extension of isometries between the unit spheres of complex lp(Γ)(p>1)spaces

Acta Mathematica Scientia 2014,34B(5):1540–1550

http://actams.wipm.ac.cn

EXTENSION OF ISOMETRIES BETWEEN THE UNIT

SPHERES OF COMPLEX lp(Γ)(p > 1) SPACES∗

Jijin YI (U7)

School of Mathematical Sciences, Xiamen University, Xiamen 361005, China

E-mail : [email protected]

Ruidong WANG (aÀ)

Department of Mathematics, Tianjin University of Technology, Tianjin 300191, China


Xiaoxiao WANG (¡¡)

School of Foreign Language Education, Liaocheng University, Liaocheng 252059, China


Abstract In this paper, we study the extension of isometries between the unit spheres

of complex Banach spaces lp(Γ) and l

p(∆)(p > 1). We first derive the representation of

isometries between the unit spheres of complex Banach spaces lp(Γ) and l

p(∆). Then we

arrive at a conclusion that any surjective isometry between the unit spheres of complex

Banach spaces lp(Γ)and l

p(∆) can be extended to be a linear isometry on the whole space.

Key words isometric mapping; isometric extension; strictly convex

2010 MR Subject Classification 46B04; 46B20; 47A67

1 Introduction

Let (E, dE) and (F, dF ) be metric spaces. A map T : E → F is said to be an isometry if

dF (Tx, T y) = dE(x, y),

whenever x, y ∈ E. In the case of normed spaces E and F , T is an isometry iff

‖Tx− Ty‖ = ‖x− y‖, ∀x, y ∈ E.

In the case where there is a surjective isometry between spaces (E, dE) and (F, dF ), they

are said to be isometric. Isometric spaces are essentially identical to metric spaces. So, the

study of isometries is conducive to understand any mathematical structure of metric spaces.

The study of the isometric theory of metric linear spaces is a very active field of research.The

reference [1] was the most important paper on theory of isometry written by Mazur and Ulum

∗Received February 27, 2013; revised December 12, 2013. The first author is supported by Higher Educa-

tional Science and Technology Program Foundation of Shandong Province (J11LA02), Young and Middle-Aged

Scientists Research Foundation of Shandong Province (BS2010SF004) and Higher Educational Science and Tech-

nology Program Foundation of Shandong Province (J10LA53).

No.5 J.J. Yi et al: ISOMETRIC EXTENSION BETWEEN COMPLEX lp(Γ) 1541

in 1932. In the paper they prove that every isometry from a normed real vector space onto

another normed real vector space is an affine (i.e., a linear transformation composed with a

translation). This is the famous Mazur-Ulum theorem.

Mankiewicz extended this result in [2] showing that an isometry which maps a connected

subset of a normed space E onto an open subset of another normed space F can be extended to

an affine isometry from E to F . Applying these results to the unit balls of E and F , it follows

that the whole spaces E and F are linearly isometric if and only if their unit balls are isometric.

On the other hand, there are subsets of Banach spaces which are isometric, but not in any

sense affinely isometric. Consider the mapping V : R → l∞2 (R2 with the max norm) given by

V (x) = (x, sinx).

Then V is an isometry, but is clearly not affine. Perhaps the reason for this is that the range

of V does not contain sufficient directions of the space. Thus, Tingley raised the following

problem in [3].

Problem Suppose that V0 : S1(E) → S1(F ) is a surjective isometric mapping, does there

exist a linear isometric mapping V : E → F such that V |S1(E) = V0?

In the same paper, Tingley proved that if E and F are finite-dimensional Banach spaces

and V : S1(E) → S1(F ) is a surjective isometry, then V (x) = −V (−x) for all x ∈ S1(E).

Some affirmative results were obtained between the same type of classical Banach spaces,

which was shown in [4–9]. The recent development on this problem you can find in [10]. In

above papers, we find all spaces are real normed spaces. In the complex spaces, the answer to

Tingley’s problem is negative. For example, we take E = F = C (complex plane) and V (x) = x.

