The extensions of isometries between the 2-dimensional normed spaces

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Available online at http://scik.org

J. Math. Comput. Sci. 7 (2017), No. 2, 347-363 ISSN: 1927-5307

THE EXTENSIONS OF ISOMETRIES BETWEEN THE 2-DIMENSIONAL NORMED SPACES

PU WANG∗, RUIDONG WANG

Department of Mathematics, Tianjin University of Technology, Tianjin 300384, China

Copyright c2017 P. Wang and R. Wang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. Isometric operator is a very significant subject in the study of space structure. In this paper, we will

give some results about the Tingley’s problem, and give sufficient conditions for isometric operatorT0, which is

between classical 2-dimensional normed sequence spaces such as`(p2)and`(∞2), to have a linear extension in some way.

Keywords:2-dimensional normed space; Tingley’s problem; linearity; isometric; extension;`(p2);`(∞2).

2010 AMS Subject Classification:46B20,47A12.

1. Introduction

In this paper,X andY are both normed spaces,S(X)andS(Y)are denoted the unit spheres ofX andY respectively, and a mappingT :X→Y is said to be an isometric operator if

kT x−Tyk=kx−yk,∀x,y∈X.

∗_{Corresponding author}

E-mail address: [email protected]

Received October 13, 2016; Published March 1, 2017

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The normed spaces X andY are said to be isometric if there exists a surjective isometric operator betweenXandY. IfX andY are isometric, then they are essentially identical as metric spaces. And we can obtain thatY is a strictly convex space ifX is strictly convex.

As the advantageous tools in the research of normed linear spaces, isometric operator and linear operator have a vital meaning in the functional analysis. Metric geometry, space structure, equivalence theory and affine motion theory are all need the support of those two tools. In 1932, Mazur and Ulam gave the famous Mazur-Ulam theorem in [1].

Theorem 1.1[1](Mazur-Ulam theorem)[1]Every surjective isometric operatorT from normed spaceX to normed spaceY is an affine. And it is also linear ifT(0) =0.

This is to say that if the normed space X and Y are isometric, then they must be linearly isometric to each other.

For all metric vector spaces, it is till open that whether the Mazur-Ulam theorem can holds. In 1953, Charzynski got an important result whenX andY are both the finite dimensional and have the same dimensions in [2].

Theorem 1.2 [2] X and Y are both the finite dimensional metric vector spaces, dim(X) =

dim(Y) =n, andT :X→Y is an isometry withT(0) =0, thenT is linear.

From then on, many researchers which both at home and abroad have tried to popularized the Mazur-Ulam theorem by different patterns, and have paid attention to weaken the condition of the Mazur-Ulam theorem from different patterns. Through the research of the space structure, many open problems were raised as well including the Alesandrov problem, the Aleksandrov-Rassias problem and the Tingley’s problem. In the consideration of the into isometric operator, Baker extended the Mazur-Ulam theorem to strictly convex normed spaces in [3], and obtained the following conclusion :

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In [4], Mankiewicz extended the Mazur-Ulam theorem, and showing that an isometric opera-tor which maps a connected subset of a normed spaceX onto an open subset of another normed spaceY can be extended to an affine operator on the whole space. Applying these results to the unit balls of X andY, we obtain that two normed spaces are linearly isometric if and only if their unit balls are isomeric.

Clearly, we have an negatively example that the subsets of Banach spaces are isometric, but not in any sense affine.

Example 1.4 A mappingT :R→`(∞2)(R2with the max norm) given by

T(x) = (x,sinx),

thenT is an isometry, but it is not affine.

Instead of connected subset, D.Tingley raised the following problem, which called Tingley’s problem in [5].

Problem 1.5 [5] Suppose that T0 :S(X)→S(Y) is a surjective isometric operator, does there exist a linear isometric operatorT :X→Y such thatT|_S(X)=T₀?

And Tingley has proved the following main result:

Theorem 1.6[5]LetX andY are both finite-dimensional normed spaces, ifT₀:S(X)→S(Y)is a surjective isometry, thenT0(−x) =−T0(x),∀x∈S(X).

