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## Function Spaces and Operators with Applications

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Research Article | Open Access

Volume 2014 |Article ID 634082 | 3 pages | https://doi.org/10.1155/2014/634082

# Isometric Reflection Vectors and Characterizations of Hilbert Spaces

Accepted18 Feb 2014
Published18 Mar 2014

#### Abstract

A known characterization of Hilbert spaces via isometric reflection vectors is based on the following implication: if the set of isometric reflection vectors in the unit sphere of a Banach space has nonempty interior in , then is a Hilbert space. Applying a recent result based on well-known theorem of Kronecker from number theory, we improve this by substantial reduction of the set of isometric reflection vectors needed in the hypothesis.

#### 1. Introduction

Throughout this paper always denotes a real Banach space with origin , unit ball , and unit sphere . The dimension of such a space is always assumed to be at least . When , is called a Minkowski plane. For two linearly independent vectors and in , we denote by the Minkowski plane spanned by and .

Following , a closed linear subspace of is said to be an isometric reflection subspace of if there exists a linear subspace of such that and that holds (i.e., is isosceles orthogonal to ; see, e.g., ) for every and every . In this situation the linear map defined by , where and , is a surjective linear isometry and is called the isometric reflection of in . In particular, if an isometric reflection subspace is spanned by a unit vector , then such a unit vector is called an isometric reflection vector. In this case the isometric reflection is simply denoted by . The set of isometric reflection vectors in is denoted by . The notion of isometric reflection vector is closely related to another type of orthogonality in normed linear spaces: a vector is said to be Roberts orthogonal to another vector (denoted by ) if holds for each real number . It is not difficult to verify the following lemma.

Lemma 1 (cf. [3, 4]). A unit vector lies in if and only if there exists a hyperplane passing through the origin such that . Moreover, if lies in , then the hyperplane is uniquely determined.

For more information concerning geometric properties of isometric reflection vectors we refer to the recent papers [3, 5, 6].

Becerra Guerrero and Rodriguez Palacios  proved the following interesting characterization of Hilbert spaces (see  for a different proof).

Theorem 2 (cf. [1, 4]). If has nonempty relative interior in , then is a Hilbert space.

Our aim is to improve Theorem 2. Before stating our main result, we still need some definitions and notation about arc lengths.

Let and be two unit vectors in a Minkowski plane . If , then the set is called the (minor) arc connecting to and denoted by . The length of is denoted by . If then is set to be , where is the circumference of . We will also make use of directed arc length. Fix an orientation on . Let and be two distinct points in . The part of connecting to in the positive orientation is called the directed arc connecting to , and its length is called the directed arc length from to and denoted by . See [7, p. 112] or the survey paper  for the definitions of arc length and circumference. For an arbitrary number we denote by and the two points in such that Moreover, we define and, for an integer ,

The following theorem is our main result.

Theorem 3. If there exists a vector such that each two-dimensional subspace of containing contains a vector such that is irrational and less than 1/4, then is a Hilbert space.

Compared with Theorem 2, the cardinality of the set of isometric reflection vectors involved in the hypothesis of Theorem 3 is substantially reduced.

#### 2. Proof of Theorem 3

The following result is well known in number theory.

Lemma 4 (Kronecker’s Theorem, cf. [9, Theorem 439, p. 376]). If is an irrational number, then the set of all numbers of the form where is the largest integer which does not exceed , is dense in .

Based on the above lemma, Martini and Wu  proved the following lemma, which is our main tool.

Lemma 5 (cf. ). Let be a Minkowski plane. Then, for any point and any irrational number , the set is dense in .

Lemma 6. Let be a unit vector in and be a linear isometry on . Then .

Proof. Since , there exists a hyperplane passing through the origin and satisfying . It is clear that is also a hyperplane passing through the origin. For each vector and each number , we have the following equalities: This implies that . Thus .

Before proving Theorem 3 we list some elementary properties of the isometric reflection.

Lemma 7. Let be an isometric reflection vector. Then the following statements hold true.(1) holds for each .(2)For each , the midpoint of and is contained in the one-dimensional space spanned by and the difference between them lies in .(3)For two unit vectors satisfying , maps to preserving length.(4)For each vector in , (5)The image of under is contained in .

Proof. By Lemma 1, there exists a unique hyperplane passing through the origin such that . Then, for each point , there exist a unique number and a unique point such that It is clear that . Thus (1) and (2) are both true. (3) follows directly from the fact that is a linear isometry. (5) is a direct corollary of Lemma 6. It remains to prove (4).
Clearly, Moreover, (3) implies that , from which (4) follows.

The following lemma is a direct corollary of (3.3′) in , and we omit the proof.

Lemma 8. Let be a Banach space. If , then is a Hilbert space.

Proof of Theorem 3. We only need to show that, under the hypothesis of the theorem, each unit vector is an isometric reflection vector.
We may assume that and are linearly independent, since otherwise there is nothing to prove. Let and fix an orientation on . Denote by the point in such that the orientation from to is . By the hypothesis of the theorem, there exists an isometric reflection vector such that is irrational and less than . Suppose that , where and are two real numbers. Then . Replacing by , which is also an isometric reflection vector, we may, if necessary, also require that . Then .
Put Then, since is the unique vector in such that , we have Since maps to and preserves arc length, we have Thus , , and are all isometric reflection vectors. We continue by setting On the one hand, since maps to , Similarly, Now and are both isometric reflection vectors of . By repeating such constructions, we can obtain a set of isometric reflection vectors of in such that, for each , From Lemma 5 it follows that the set is dense in . Thus lies in the closure of . Since the set of isometric reflection vectors in is closed, this implies that itself is an isometric reflection vector. This completes the proof.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publications of this paper.

#### Acknowledgments

The corresponding author is partially supported by a 973 program (Grant no. 2013CB834201) and both of the authors are supported by a foundation from the Ministry of Education of Heilongjiang Province (Grant no. 1251H013), the Nature Science Foundation of Heilongjiang Province (Grant no. A201011), the National Nature Science Foundation of China (Grant nos. 11371114 and 11171082), China Postdoctoral Science Foundation (Grant nos. 2012M520097 and 2013T60019), and by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

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