But we know the fact complex spaces are also real spaces. When we take E = F = C

(complex plane) and V (x) = x and regard E = F = C as real normed spaces, then Tingley’s

problem is right. So when we study Tingley’s problem, we’ll take complex normed spaces as

real normed spaces. In [4], we studied Tingley’s problem between Banach spaces lp(Γ) (Γ is an

index set) and requires that all components of the elements of lp(Γ) are real numbers. That is

lp(Γ) =

x|x =∑

γ∈Γ

ξγeγ , ξγ ∈ R

,

where eγ = ξγ′ |ξγ = 1, ξγ′ = 0, γ 6= γ′, γ′ ∈ Γ, ∀γ ∈ Γ. When we take complex Banach

spaces lp(Γ) as real spaces, we should allow components of the elements of lp(Γ) to be complex

numbers. Under such a situation, we find it is not easy to get the similar results as those in

[4] when we think about this question deeply. That all components of the elements of lp(Γ)

are real numbers makes the representation of isometries between the unit spheres be obtained

more easily. When we allow components of the elements of lp(Γ) to be complex numbers, the

representation of isometries between the unit spheres will be very complex. You can see this

from Lemma 3.4 in [4] and Lemma 3.1 in this paper.

Why is there such a question? Because we can’t solve Tingley’s problem completely, and

we want to solve Tingley’s problem between some specific space. When we discuss Tingley’s

problem between specific spaces we must distinguish between real and complex space carefully.

By this paper we can see this problem can be solved perfectly when the specific space is complex

lp(Γ)(p > 1).

1542 ACTA MATHEMATICA SCIENTIA Vol.34 Ser.B

In this paper, we allow components of the elements of lp(Γ) and lp(∆)(p > 1, p 6= 2) to be

complex numbers and we call such lp(Γ) and lp(∆)(p > 1, p 6= 2) complex. If components of the

elements of lp(Γ) and lp(∆)(p > 1, p 6= 2) are all real numbers, we can see the results in [4]. we

shall first derive the representation of isometries between the unit spheres of complex Banach

spaces lp(Γ) and lp(∆). Then we show that any surjective isometric mapping T between the

unit spheres of complex normed space lp(Γ) and lp(∆)(p > 1, p 6= 2) can be extended to a real

linear isometry on the whole space. In the whole paper we let p > 1, p 6= 2. If p = 2, there was

an affirmative answer to Tingley’s problem in Bibliography [7].

The results in this paper cover the contents of the reference [4] completely, which makes the

results of reference [4] become a special case of this paper, and you can find this from Lemma

3.3 and Remark 3.5 in this paper.

The core point of this paper is that surjective isometric mapping V0 between the unit

spheres of complex normed space lp(Γ) and lp(∆)(p > 1, p 6= 2) makes V0(αeγ) = α′dσ(γ) (you

can find it in Lemma 3.1 in this paper), and it also gets another surjective isometric mapping

between S1(C) (you can find it in Lemma 3.6 in this paper). To get the final result, we need

Lemma 3.11. Lemma 3.11 can be proved by Lemma 3.9 [7].

2 Some Lemmas

First, let us introduce some lemmas.

Lemma 2.1 ([7]) Let E,F be normed linear spaces, and E be strictly convex, and V0 be

a mapping from the unit sphere S1(E) into S1(F ). If −V0[S1(E)] ⊂ V0[S1(E)] and

‖V0(x1) − V0(x2)‖ ≤ ‖x1 − x2‖, ∀x1, x2 ∈ S1(E),

then V0 is one-to-one, and V0(−x) = −V0(x) for all x ∈ S1(E).

Remark 2.2 From the proof of Lemma 2.1 we can find that it is correct for complex or

real normed space F and complex or real strictly convex space E.

