then prove l = p . It is the same an Lemma 2.2.

Finally, we assert that, there exists x ( k ) such that V 0 ( ± λ k e x ( k ) ) = ± λ k e x ( k ) . Indeed, if V 0 ( λ k e x ( k ) ) = λ l e x ( l ) , by the result in the last step, we have V 0 ( λ k e x ( k ) ) = λ l e x ( l ) ,

{ 0 , 1 2 k 2 λ k 1 + 2 λ k } = { V 0 ( λ k e x ( k ) ) V 0 ( λ k e x ( k ) ) , V 0 ( λ k e x ( k ) ) + V 0 ( λ k e x ( k ) ) } = { λ l e x ( l ) λ l e x ( l ) , λ l e x ( l ) λ l e x ( l ) } = { 1 2 l λ l x ( l ) x ( l ) 1 + λ l x ( l ) x ( l ) , 1 2 l λ l x ( l ) + x ( l ) 1 + λ l x ( l ) + x ( l ) }

Therefore,

1 2 k 2 λ k 1 + 2 λ k = 1 2 l λ l x ( l ) x ( l ) 1 + λ l x ( l ) x ( l ) 1 2 l 2 λ l 1 + 2 λ l

or

1 2 k 2 λ k 1 + 2 λ k = 1 2 l λ l x ( l ) + x ( l ) 1 + λ l x ( l ) + x ( l ) 1 2 l 2 λ l 1 + 2 λ l (10)

So, we can get k = l . And x ( l ) x ( l ) = x ( l ) + x ( l ) = 2 , that means x ( l ) = ± x ( l ) .

Lemma 3.4. Let X = s n ( H ) and Y = s n ( H ) . Suppose that V : X Y is a surjective mapping satisfying Equation (1). there exist two real numbers α and β with absolute 1 such that

V ( x + y ) = α V ( x ) + β V (y)

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

V { x ( i ) } = i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) ,

V { y ( i ) } = i = 1 n y ( i ) μ i V ( μ i e y ( i ) y ( i ) )

V { x ( i ) + y ( i ) } = i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) + i = 1 n y ( i ) μ i V ( μ i e y ( i ) y ( i ) ) ,

where x ( i ) = x ( i ) and y ( i ) = y ( i ) . We infer from Equation (1) that

{ 2 x + y , y } = { V { x ( i ) + y ( i ) } + V { x ( i ) } , V { x ( i ) + y ( i ) } + V { y ( i ) } } = { i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) + i = 1 n y ( i ) μ i V ( μ i e y ( i ) y ( i ) ) + i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) , i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) + i = 1 n y ( i ) μ i V ( μ i e y ( i ) y ( i ) ) + i = 1 n y ( i ) μ i V ( μ i e y ( i ) y ( i ) ) } = { i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) + i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) + { y ( i ) } , i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) i = 1 n x ( i ) λ i V ( λ i e x ( i ) x ( i ) ) + { y ( i ) } }

Through the above equation we can get x ( i ) = x ( i ) or x ( i ) = x ( i ) . The proof is complete.,

Lemma 3.5. Let X = s n ( H ) and Y = s n ( H ) . Suppose that V : X Y is a surjective mapping satisfying Equation (1). Then V is injective and V ( x ) = V ( x ) for all x X .

Proof: Suppose that V is surjective and V ( x ) = V ( y ) for some x , y X . Putting y = x in the Equation (1), this yields

{ 2 V ( x ) , 0 } = { 2 x , 0 }

V ( x ) = 0 if and only if x = 0 . Assume that V ( x ) = V ( y ) 0 choose z X such that V ( z ) = V ( x ) , using the Equation (1) for x , y , z , we obtain

{ x + y , x y } = { V ( x ) + V ( y ) , V ( x ) V ( y ) } = { 2 V ( x ) , 0 }

{ x + z , x z } = { V ( x ) + V ( z ) , V ( x ) V ( z ) } = { 2 V ( x ) , 0 }

This yields y , z { x , x } . If z = x , then V ( x ) = V ( x ) = 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 ( x ) = V ( y ) = V ( z ) = V ( x ) = 0

This lead to the contradiction that V ( x ) 0 .

