# On Isometric Extension in the Spacesn(H)

Authors: Xiaohong Fu
Journal: Journal of Function Spaces
 (2014)

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## Abstract

We study the problem of isometric extension on a sphere of the spacesn(H). We give an affirmative answer to Tingley’s problem in the space sn(H).


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## Body

## 1. Introduction

LetE and F be metric linear spaces. A mapping V:E→F is called an isometry if dF(Vx,Vy)=dE(x,y) for all x,y∈E. The classical Mazur-Ulam theorem in [1] describes the relation between isometry and linearity and states that every onto isometry V between two normed spaces with V(0)=0 is linear. In 1987, Tingley [2] posed the problem of extending an isometry between unit spheres as follows.LetE and F be two real Banach spaces. Suppose that V0 is a surjective isometry between the two unit spheres S(E) and S(F). Is V0 necessarily a restriction of a linear or affine transformation to S(E)?It is very difficult to answer this question, even in two-dimensional cases. In the same paper, Tingley proved that ifE and F are finite-dimensional Banach spaces and V0:S(E)→S(F) is a surjective isometry, then V0(x)=-V0(-x) for all x∈S(E). In [3], Ding gave an affirmative answer to Tingley’s problem, when E and F are Hilbert spaces. Kadets and Martín in [4] proved that any surjective isometry between unit spheres of finite-dimensional polyhedral Banach spaces has a linear isometric extension on the whole space. In the case when E and F are some metric vector spaces, the corresponding extension problem was investigated in [5, 6]. See also [7–14] for some related results.We introduce a new spaces(H) which consists of all H-valued sequences, where H is a Hilbert space, and, for each element x={x(k)}, the F-norm of x is defined by ∥x∥=∑k=1∞(1/2k)(∥x(k)∥/(1+∥x(k)∥)). Let sn(H) denote the set of all elements of the form x=(x(1),…,x(n)) with ∥x∥=∑k=1n(1/2k)(∥x(k)∥/(1+∥x(k)∥)), where x(i)(i=1,…,n) is an element in the Hilbert space H.In this paper, we study the problem of isometric extension on a sphereSr0(sn(H)) with radius r0 and center 0 in sn(H). We prove that if V0 is an isometric mapping from Sr0(sn(H)) onto itself, then it can be extended to an isometry on the whole space sn(H).Here is a notation used throughout this paper:(1)ex(k)=(0,…,x(k),…,0)︸kthplace∈sn(H),
where ∥x(k)∥=1. Particularly, when ∥x(k)∥=0, we define ex(k)/∥x(k)∥=(0,…,0).