Lemma 2.3 Let E,F be normed linear spaces, and E be strictly convex, and V0 be a

mapping from the unit sphere S1(E) into S1(F ). If −V0[S1(E)] ⊂ V0[S1(E)] and

‖V0(x1) − V0(x2)‖ ≤ ‖x1 − x2‖, ∀x1, x2 ∈ S1(E),

then V0 is one-to-one, and V0(−x) = −V0(x) for all x ∈ S1(E).

Lemma 2.4 Assuming that x, y ∈ lp(Γ), we have

(supp x) ∩ (supp y) = ∅ ⇔ ‖x+ y‖p + ‖x− y‖p = 2(‖x‖p + ‖y‖p)

(supp x = γ|x(γ) 6= 0, γ ∈ Γ, p ≥ 1, p 6= 2).

Lemma 2.5 Suppose that V0 is an isometric mapping from the unit sphere S1[lp(Γ)] onto

S1[lp(∆)] (Γ,∆ are two index sets), p > 1, p 6= 2. Then

(supp x) ∩ (supp y) = ∅ ⇔ (supp V0(x)) ∩ supp (V0(y)) = ∅.

Lemma 2.6 ([4]) Let V0 be the same as in Lemma 2.5. Then⋃

γ∈Γ

supp V0(eγ) = ∆,

where eγ = ξγ′ |ξγ = 1, ξγ′ = 0, γ 6= γ′, γ′ ∈ Γ, ∀γ ∈ Γ.


Remark 2.7 In [4], from the proof of Lemmas 2.4, 2.5, 2.6, we can find that they are

correct for complex normed spaces lp(Γ) and lp(∆).

3 Main Results

In this paper, we let C denote complex number field, S1(C) denote the unit sphere of C(in

space C, we use the modulus of a complex number as norm), and R denotes real number field,

S1(R) denote the unit sphere of R (in space R, we use the absolute value of a real number as

norm). S1(R2) denote the unit sphere of R2, and we take ‖(x, y)‖ = 2

√

x2 + y2 as the norm in

R2. Supposing lp(Γ) and lp(∆)(p > 1, p 6= 2) are complex.

Lemma 3.1 Suppose that lp(Γ) is complex and V0 is an isometric mapping from the unit

sphere S1[lp(Γ)] onto S1[l

p(∆)] , p > 1, p 6= 2. Then there is a bijection σ : Γ → ∆ and we have

V0(αeγ) = α′dσ(γ),

where α, α′ ∈ S1(C), dγ = ηγ′ |ηγ = 1, ηγ′ = 0, γ 6= γ′, γ′ ∈ Γ, ∀γ ∈ Γ.

Proof For ∀α ∈ S1(C), V0(αeγ) ∈ S1[lp(∆)], we know that supp V0(αeγ) 6= ∅. If there

exists some δ ∈ ∆ satisfying δ ∈ supp V0(αeγ), then

V0(αeγ) = V0(αeγ)|δ + V0(αeγ)|∆\δ.

Let

y =V0(αeγ)|δ

‖V0(αeγ)|δ‖.

Then obviously y ∈ S1[lp(∆)]. Since V0 is surjective, we can find x ∈ S1[l

p(Γ)] such that

V0(x) = y. Then according to Lemma 2.5, we have

[supp V0(x)]⋂

[supp V0(eγ′)] = (supp y)⋂

[supp V0(eγ′)]

⊂ [supp V0(αeγ)]⋂

[supp V0(eγ′)] = ∅,

∀γ 6= γ′, γ′ ∈ Γ.

By Lemma 2.5, we get

(supp x)⋂

(supp eγ′) = ∅, ∀γ 6= γ′, γ′ ∈ Γ.

So supp x = γ. And x ∈ S1[lp(Γ)], we can get x = α1eγ . Then

y =V0(αeγ)|δ

‖V0(αeγ)|δ‖= V0(x) = V0(α1eγ).

If there is another δ′ ∈ ∆, δ′ 6= δ satisfying δ′ ∈ supp V0(αeγ), using the same way as above we

can find

y′ =V0(αeγ)|δ′

‖V0(αeγ)|δ′‖= V0(x) = V0(α2eγ).