In this case, we call theT₀preserves anti-polar points.

For this question, we only considerate the real normed space, because the Tingley’s problem is negative when X andY are complex normed spaces, like X =Y =_C and T(x) =x¯ for all

x∈_Cwith|x|=1. It’s hard to make an affirmative answer because there is no linear or even metrically convex structure on this unit sphere. Even for 2-dimensional normed space, Tingley’s problem is not salved until recently.

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isometric operator between unit spheres can’t be linear extended. In [15], L. Zhang obtained an counterexample which a operator fromS₁(`(∞2))intoS1(`

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∞ )shows that the isometric extension

problem fails.

Example 1.7[15]IsometryT₀which fromS₁(`(∞2))intoS1(`

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∞ )is given by

T₀[(ξ1,ξ2)] =

             

            

(1,3₄ξ2,ξ2), ξ1=1,ξ2≥0;

(−1,ξ2,3₄ξ2), ξ1=−1,ξ2≥0;

(ξ1,1−1₄ξ1,1), ξ2=1,ξ1≥0;

(ξ1,1,1+1₄ξ1), ξ2=1,ξ1<0;

(ξ1,ξ2,ξ2), ξ2<0.

thenT₀is an isometry fromS₁(`(∞2))toS1(`(∞3)), but it can’t be extended to an isometric operator

or linear operator on whole space`(∞2).

From now on, we can say that the question of whether an isometry fromS(X)intoS(Y)can be extended an isometric and linear operator for all Banach spaceX andY is negative. Some affirmative conclusions were obtained between the same type and different type of classical Banach spaces which can be found in [7-13].

We notice that, for finite-dimensional normed spaces X and Y, if dim(X)> dim(Y) then there is no linear isometry fromX toY. Since whendim(X) =1 is so trivial, in this paper, we consider the Tingley’s problem between the 2-dimensional normed spaces, and we obtain that whendim(Y) =2 the dimension ofX should be less then or equal to 2, so we just consider the problem whendim(X) =dim(Y) =2.

And for 2-dimensional normed spaces, Wang R D and Wang P has obtained an useful result to help extending the isometry on unit spheres in[19].

Theorem 1.8[19] LetX andY are both two dimensional normed space ,T₀:S(X)→S(Y)is a surjective isometric operator. If

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kT₀(x)−(kT₀(x) +T_o(y)k −1)T_o(y)k=kx−(kx+yk −1)yk,

for∀x,y∈S(X),kx+yk ≥1, thenT₀can extended to be an linear isometric operator onX. In this paper, we let 1≤ p≤∞, so we consider the space `(p2)(Γ) and `

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∞ (Γ) together by

symbol `(p2)(Γ) (1≤ p≤∞). First, we give some metric property of the unit sphere of

2-dimensional normed space`(p2)(Γ) (1≤p≤∞). Then we study the Tingley’s problem between

`(p2)(Γ) (1≤ p≤∞), we obtain that surjective isometric operator formS[`(p2)(Γ)]toS[`(p2)(Γ)]

can be extended to the whole space`(p2)(Γ)linearly.

2. Some Properties of the

S

[

`

(p2)

(

Γ

)] (1

≤

p

≤

∞

)

By using the characteristics of norm of sequence spaces`p(Γ) (1≤p<∞)and`∞_(Γ)_{, Ding}

G G obtained the representation theorem of the surjective isometric operator of unit spheres of this type spaces in[13,14]. For Hilbert spaces, we have got a good answer about the Tingley’y problem in [18], when p=2, as a special case we also solved the extension of isometries between two`(₂2), so we no longer discuss this problem when p=2.

Theorem 2.1[13,14]Suppose that

T₀:S[`p(Γ)]→S[`p(∆)] (1≤p≤∞, p6=2)

is an surjective isometric operator. Then there exist a 1-1 permutation mappingπ:∆→Γand

the number set{θσ}σ∈∆ with|θσ |=1 for allσ∈∆, such that

T₀(x) =

_∑

σ∈∆

θσ·ξπ(σ)dσ, ∀x=

∑

γ∈Γ

ξγeγ ∈S[`p(Γ)].

