Abstract
We study the problem of isometric extension on a sphere of the space . We give an affirmative answer to Tingley’s problem in the space .
1. Introduction
Let and be metric linear spaces. A mapping is called an isometry if for all . The classical Mazur-Ulam theorem in [1] describes the relation between isometry and linearity and states that every onto isometry between two normed spaces with is linear. In 1987, Tingley [2] posed the problem of extending an isometry between unit spheres as follows.
Let and be two real Banach spaces. Suppose that is a surjective isometry between the two unit spheres and . Is necessarily a restriction of a linear or affine transformation to ?
It is very difficult to answer this question, even in two-dimensional cases. In the same paper, Tingley proved that if and are finite-dimensional Banach spaces and is a surjective isometry, then for all . In [3], Ding gave an affirmative answer to Tingley’s problem, when and 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 and 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 space which consists of all -valued sequences, where is a Hilbert space, and, for each element , the -norm of is defined by . Let denote the set of all elements of the form with , where is an element in the Hilbert space .
In this paper, we study the problem of isometric extension on a sphere with radius and center 0 in . We prove that if is an isometric mapping from onto itself, then it can be extended to an isometry on the whole space .
Here is a notation used throughout this paper: where . Particularly, when , we define .
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. If , then where .
Proof. The sufficiency is trivial. In the following, we prove the necessity.
Suppose that and are elements in and that . Then
In view of (3), it is sufficient to show that
and the equality holds if and only if .
Indeed, since is strictly increasing on , this lemma is proven.
Lemma 2. Let be a sphere with radius and center 0 in . Suppose that is a surjective isometry; then if and only if .
Proof. Necessity. Take any two disjoint elements and in . Let and .
Since is an isometry, we have by Lemma 1 and (4) that
Thus,
According to Lemma 1 again, we obtain
The proof of sufficiency is similar to that of necessity because is also an isometry from onto itself.
Remark 3. The space in Lemmas 1 and 2 can be replaced by the space .
Lemma 4. Let be a sphere with radius in the space , where . Suppose that is an isometry, , and . Then there exists such that and .
Proof. We prove first that, for any , there exist and such that (notice that the assumption of implies . To this end, suppose on the contrary that and , . In view of Lemma 2, we have
Hence, by the “pigeon nest principle” there must exist such that , which leads to a contradiction.
Next, we prove that if , , then .
Indeed, if , we have
and this contradiction implies .
Finally, we assert that, for any , there exists such that and that .
Indeed, if , by the result in the last step, we have ; thus
Therefore,
It follows that
and so . Applying the “pigeon nest principle” again, we have . Thus . Since and are elements in Hilbert space , we have .
Lemma 5. Suppose that and are elements in the Hilbert space H, and are some nonzero real numbers, and , , and . Then .
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. Let be a sphere with radius in the space , where . Suppose that is a surjective isometry. Then can be extended to an isometry on the whole space .
Proof. Let . For and are points in such that , it follows from Lemma 2 that
Since is surjective, there is an element such that
If , then and . In the following, we explain why the right-hand side of (14) is norm .
By Lemma 4, we can see that, for any , there exists such that and
So
Since is an isometry, we have
On the other hand
Given (17) and (18) and the fact that is decreasing on , we have
By (14) and Lemma 4, we get
Combining (19) and (20) implies
It follows from (17), (18), and (21) that
and consequently
That is,
We now define a mapping on the space as follows:
for all . If , then .
Suppose that and are elements in . By Lemma 4, (24), and (25), we can assume
where and .
To prove
we proceed as follows.
Since is an isometry, it follows from Lemma 4 that
On the other hand
If , then and . It follows from (28) and (29) that
Notice that and and notice (30); it follows from Lemma 5 that
Since
Equations (31) and (32) assure that (27) holds. That is, we have obtained an isometry on the space and it is the extension of .
Conflict of Interests
The author declares that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
The helpful suggestions from the referees are appreciated. This work is supported by the Project of Department of Education of Guangdong Province (Grant no. 2013KJCX0170).