Using Lemma 2.5 again, we get

(supp y)⋂

(supp y′) = ∅ ⇒ (supp α1eγ)⋂

(supp α2eγ) = ∅.

This is obviously a contradiction. So supp V0(αeγ) is a single point set. Because V0 is an onto

isometry between the unit spheres, ‖V0(αeγ)‖ = 1 = ‖V0(αeγ)|δ‖. Then we have V0(αeγ) =

V0(αeγ)|δ = y = α′dδ.


For ∀α ∈ S1(C), we get V0(αeγ) = α′dδ. By this we define a mapping

σ : Γ → ∆,

γ 7→ δ,

σ(γ) = δ.

From Lemma 2.6, we know σ is a surjection. Then we only need to verify that it is an injective

mapping. In fact, if γ1 6= γ2, σ(γ1) = σ(γ2) = δ, then by Lemma 2.5 we can get

(supp eγ1)⋂

(supp eγ2) = ∅ ⇒ [supp V0(eγ1

)]⋂

[supp V0(eγ2)] = ∅.

This is a contradiction. So σ : Γ → ∆ is a bijection. Thus we get

V0(αeγ) = α′dσ(γ).

Obviously when α ∈ S1(C), then α′ ∈ S1(C). In fact, V0 is an isometric mapping from the

unit sphere S1[lp(Γ)] onto S1[l

p(∆)] and V0(αeγ) = α′dσ(γ). If α′ ∈ S1(R) at all the time, we

can get contradiction easily.

Remark 3.2 Lemma 3.1 indicates that V0 maps each element of αeγ ∈ S1[lp(Γ)], α ∈

S1(C), γ ∈ Γ ⊂ S1[lp(Γ)] into S1[l

p(∆)] still having single-point support set. And all elements

of V0(αeγ), α ∈ S1(C) have the same single-point support set. But we don’t know the precise

relation between α and α′. Maybe you think that α = α′. When you see the preceding example

E = F = C (complex plane) and V (x) = x, you will give up this idea.

If we only allow components of the elements of lp(Γ) (or lp(∆)(p > 1, p 6= 2)) to be real

number, then we can get the following lemma.

Lemma 3.3 Suppose that V0 is an isometric mapping from the unit sphere S1[lp(Γ)] onto

S1[lp(∆)] , p > 1, p 6= 2. Then there is a bijection σ : Γ → ∆ and we have

V0(αeγ) = α′dσ(γ),

where α, α′ ∈ S1(R).

In fact, this lemma is the result in [4] as following.

Lemma 3.4 ([4]) Suppose that V0 is an isometric mapping from the unit sphere S1[lp(Γ)]

onto S1[lp(∆)], p > 1, p 6= 2. Then there is a bijection σ : Γ → ∆ and a set of reals θγ with

absolute value 1, satisfying

V0(eγ) = θγdσ(γ).

Remark 3.5 Although Lemma 3.4 seems the same as Lemma 3.3, they are different in the

process of obtaining. In Lemma 3.4, in fact, we suppose that components of the elements of lp(Γ)

and lp(∆)(p > 1, p 6= 2) are all real numbers. However in Lemma 3.3 we just allow components

of the elements of lp(Γ) (or lp(∆)(p > 1, p 6= 2)) are real numbers. By Lemma 3.1 and Lemma 3.3

we can see when we allow components of the elements of lp(Γ) (or lp(∆)(p > 1, p 6= 2)) are real

numbers, the components of the elements of the other space must be real numbers. And when

we allow components of the elements of lp(Γ) (or lp(∆)(p > 1, p 6= 2)) are complex numbers, the

components of the elements of the other space must be complex numbers (surjective isometries

is a key condition).


Based on the above results, in the following discussion, we suppose lp(Γ) and lp(∆)(p > 1)

are all complex space.