With the representation theorem of the surjective isometric operator, we can easily get the affirmative answer to the Tingley’s question in the`p type space, but we can get more explicit

representation theorem of the surjective isometric operator on `(p2) by letting the index sets Γ=∆={1,2}in the third chapter.

Definition 2.2 For∀x,y∈`(p2)(1≤ p≤∞,p6=2), we say thatxis orthogonal toy(denoted by

x⊥y) if

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hold for all scalarsλ.

Definition 2.3 For∀x,y∈`(p2)(1≤p≤∞,p6=2), we say that{e1,e2}is a normalized orthog-onal basis of`(p2)if

ke₁k=ke₂k=1 and e₁⊥e₂.

Lete₁= (1,0),e₂= (0,1)∈S[`(p2)] (1≤ p≤∞,p=6 2), it is easy to prove that{e1,e2}is a normalized orthogonal basis, and for∀x= (ξ1,ξ2)∈`

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p , we can letx=ξ1e1+ξ2e2.

Definition 2.4 For∀x,y∈`(p2)(1≤p≤∞,p6=2),{e1,e2}is a normalized orthogonal basis of `(p2), for∀x= (ξ1,ξ2) =ξ1e1+ξ2e2∈`

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p ,|θ1|=|θ2|=1, we rememberθ1ξ1e1+θ2ξ2e2as ˜x, rememberθ2ξ2e1+θ1ξ1e2as ¯x.

Proposition 2.5 For`(p2)(1≤p≤∞,p6=2), ˜x,x¯∈S[` (2)

p ]if and only ifx∈S[` (2)

p ], and

kxk=kx˜k=kx¯k.

Proof. For`(p2)(1≤p≤∞,p6=2),∀x= (ξ1,ξ2),

When 1≤p<∞,

x∈S[`(p2)] ⇔ kxk=1 ⇔ |ξ1|p+|ξ2|p=1 ⇔ |θ1ξ1|p+|θ2ξ2|p=1

⇔ k_exk=kxk=1 ⇔ x˜,x¯∈S[`(p2)];

When p=∞,

x∈S[`(p2)] ⇔ kxk=1 ⇔ max{|ξ1|,|ξ2|}=1 ⇔ max{|θ1ξ1|,|θ2ξ2|}=1 ⇔ k_exk=kxk=1 ⇔ x˜,x¯∈S[`(p2)].

3. The Extensions of Isometries between the 2-dimensional Normed Spaces

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Proposition 3.1 For`(p2), when 1≤p<∞,p=6 2,∀x,y∈S[`(p2)], then

supp(x)∩supp(y) =_∅ ⇔ kx±ykp=kxkp+kykp.

Proof. For∀x,y∈S[`(p2)], letx= (ξ1,ξ2)y= (η1,η2).

0_⇒0

Ifsupp(x)∩supp(y) =_∅, without loss of generality, letsupp(x) ={1}, andsupp(y) ={2}. Sox= (ξ1,0)y= (0,η2), then

kx±ykp=k(ξ1,±η2)kp=|ξ1|p+|η2|p=kxkp+kykp.

0_⇐0

Ifkx±ykp₌_k_x_kp₊_k_y_kp_{, then}_k_x₊_y_kp₊_k_x₋_y_kp₌₂₍_k_x_kp₊_k_y_kp₎_,

|ξ1+ξ2|p+|η1+η2|p+|ξ1−ξ2|p+|η1−η2|p=2(|ξ1|p+|ξ2|p+|η1|p+|η2|p).(∗)

Following the famous complex inequalities(f or ∀ξ 6=0,η 6=0)

|ξ+η|p+|ξ−η|p<2(|ξ |p+|η|p),(1≤p<2)

and

|ξ+η|p+|ξ−η|p>2(|ξ |p+|η|p),(p>2)

with the(∗), we can obtain that

ξ1·η1=ξ2·η2=0, thensupp(x)∩supp(y) =_∅.