## 2. Main Results and Proofs

In this section, we give our main results. For this purpose, we need some lemmas that will be used in the proofs of our main results. We begin with the following result.Lemma 1.
Ifx,y∈s(H), then
(2)∥x-y∥=∥x∥+∥y∥⟺supp⁡x∩supp⁡y=∅,
where supp⁡x={n:x(n)≠0,n∈N}.Proof.
The sufficiency is trivial. In the following, we prove the necessity.
Suppose thatx={x(n)} and y={y(n)} are elements in s(H) and that ∥x-y∥=∥x∥+∥y∥. Then
(3)∑n=1∞‍12n∥x(n)-y(n)∥1+∥x(n)-y(n)∥=∑n=1∞‍12n∥x(n)∥1+∥x(n)∥+∑n=1∞‍12n∥y(n)∥1+∥y(n)∥.
In view of (3), it is sufficient to show that
(4)∥x(n)-y(n)∥1+∥x(n)-y(n)∥≤∥x(n)∥1+∥x(n)∥+∥y(n)∥1+∥y(n)∥,
and the equality holds if and only if ∥x(n)∥∥y(n)∥=0.
Indeed, sincex/(1+x) is strictly increasing on [0,∞), this lemma is proven.Lemma 2.
LetSr0(s(H)) be a sphere with radius r0 and center 0 in s(H). Suppose that V0:Sr0(s(H))→Sr0(s(H)) is a surjective isometry; then (supp⁡x)∩(supp⁡y)=∅ if and only if (supp⁡V0x)∩(supp⁡V0y)=∅.Proof.
Necessity. Take any two disjoint elements x and y in Sr0(s(H)). Let V0(x)={x′(n)} and V0(y)={y′(n)}.
SinceV0 is an isometry, we have by Lemma 1 and (4) that
(5)2r0=∥x∥+∥y∥=∥x-y∥=∥V0x-V0y∥=∑n=1∞‍12n∥x′(n)-y′(n)∥1+∥x′(n)-y′(n)∥≤∑n=1∞‍12n∥x′(n)∥1+∥x′(n)∥+∑n=1∞‍12n∥y′(n)∥1+∥y′(n)∥=2r0.
Thus,
(6)∥V0x-V0y∥=∥V0x∥+∥V0y∥.
According to Lemma 1 again, we obtain
(7)(supp⁡V0x)∩(supp⁡V0y)=∅.
The proof of sufficiency is similar to that of necessity becauseV0-1 is also an isometry from Sr0(s(H)) onto itself.Remark 3.
The spaces(H) in Lemmas 1 and 2 can be replaced by the space sn(H).Lemma 4.
LetSr0(sn(H)) be a sphere with radius r0 in the space sn(H), where r0<1/2n. Suppose that V0:Sr0(sn(H))→Sr0(sn(H)) is an isometry, λk=2kr0/(1-2kr0)(k∈N,1≤k≤n), and ex(k)∈sn(H)(∥x(k)∥=1). Then there exists x′(k)∈H(∥x′(k)∥=1) such that V0(λkex(k))=λkex′(k) and V0(-λkex(k))=-λkex′(k).Proof.
We prove first that, for anyk(1≤k≤n), there exist l(1≤l≤n) and x′(l)(∥x′(l)∥=1) such that V0(λkex(k))=λlex′(l) (notice that the assumption of λk implies λkex′(k)∈Sr0(sn(H))). To this end, suppose on the contrary that V0(λk0ex(k0))=∑k=1nηkex′(k) and ηk1≠0, ηk2≠0. In view of Lemma 2, we have
(8)[supp⁡V0(λk0ex(k0))]∩[supp⁡V0(λkex(k))]=∅,hhhhhhhhhhhhhhhhhhhhhhhhhhhh∀k≠k0.
Hence, by the “pigeon nest principle” there must exist ki0(1≤ki0≤n) such that V0(λki0ex(ki0))=0, which leads to a contradiction.