Lemma 3.6 Let V0 be the same as in Lemma 3.1. Then for every γ ∈ Γ there is an

isometric surjection

vγ : S1(C) → S1(C),

vγ(α) = α′,

where α, α′ ∈ S1(C) and V0(αeγ) = α′dσ(γ).

Proof First vγ(α) = α′ is a mapping. In fact, if α1 = α2, then we have V0(α1eγ) =

V0(α2eγ) = α′dσ(γ).

And it is also an injective mapping. If α1 6= α2, we have vγ(α1) = vγ(α2) = α′. Then

V0(α1eγ) = V0(α2eγ) = α′dσ(γ). Because V0 is an onto isometry between the unit spheres, we

have

‖V0(α1eγ) − V0(α2eγ)‖ = ‖(α1 − α2)eγ‖ 6= 0.

This is a contradiction.

Then vγ(α) = α′ is a surjection. This result can be obtained in the condition V0 is an

onto isometry between the unit spheres. For ∀α′ ∈ S1(C), α′dσ(γ) ∈ S1[lp(∆)]. Because V0 is

an isometric mapping from the unit sphere S1[lp(Γ)] onto S1[l

p(∆)], p > 1, p 6= 2, V −10 is an

isometric mapping from the unit sphere S1[lp(∆)] onto S1[l

p(Γ)]. So V −10 (α′dσ(γ)) ∈ S1[l

p(Γ)].

By Lemma 3.1 there is α ∈ S1(C), so as to V −10 (α′dσ(γ)) = αeγ′ . If γ′ 6= γ, then we have

(supp eγ)⋂

(supp eγ′) = ∅ ⇒ [supp V0(eγ)]⋂

[supp V0(eγ′)] = ∅.

This contradicts V −10 (α′dσ(γ)) = αeγ′ and V0(βeγ) = β′dσ(γ). So we have V −1

0 (α′dσ(γ)) = αeγ .

Finally, we prove vγ(α) = α′ is an isometric mapping. For ∀α1, α2 ∈ S1(C), we have

vγ(α1) = α′1, vγ(α2) = α′

2.

Then we get

V0(α1eγ) = α′1dσ(γ), V0(α2eγ) = α′

2dσ(γ).

Since V0 is an onto isometry between the unit spheres, we have

‖V0(α1eγ) − V0(α2eγ)‖ = ‖α1eγ − α2eγ‖ = ‖α′1dσ(γ) − α′

2dσ(γ)‖.

From this equation, we obtain

|α1 − α2| = |α′1 − α′

2| = |vγ(α1) − vγ(α2)|.

Corollary 3.7 Let vγ be the same as in Lemma 3.6. Then we have an isometric surjection

v′γ : S1(R2) → S1(R

2)),

v′γ(x, y) = (x′, y′),

where x+ iy = α, x′ + iy′ = α′ ∈ S1(C) and S1(R2)) denotes the unit sphere of R2().


Proof When we take C as a real normed space, then its elements α = x + yi have the

form (x, y), with the norm

‖(x, y)‖ = 2√

x2 + y2.

Under this situation, by the definition of vγ , we get v′γ is an isometric surjection between the

unit spheres of S1(R2).

Next we shall give the representation theorem of onto isometric mapping between the unit

spheres of complex Banach spaces lp(Γ) and lp(∆).

Theorem 3.8 If V0 is an isometric mapping from the unit sphere S1[lp(Γ)] onto S1[l

p(∆)],

(p > 1, p 6= 2), then there must exist a bijection π : ∆ → Γ such that V0(x) =∑

δ∈∆

ξπ(δ)α′π(δ)dδ,

∀x =∑

γ∈Γ

ξγαγeγ ∈ S1[lp(Γ)]. Where ξγ ≥ 0 and αγ ∈ S1(C) for all γ ∈ Γ.

Proof For any x =∑

γ∈Γ

γeγ ∈ S1[lp(Γ)], γ ∈ C,

x =∑

γ∈Γ

γeγ =∑

γ∈Γ

‖γ‖γ

‖γ‖eγ .