Proposition 3.2 For`(p2), whenp=∞,∀x,y∈S[` (2) ∞ ], then

supp(x)∩supp(y) =_∅ ⇔ kx±yk=max{kxk,kyk}.

Proof. For∀x,y∈S[`(∞2)], letx= (ξ1,ξ2)y= (η1,η2).

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Ifsupp(x)∩supp(y) =_∅, without loss of generality, letsupp(x) ={1}, andsupp(y) ={2}, that isx= (ξ1,0),y= (0,η2), then

kx±yk=k(ξ1,±η2)k=max{|ξ1|,| ±η2|}=max{kxk,kyk}.

0_⇐0

For∀x,y∈S[`(∞2)], ifkx±yk=max{kxk,kyk}, that is

kx±yk=k(ξ1±η1,ξ2±η2)k

=max{kxk,kyk}

=max{max{|ξ1|,|ξ2|},max{|η1|,|η2|}}

=max{|ξ1|,|ξ2|,|η1|,|η2|}

=1,

max{|ξ1+η1|,|ξ2+η2|}=max{kxk,kyk}=max{|ξ1|,|ξ2|,|η1|,|η2|}, (∗∗) and

max{|ξ1−η1|,|ξ2−η2|}=max{kxk,kyk}=max{|ξ1|,|ξ2|,|η1|,|η2|}, (∗ ∗ ∗) Ifsupp(x)∩supp(y)6=_∅.

Case I supp(x)∩supp(y)has only one point.

Without loss of generality,let η2 =0, that is supp(x)∩supp(y) ={1}, x= (ξ1,ξ2), y=

(η1,0).

kyk=1 becausey∈S[`(∞2)], that is

max{|η1|,|0|}=|η1|=1. It is clearly that

max{|ξ1+η1|,|ξ1−η1|}>|η1|=1, but following the(∗∗)and(∗ ∗ ∗), we obtain that

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Case II supp(x)∩supp(y) has two points, namely that supp(x)∩supp(y) ={1,2},

ξ1,ξ2,η1,η2are all not equal to 0. In this case, it is clearly that

max{|ξ1+η1|,|ξ1−η1|}>|ξ1|, max{|ξ1+η1|,|ξ1−η1|}>|η1|,

max{|ξ2+η2|,|ξ1−η1|}>|ξ2|, max{|ξ2+η2|,|ξ1−η1|}>|η2|, but this is contradictory with the(∗∗)and(∗ ∗ ∗).

Corollary 3.3 For`(p2)(1≤p≤∞,p6=2),T0:S[`(p2)]→S[`(p2)]is a surjective isometry, ∀x,y∈

S[`(p2)], then

supp(x)∩supp(y) =_∅ ⇔ supp[T₀(x)]∩supp[T₀(y)] =_∅.

Proof. For∀x,y∈S[`(p2)],T0(x),T0(y)∈S[`

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p ], letx= (ξ1,ξ2)y= (η1,η2).

0_⇒0 _{When 1}_≤ _p_<_∞_{, and} _p₌₆ _{2, if}_supp₍_x₎_∩_supp₍_y_{) =}

∅, then by the proposition 3.1,

we have

kx±ykp=kxkp+kykp=2, and

kT0(x)±T0(y)kp=kx±ykp=kxkp+kykp=2=kT0(x)kp+kT0(y)kp, then

supp[T₀(x)]∩supp[T₀(y)] =_∅.

Whenp=∞, for`(∞2), ifsupp(x)∩supp(y) =∅, then following the proposition 3.2, we have

kx±yk=max{kxk,kyk}=1, and

kT₀(x)±T₀(y)k=kx±yk=max{kxk,kyk}=1=max{kT₀(x)k,kT₀(y)k}, then

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0_⇐0 _{When 1}_≤ _p_<

∞, and p6=2, if supp[T0(x)]∩supp[T0(y)] =∅, then following the proposition 3.1, we have

kT₀(x)±T₀(y)kp=kT₀(x)kp+kT₀(y)kp=2, and

kx±ykp=kT₀(x)±T₀(y)kp=kT₀(x)kp+kT₀(y)kp=2=kxkp+kykp, then

supp(x)∩supp(y) =_∅.