Next, we prove that ifV0(λkex(k))=λlex′(l), V0(-λkex(k))=λpex′′(p), then l=p.
Indeed, ifl≠p, we have
(9)∥V0(λkex(k))-V0(-λkex(k))∥=∥2λkex(k)∥=12k2λk1+2λk≠2r0,∥V0(λkex(k))-V0(-λkex(k))∥=∥λlex′(l)-λpex′′(p)∥=2r0,
and this contradiction implies l=p.
Finally, we assert that, for anyk(1≤k≤n), there exists x′(k) such that V0(λkex(k))=λkex′(k) and that V0(-λkex(k))=-λkex′(k).
Indeed, ifV0(λkex(k))=λlex′(l), by the result in the last step, we have V0(-λkex(k))=λlex′′(l); thus
(10)∥V0(λkex(k))-V0(-λkex(k))∥=∥2λkex(k)∥=12k2λk1+2λk,∥V0(λkex(k))-V0(-λkex(k))∥=∥λlex′(l)-λlex′′(l)∥=12lλl∥x′(l)-x′′(l)∥1+λl∥x′(l)-x′′(l)∥.
Therefore,
(11)12k2λk1+2λk=12lλl∥x′(l)-x′′(l)∥1+λl∥x′(l)-x′′(l)∥≤12l2λl1+2λl.
It follows that
(12)2r01+2kr0≤2r01+2lr0,
and so k≥l. Applying the “pigeon nest principle” again, we have k=l. Thus ∥x′(l)-x′′(l)∥=2. Since x′(l) and x′′(l) are elements in Hilbert space H, we have x′(l)=-x′′(l).Lemma 5.
Suppose thatx1 and y1 are elements in the Hilbert space H, λ and μ are some nonzero real numbers, and ∥λx1±μy1∥=∥λx2±μy2∥, ∥x1∥=∥x2∥, and ∥y1∥=∥y2∥. Then ∥x1-y1∥=∥x2-y2∥.Proof.
It is easy to prove this lemma by the parallelogram law.Now we are in a position to state the main result and proof in this paper.Theorem 6.
LetSr0(sn(H)) be a sphere with radius r0 in the space sn(H), where r0<1/2n. Suppose that V0:Sr0(sn(H))→Sr0(sn(H)) is a surjective isometry. Then V0 can be extended to an isometry on the whole space sn(H).Proof.
Letx={x(i)}i=1n∈Sr0(sn(H)). For i and j are points in supp⁡x such that i≠j, it follows from Lemma 2 that
(13)supp⁡V0(λiex(i)/∥x(i)∥)⋂‍supp⁡V0(λjex(j)/∥x(j)∥)=∅.
Since V0 is surjective, there is an element z={z(i)}i=1n∈Sr0(sn(H)) such that
(14)V0(z)=∑i=1n‍∥x(i)∥λiV0(λiex(i)/∥x(i)∥).
If ∥x(i)∥=0, then V0(λiex(i)/∥x(i)∥)=def0 and ex(i)/∥x(i)∥=def0. In the following, we explain why the right-hand side of (14) is norm r0.
By Lemma4, we can see that, for any i(1≤i≤n), there exists x′(i) such that ∥x(i)∥=∥x′(i)∥ and
(15)V0(λiex(i)/∥x(i)∥)=λiex′(i)/∥x′(i)∥.
So
(16)∥∑i=1n‍∥x(i)∥λiV0(λiex(i)/∥x(i)∥)∥=∥∑i=1n‍∥x(i)∥λiλiex′(i)/∥x′(i)∥∥=∥∑i=1n‍ex′(i)∥=∑i=1n‍12i∥x′(i)∥1+∥x′(i)∥=r0.
SinceV0 is an isometry, we have