So it must have the form

x =∑

γ∈Γ

ξγαγeγ ∈ S1[lp(Γ)], ξγ ≥ 0, αγ ∈ S1(C).

Let

V0(x) =∑

δ∈∆

ζδβδdδ, ζδ ≥ 0, βδ ∈ S1(C).

By Lemma 3.1 and Lemma 2.1, we have

‖V0(x) + V0(αγ1eγ1

)‖p =

∥

∥

∥

∥

∑

δ∈∆

ζδβδdδ + α′γ1dσ(γ1)

∥

∥

∥

∥

p

=

∥

∥

∥

∥

∑

γ∈Γ

ζσ(γ)βσ(γ)dσ(γ) + α′γ1dσ(γ1)

∥

∥

∥

∥

p

=

∥

∥

∥

∥

∑

γ 6=γ1

ζσ(γ)βσ(γ)dσ(γ) + (ζσ(γ1)βσ(γ1) + α′γ1

)dσ(γ1)

∥

∥

∥

∥

p

= (1 − |ζσ(γ1)βσ(γ1)|p) + |ζσ(γ1)βσ(γ1) + α′

γ1|p

= (1 − |ζσ(γ1)|p) + |ζσ(γ1)βσ(γ1) + α′

γ1|p.

Because V0 is an isometry, we have

‖V0(x) + V0(αγ1eγ1

)‖p = ‖x+ αγ1eγ1

‖p = (1 − |ξγ1|p) + |ξγ1

αγ1+ αγ1

|p.

Then we have an equation

(1 − |ξγ1|p) + |ξγ1

αγ1+ αγ1

|p = (1 − |ζσ(γ1)|p) + |ζσ(γ1)βσ(γ1) + α′

γ1|p.

Immediately we get

−ξpγ1

+ (ξγ1+ 1)p = −ζp

σ(γ1) + |ζσ(γ1)βσ(γ1) + α′γ1|p.

By ‖V0(x)‖p = ‖x‖p we have

∑

γ∈Γ

ξpγ =

∑

δ∈∆

ζpδ .


If ξγ1< ζσ(γ1), then there exists some γ2 such that ξγ2

> ζσ(γ2). Since γ1 is arbitrary, for γ2,

we also have

−ξpγ2

+ (ξγ2+ 1)p = −ζp

σ(γ2) + |ζσ(γ2)βσ(γ2) + α′γ2|p.

Thus

−ξpγ2

+ (ξγ2+ 1)p = −ζp

σ(γ2)+ |ζσ(γ2)βσ(γ2) + α′

γ2|p ≤ −ζp

σ(γ2) + (1 + ζσ(γ2))p.

Since f(t) = (1 + t)p − tp is strictly increasing when p > 1, from ξγ2> ζσ(γ2), we can get

−ξpγ2

+ (ξγ2+ 1)p > −ζp

σ(γ2) + (1 + ζσ(γ2))p.

This is a contradiction. Because γ1 is arbitrary, ξγ ≥ ζσ(γ) for all γ ∈ Γ. If there exists some γ3

such that ξγ3> ζσ(γ3), we’ll get a contradiction. Thus for all γ ∈ Γ, we have ξγ = ζσ(γ). And

we have obtained for arbitrary γ1 ∈ Γ,

−ξpγ1

+ (ξγ1+ 1)p = −ζp

σ(γ1) + |ζσ(γ1)βσ(γ1) + α′γ1|p.

Thus we get

ζσ(γ1) + 1 = ξγ1+ 1 = |ζσ(γ1)βσ(γ1) + α′

γ1| =

∣

∣

∣

∣

βσ(γ1)

α′γ1

ζσ(γ1) + 1

∣

∣

∣

∣

.

Then we getβσ(γ1)

α′γ1

= 1.

So we get

ζσ(γ1)βσ(γ1) = ξγ1α′

γ1

for arbitrary γ1 ∈ Γ. Let π = σ−1, and then we get ζδβδ = ξπ(δ)α′π(δ) for all δ ∈ ∆. Thus we

complete the proof of Theorem 3.8.