When p=∞, for `(∞2), ifsupp[T0(x)]∩supp[T0(y)] =∅, then following the proposition 3.2,

we have

kT₀(x)±T₀(y)k=max{kT₀(x)k,kT₀(y)k}=1, and

kx±yk=kT0(x)±T0(y)k=max{kT0(x)k,kT0(y)k}=1=max{kxk,kyk}, then

supp(x)∩supp(y) =_∅.

Theorem 3.4 For `(p2)(1 ≤ p ≤∞,p 6=2), T0 :S[`

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p ]→S[` (2)

p ] is a surjective isometry,

e1= (1,0),e2= (0,1),{e1,e2}is a normalized orthogonal basis of`(p2), then

T₀(e₁) =e˜₁, T₀(e₂) =e˜₂ or

T0(e1) =e¯1, T0(e2) =e¯2.

Proof. Fore₁,e₂∈S[`(p2)],T0is a surjective isometry onS[`

(2) p ], so

T0(e1), T2(e2)∈S[`(p2)].

Following the proposition 3.1, 3.2 and 3.3, we obtain that

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andkT₀(e₁)k=kT₀(e₂)k=1, so{|T₀(e₁)|,|T₀(e₂)|}={|e₁|,|e₂|}, then there are|θ1|=|θ2|=1 and

T0(e1) =θ1e1=e˜1, T0(e2) =θ2e2=e˜2

or

T₀(e₁) =θ1e2=e¯1, T0(e2) =θ2e1=e¯2.

Corollary 3.5 For `(p2)(1≤ p≤ ∞,p6= 2), T0 :S[`

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p ]→ S[`(p2)] is a surjective isometry,

{e₁,e₂}is a normalized orthogonal basis of`(p2), then

T₀(x) =x˜ or T₀(x) =x¯ f or∀x∈S[`(p2)].

Proof. For∀x= (ξ1,ξ2)∈S[`

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p ], letx6=±e, T0(x) =x

0

= (ξ₁0,ξ₂0)∈S[`(p2)], soT0(x)6=±e, by the theorem 3.4, there are only two cases.

Case I T₀(e₁) =e˜₁,T₀(e₂) =e˜₂. When 1≤p<∞, p6=2,x,x

0

∈S[`(p2)], that is

kxkp=1=|ξ1|p+|ξ2|p ⇒ 0<|ξ1|<1, 0<|ξ2|<1, kx0kp=1=|ξ

0

1|p+|ξ

0

2|p ⇒ 0<|ξ

0

1|<1, 0<|ξ

0

2|<1. Then

kT0(x)±T0(e1)kp=kx±e1kp=kx

0

±e˜1kp=kx

0

±θ1e1kp, (3) kT0(x)±T0(e2)kp=kx±e2kp=kx

0

±e˜2kp=kx

0

±θ2e2kp, (4) that is

k(ξ1±1,ξ2)kp=k(ξ

0

1±θ1,ξ

0

2)kp,

k(ξ1,ξ2±1)kp=k(ξ

0

1,ξ

0

2±θ2)kp.

kx±e₁kp=k(ξ1±1,ξ2)kp

=|ξ1±1|p+|ξ2|p

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kx0±e˜₁kp=k(ξ

0

1±θ1,ξ

0

2)kp

=|ξ

0

1±θ1|p+|ξ

0

2|p

=|ξ

0

1±θ1|p+1− |ξ

0

1|p, so by the (3) we get

|ξ1+1|p− |ξ1|p=|ξ

0

1+θ1|p− |ξ

0

1|p, (5) |ξ1−1|p− |ξ1|p=|ξ

0

1−θ1|p− |ξ

0

1|p. (6)