(17)∥z-λiex(i)/∥x(i)∥∥=∥V0(z)-V0(λiex(i)/∥x(i)∥)∥=∥∑i=1n‍∥x(i)∥λiV0(λiex(i)/∥x(i)∥)-V0(λiex(i)/∥x(i)∥)∥=∥∑j≠i‍∥x(j)∥λjV0(λjex(j)/∥x(j)∥)kkk∑j≠i‍+(∥x(i)∥λi-1)V0(λiex(i)/∥x(i)∥)∥=∥∑j≠i‍∥x(j)∥λjλjex′(j)/∥x′(j)∥kkkkkk∑j≠i‍+(∥x(i)∥λi-1)λiex′(i)/∥x′(i)∥∥=r0-12i∥x′(i)∥1+∥x′(i)∥+12iλi-∥x(i)∥1+λi-∥x(i)∥=r0-12i∥x(i)∥1+∥x(i)∥+12iλi-∥x(i)∥1+λi-∥x(i)∥.
On the other hand
(18)∥z-λiex(i)/∥x(i)∥∥=∑j≠i‍12j∥z(j)∥1+∥z(j)∥+∥∥z(i)∥ez(i)/∥z(i)∥-λiex(i)/∥x(i)∥∥≥r0-12i∥z(i)∥1+∥z(i)∥+12iλi-∥z(i)∥1+λi-∥z(i)∥.
Given (17) and (18) and the fact that f(x)=-(x/(1+x))+(λi-x)/(1+λi-x) is decreasing on [0,+∞), we have
(19)∥x(i)∥≤∥z(i)∥.
By (14) and Lemma 4, we get
(20)∑i=1n‍12i∥z(i)∥1+∥z(i)∥=∥z∥=∥V0z∥=∥∑i=1n‍∥x(i)∥λiV0(λiex(i)/∥x(i)∥)∥=∑i=1n‍12i∥x(i)∥1+∥x(i)∥.
Combining (19) and (20) implies
(21)∥x(i)∥=∥z(i)∥.
It follows from (17), (18), and (21) that
(22)∥∥z(i)∥ez(i)/∥z(i)∥-λiex(i)/∥x(i)∥∥=12iλi-∥z(i)∥1+λi-∥z(i)∥
and consequently
(23)x(i)=z(i).
That is,
(24)V0x=∑i=1n‍∥x(i)∥λiV0(λiex(i)/∥x(i)∥).
We now define a mapping on the spacesn(H) as follows:
(25)V({x(i)}i=1i=n)=def∑i=1n‍∥x(i)∥λiV0(λiex(i)/∥x(i)∥)
for all {x(i)}i=1i=n∈sn(H). If ∥x(i)∥=0, then V0(λiex(i)/∥x(i)∥)=def0.
Suppose that{x(i)}i=1i=n and {y(i)}i=1i=n are elements in sn(H). By Lemma 4, (24), and (25), we can assume
(26)V({x(i)}i=1i=n)={x′(i)}i=1i=n,V({y(i)}i=1i=n)={y′(i)}i=1i=n,
where ∥x(i)∥=∥x′(i)∥ and ∥y(i)∥=∥y′(i)∥.
To prove(27)∥V({x(i)}i=1i=n)-V({y(i)}i=1i=n)∥=∥{x(i)}i=1i=n-{y(i)}i=1i=n∥,
we proceed as follows.
SinceV0 is an isometry, it follows from Lemma 4 that
(28)∥V0(λiex(i)/∥x(i)∥)±V0(λiey(i)/∥y(i)∥)∥=∥λiex(i)/∥x(i)∥±λiey(i)/∥y(i)∥∥=12iλi∥x(i)/∥x(i)∥±y(i)/∥y(i)∥∥1+λi∥x(i)/∥x(i)∥±y(i)/∥y(i)∥∥.
On the other hand
(29)∥V0(λiex(i)/∥x(i)∥)±V0(λiey(i)/∥y(i)∥)∥=∥λiex′(i)/∥x′(i)∥±λiey′(i)/∥y′(i)∥∥=12iλi∥x′(i)/∥x′(i)∥±y′(i)/∥y′(i)∥∥1+λi∥x′(i)/∥x′(i)∥±y′(i)/∥y′(i)∥∥.
If ∥x(i)∥=0, then x(i)/∥x(i)∥=def0 and x′(i)/∥x′(i)∥=def0. It follows from (28) and (29) that
(30)∥x(i)∥x(i)∥±y(i)∥y(i)∥∥=∥x′(i)∥x′(i)∥±y′(i)∥y′(i)∥∥.
Notice that∥x(i)∥=∥x′(i)∥ and ∥y(i)∥=∥y′(i)∥ and notice (30); it follows from Lemma 5 that
(31)∥x(i)-y(i)∥=∥x′(i)-y′(i)∥.
Since(32)∥V({x(i)}i=1i=n)-V({y(i)}i=1i=n)∥=∑i=1n‍12i∥x′(i)-y′(i)∥1+∥x′(i)-y′(i)∥,∥{x(i)}i=1i=n-{y(i)}i=1i=n∥=∑i=1n‍12i∥x(i)-y(i)∥1+∥x(i)-y(i)∥.
Equations (31) and (32) assure that (27) holds. That is, we have obtained an isometry on the space sn(H) and it is the extension of V0.



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