Share This Article:

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

Abstract Full-Text HTML XML Download Download as PDF (Size:448KB) PP. 209-221
DOI: 10.4236/ajcm.2018.83017    340 Downloads   571 Views  

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.

Conflicts of Interest

The authors declare no conflicts of interest.

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.

References

[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.

  
comments powered by Disqus

Copyright © 2020 by authors and Scientific Research Publishing Inc.

Creative Commons License

This work and the related PDF file are licensed under a Creative Commons Attribution 4.0 International License.