Whenθ1=1, by (5) and f(t) =|t+1|p− |t|p is a strictly increasing mapping when −1<

t<1, we obtain thatξ₁0 =ξ1=θ1ξ1, and similarly we obtain thatξ

0

2=ξ2=θ2ξ2whenθ2=1. Whenθ1=−1, by (5) we get

|ξ1+1|p− |ξ1|p=|ξ

0

1−1|p− |ξ

0

1|p=|1−ξ

0

1|p− |ξ

0

1|p,

by f(t) =|t+1|p_{− |}_t_|p _{we obtain that}_ξ0

1<0 and ξ

0

1=−ξ1=θ1ξ1, and similarly we obtain thatξ₂0 =−ξ2=θ2ξ2whenθ2=−1, that is

T0(x) =x˜. When p=∞, for`(∞2),x,x

0

∈S[`(∞2)], that is

kxk=1=max{|ξ1,|ξ2|} ⇒ 0<|ξ1| ≤1, 0<|ξ2| ≤1,

kx0k=1=max{|ξ

0

1|,|ξ

0

2|} ⇒ 0<|ξ

0

1| ≤1, 0<|ξ

0

2| ≤1.

kT₀(x)±T₀(e₁)k=kx±e₁k=kx0±e˜₁k, kT0(x)±T0(e2)k=kx±e2k=kx

0

±e˜2k, that is

k(ξ1+1,ξ2)k=max{|ξ1+1|,|ξ2|}=k(ξ

0

1+θ1,ξ

0

2)k=max{|ξ

0

1+θ1|,|ξ

0

2|},

k(ξ1−1,ξ2)k=max{|ξ1−1|,|ξ2|}=k(ξ

0

1−θ1,ξ

0

2)k=max{|ξ

0

1−θ1|,|ξ

0

2|},

k(ξ1,ξ2+1)k=max{|ξ1|,|ξ2+1|}=k(ξ

0

1,ξ

0

2+θ2)k=max{|ξ

0

1|,|ξ

0

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and

k(ξ1,ξ2−1)k=max{|ξ1|,|ξ2−1|}=k(ξ

0

1,ξ

0

2−θ2)k=max{|ξ

0

1|,|ξ

0

2−θ2|},

Whenξ1>0, we have|ξ1+1|>1,|ξ2| ≤1 and|ξ

0

2| ≤1, so max{|ξ1+1|,|ξ2|}=|ξ1+1|=max{|ξ

0

1+θ1|,|ξ

0

2|}=|ξ

0

1+θ1|.

We can get |ξ1+1|=|ξ

0

1+θ1|, and by the mapping g(t) =|t+1| is strictly increasing when 0≤t ≤1, we obtain thatξ₁0 =θ1ξ1.

Similarly we also obtain thatξ₂0 =θ2ξ2whenξ2>0. Whenξ1<0, we have|ξ1−1|>1,|ξ2| ≤1 and|ξ

0

2| ≤1, so max{|ξ1−1|,|ξ2|}=|ξ1−1|=max{|ξ

0

1−θ1|,|ξ

0

2|}=|ξ

0

1−θ1|.

We can get|ξ1−1|=|ξ

0

1−θ1|, and by the mapping g(t) =|t−1|is strictly decreasing when −1≤t ≤0, we obtain thatξ₁0 =θ1ξ1.

Similarly we also obtain thatξ

0

2=θ2ξ2whenξ2<0, that is

T₀(x) =x˜.