Theorem 3.6. Let X = s n ( H ) and Y = s n ( H ) . Suppose that V : X Y is a surjective mapping satisfying Equation (1). Then V is phase equivalent to a linear isometry J.

Proof: Fix γ 0 Γ , and let Z = { z X : z e x ( γ 0 ) γ 0 } . By Lemma 3.4 we can write

V ( z + λ e x ( γ 0 ) γ 0 ) = α ( z , λ ) V ( z ) + β ( z , λ ) V ( λ e x ( γ 0 ) γ 0 ) , | α ( z , λ ) | = | β ( z , λ ) | = 1

for any z Z . Then, we can define a mapping J : s n ( H ) s n ( H ) as follows:

J ( z + λ e x ( γ 0 ) γ 0 ) = α ( z , λ ) β ( z , λ ) V ( z ) + V ( λ e x ( γ 0 ) γ 0 )

J ( λ z ) = α ( z , λ ) β ( z , λ ) V (λz)

J ( e x ( γ 0 ) γ 0 ) = V ( e x ( γ 0 ) γ 0 ) , J ( e x ( γ 0 ) γ 0 ) = V ( e x ( γ 0 ) γ 0 )

for 0 λ . The J is phase equivalent to V. So it is easily to know that J satisfies functional Equation (1). For any z Z , and 0 λ ,

{ 2 z + 1 2 γ 0 | 1 + λ | 1 + | 1 + λ | , 1 2 γ 0 | 1 λ | 1 + | 1 λ | } = { J ( z + e x ( γ 0 ) γ 0 ) + J ( z + λ e x ( γ 0 ) γ 0 ) , J ( z + e x ( γ 0 ) γ 0 ) J ( z + λ e x ( γ 0 ) γ 0 ) } = { α ( z , 1 ) β ( z , 1 ) V ( z ) + α ( z , λ ) β ( z , λ ) V ( z ) + V ( e x ( γ 0 ) γ 0 ) + V ( λ e x ( γ 0 ) γ 0 ) , α ( z , 1 ) β ( z , 1 ) V ( z ) α ( z , λ ) β ( z , λ ) V ( z ) + V ( e x ( γ 0 ) γ 0 ) V ( λ e x ( γ 0 ) γ 0 ) } = { | α ( z , 1 ) β ( z , 1 ) + α ( z , λ ) β ( z , λ ) | V ( z ) + 1 2 γ 0 | 1 + λ | 1 + | 1 + λ | , | α ( z , 1 ) β ( z , 1 ) α ( z , λ ) β ( z , λ ) | V ( z ) + 1 2 γ 0 | 1 + λ | 1 + | 1 + λ | }

That means α ( z , 1 ) β ( z , 1 ) = α ( z , λ ) β ( z , λ ) , J ( z + λ e x ( γ 0 ) γ 0 ) = J ( z ) + V ( λ e x ( γ 0 ) γ 0 ) for any z Z , and 0 λ .

That yields

{ J ( z ) + J ( z ) , J ( z ) J ( z ) + 2 V ( e x ( γ 0 ) γ 0 ) } = { J ( z + e x ( γ 0 ) γ 0 ) + J ( z e x ( γ 0 ) γ 0 ) , J ( z + e x ( γ 0 ) γ 0 ) J ( z e x ( γ 0 ) γ 0 ) } = { 0 , 2 ( z + e x ( γ 0 ) γ 0 ) }

That means J ( z ) = J ( z ) . On the other hand,

{ z 1 + z 2 + 2 3 1 2 γ 0 , z 1 z 2 } = { J ( z 1 + e x ( γ 0 ) γ 0 ) + J ( z 2 + e x ( γ 0 ) γ 0 ) , J ( z 1 + e x ( γ 0 ) γ 0 ) J ( z 2 + e x ( γ 0 ) γ 0 ) } = { J ( z 1 ) + J ( z 2 ) + 2 3 1 2 γ 0 , J ( z 1 ) J ( z 2 ) }

for z 1 , z 2 Z , It follows that J ( x ) J ( y ) = x y for all x , y 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 ( H ) .

Acknowledgements

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

Cite this paper: 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.
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