In [4], the author requires all components of the elements of those spaces are real numbers.

Under this requirement, ∀x =∑

γ∈Γ

ξγeγ ∈ S1[lp(Γ)] has a simple representation form. And the

isometric mapping V0(x) just makes the coefficients be rearranged and each coefficient multiply

by a real number with absolute value 1.

In Lemma 3.1, we don’t know the precise relation between α and α′. And this relation is

dependent on every γ ∈ Γ, so there are a lot of relations between α(γ) and α′(γ). This thing

will produce the crucial difficulties during solving Tingley’s problem. Corollary 3.7 is prepared

for solving these difficulties. We also need another lemma.

Lemma 3.9 ([7]) Let E,F be Hilbert spaces, and V0 be an isometric mapping from S1(E)

into S1(F ). Then V0 can be extended to a real linear isometric mapping from E into F .

Remark 3.10 This Lemma doesn’t tell us E,F are real or complex. By the preceding

explanation, we always take them as real spaces. So we have the following lemma.

Lemma 3.11 Let vγ be the same as in Lemma 3.6, π be the same as in Theorem 3.8.

Then

|µ · vπ(δ)(α(1)π(δ)) − ν · vπ(δ)(α

(2)π(δ))| = |µ · α

′(1)π(δ) − ν · α

′(2)π(δ)| = |µ · α

(1)π(δ) − ν · α

(2)π(δ)|,

∀µ, , ν ∈ R, δ ∈ ∆.


Proof By the definition of vγ , we have

vπ(δ)(α(1)π(δ)) = α

′(1)π(δ), vπ(δ)(α

(2)π(δ)) = α

′(2)π(δ).

We suppose that

α(1)π(δ) = x1 + iy1, α

(2)π(δ) = x2 + iy2, α

′(1)π(δ) = x′1 + iy′1, α

′(2)π(δ) = x′2 + iy′2.

In the proof of this lemma, ‖ · ‖ denotes the nomed on R2 with the form

‖(x, y)‖ = 2√

x2 + y2.

Then we have

|µ · vπ(δ)(α(1)π(δ)) − ν · vπ(δ)(α

(2)π(δ))| = |µ · α

′(1)π(δ) − ν · α

′(2)π(δ)|

= ‖µ · (x′1, y′1) − ν · (x′2, y

′2)‖,

|µ · vπ(δ)(α(1)π(δ)) − ν · vπ(δ)(α

(2)π(δ))| = |µ · α

′(1)π(δ) − ν · α

′(2)π(δ)|

= ‖µ · (x′1, y′1) − ν · (x′2, y

′2)‖

= ‖µ · v′γ(x1, y1) − ν · v′γ(x2, y2)‖.

From Corollary 3.7, v′γ is an isometric mapping between the unit spheres of S1(R2). By Lemma

3.9, v′γ can be extended to a real linear isometric surjection ψ′γ from R2 onto R2. And we have

ψ′γ |S1(R2) = v′γ . Thus we can obtain

|µ · vπ(δ)(α(1)π(δ)) − ν · vπ(δ)(α

(2)π(δ))| = ‖µ · v′γ(x1, y1) − ν · v′γ(x2, y2)‖

= ‖µ · ψ′γ(x1, y1) − ν · ψ′

γ(x2, y2)‖

= ‖ψ′γ [µ · (x1, y1)] − ψ′

γ [ν · (x2, y2)]‖

= ‖µ · (x1, y1) − ν · (x2, y2)‖

= |µ · α(1)π(δ) − ν · α

(2)π(δ)|.

Theorem 3.12 If V0 is an isometric mapping from the unit sphere S1[lp(Γ)] onto S1[l

p(∆)]

(p > 1, p 6= 2), then V0 can be extended to a real linear isometry from lp(Γ) onto lp(∆).