Case II T₀(e₁) =e¯₁,T₀(e₂) =e¯₂. When 1≤p<∞, p6=2,

kx+e₁kp=|ξ1+1|p+1− |ξ1|p, kx0+e¯₁kp=|ξ

0

2+θ1|p+1− |ξ

0

2|p,

kx−e1kp=|ξ1−1|p+1− |ξ1|p, kx0−e¯1kp=|ξ

0

2−θ1|p+1− |ξ

0

2|p,

kT₀(x) +T₀(e₁)kp=kx+e₁kp=kx0+θ1e2kp,

kT₀(x)−T₀(e₁)kp=kx−e₁kp=kx0−θ1e2kp,

like the above proof about case I, we obtain that ξ

0

2=θ1ξ1, and similarly we also obtain that

ξ₁0 =θ2ξ2, that is

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When p=∞, for`(∞2),

kT0(x)±T0(e1)k=kx±e1k=kx

0

±θ1e2k,

kT0(x)±T0(e2)k=kx±e2k=kx

0

±θ2e1k,

that is

k(ξ1+1,ξ2)k=max{|ξ1+1|,|ξ2|}=k(ξ

0

1,ξ

0

2+θ1)k=max{|ξ

0

1|,|ξ

0

2+θ1|},

k(ξ1−1,ξ2)k=max{|ξ1−1|,|ξ2|}=k(ξ

0

1,ξ

0

2−θ1)k=max{|ξ

0

1|,|ξ

0

2−θ1|},

k(ξ1,ξ2+1)k=max{|ξ1|,|ξ2+1|}=k(ξ

0

1+θ2,ξ

0

2)k=max{|ξ

0

1+θ2|,|ξ

0

2|},

and

k(ξ1,ξ2−1)k=max{|ξ1|,|ξ2−1|}=k(ξ

0

1−θ2,ξ

0

2)k=max{|ξ

0

1−θ2|,|ξ

0

2|},

like the above proof about case I, we getξ₁0 =θ2ξ2andξ

0

2=θ1ξ1, that is

T₀(x) =x¯.

Theorem 3.6 For`(p2) (1≤ p≤∞, p6=2), T0:S[`(p2)]→S[`(p2)]is a surjective isometry, then

T₀can be linearly extended to an real isometry from`(p2)to` (2) p .

Proof. For`(p2), e1= (1,0), e2= (0,1), {e1,e2}is a normalized orthogonal basis, for∀x∈

`(p2), let

T(x) =

  

 

kxkT0(_kxkx ), if x6=θ;

θ, if x=θ.

For∀x= (ξ1,ξ2),y= (η1,η2)∈`

(2)

p ,x6=0, andy6=0, there are only two cases, Case I T₀(e₁) =e˜₁,T₀(e₂) =e˜₂.

In this case, by theorem 3.5 and proposition 2.5, we obtain that

T₀( x

kxk) =

gx

kxk = ˜

x

kxk, T0(

y

kyk) =

gy

kyk = ˜

y

kyk,

T(x) =kxkT₀( x

kxk) =kxk · ˜

( x

(15)

kT(x)−T(y)k=kkxkT₀( x

kxk)− kykT0(

y

kyk)k

=kkxk ·( x˜

kxk)− kyk · ˜

( y

kyk)k

=kx˜−y˜k

=k(^x−y)k

=kx−yk,

T(αx+βy) =kαx+βykT0(

αx+βy

kαx+βyk)

=kαx+βyk ·

^ ( αx+βy

kαx+βyk)

=(α^x+βy)

=αx˜+βy˜

=αT(x) +βT(y),

thenT is the linearly isometric extension ofT₀to whole space`(p2). Case II T0(e1) =e¯2,T0(e2) =e¯1.

In this case, by theorem 3.5 and proposition 2.5, we obtain that

T₀( x

kxk) = (

x

kxk) = 1

kxk·x¯, T0(

y

kyk) = (

y

kyk) = 1 kyk·y¯,

T(x) =kxkT₀( x

kxk) =kxk ·( 1

kxk)·x¯=x¯,

kT(x)−T(y)k=kkxkT0(

x

kxk)− kykT0(

y

kyk)k

=kkxk ·( 1

kxk)·x¯− kyk ·( 1 kyk)·y¯k

=kx¯−y¯k

=k(x−y)k

(16)

T(αx+βy) =kαx+βykT0(

αx+βy

kαx+βyk)

=kαx+βyk ·( 1

kαx+βyk)·α

x+βy

=αx¯+βy¯

=αT(x) +βT(y),

thenT is the linearly isometric extension ofT0to whole space`(p2).

SinceT(θ) =θ, the linearity about theT can also be proved by the Mazur-Ulam theorem.

Conflict of Interests

The authors declare that there is no conflict of interests.

Acknowledgements

This work is supported by NSFC (11301384).

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