Proof For each x ∈ lp(Γ), let

V (x) =

‖x‖V0

(

x

‖x‖

)

, if x 6= θ,

θ, if x = θ.

Then V is an extension of V0.

For ∀x1, x2 ∈ lp(Γ), let x1

‖x1‖ =∑

γ∈Γ

ξ(1)γ α

(1)γ eγ and x2

‖x2‖ =∑

γ∈Γ

ξ(2)γ α

(2)γ eγ . Where ξ

(1)γ , ξ

(2)γ ≥

0, α(1)γ , α

(2)γ ∈ S1(C). From Theorem 3.8, we can get that

V

(

xi

‖xi‖

)

= V0

(

xi

‖xi‖

)

=∑

δ∈∆

ξ(i)π(δ)α

′(i)π(δ)dδ, i = 1, 2.

So we have

‖V (x1) − V (x2)‖p =

∥

∥

∥

∥

‖x1‖V0

(

x1

‖x1‖

)

− ‖x2‖V0

(

x2

‖x2‖

)∥

∥

∥

∥

p


= ‖x1‖p ·

∥

∥

∥

∥

V0

(

x1

‖x1‖

)

−‖x2‖

‖x1‖V0

(

x2

‖x2‖

)∥

∥

∥

∥

p

= ‖x1‖p ·

∥

∥

∥

∥

∑

δ∈∆

ξ(1)π(δ)α

′(1)π(δ)dδ −

‖x2‖

‖x1‖

∑

δ∈∆

ξ(2)π(δ)α

′(2)π(δ)dδ

∥

∥

∥

∥

p

= ‖x1‖p ·

∥

∥

∥

∥

∑

δ∈∆

(ξ(1)π(δ)α

′(1)π(δ) −

‖x2‖

‖x1‖ξ(2)π(δ)α

′(2)π(δ))dδ

∥

∥

∥

∥

p

= ‖x1‖p ·

∑

δ∈∆

∣

∣

∣

∣

ξ(1)π(δ)α

′(1)π(δ) −

‖x2‖

‖x1‖ξ(2)π(δ)α

′(2)π(δ)

∣

∣

∣

∣

p

.

By Lemma 3.11,

‖V (x1) − V (x2)‖p = ‖x1‖

p ·∑

δ∈∆

∣

∣

∣

∣

ξ(1)π(δ)α

′(1)π(δ) −

‖x2‖

‖x1‖ξ(2)π(δ)α

′(2)π(δ)

∣

∣

∣

∣

p

= ‖x1‖p ·

∑

δ∈∆

∣

∣

∣

∣

ξ(1)π(δ)α

(1)π(δ) −

‖x2‖

‖x1‖ξ(2)π(δ)α

(2)π(δ)

∣

∣

∣

∣

p

= ‖x1‖p ·

∑

γ∈Γ

∣

∣

∣

∣

ξ(1)γ α(1)γ −

‖x2‖

‖x1‖ξ(2)γ α(2)

γ

∣

∣

∣

∣

p

= ‖x1‖p ·

∥

∥

∥

∥

x1

‖x1‖−

‖x2‖

‖x1‖·x2

‖x2‖

∥

∥

∥

∥

p

= ‖x1 − x2‖p.

Thus V is an isometry from lp(Γ) onto lp(∆). From Mazur-Ulam theorem [1], we also know

that V must be real linear. Now the proof is completed.

When p = 2, it is a simple conclusion of Lemma 3.9.

Remark 3.13 We confirm that all the same questions can be solved by one kind of

common means in the near future. If we can solve Tingley’s problem completely we can avoid

discussing Tingley’s problem between specific spaces. When we study Tingley’s problem, we

can also discuss some kinds of space (such as reflexive space), and you can find some results in

[18, 19].

Acknowledgements We express our deep-hearted gratitude to functional analysis sem-

inar of Xiamen University and Professor Ding Guanggui.

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extension of isometries between the unit spheres of complex lp(Γ)(p>1)spaces

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