Isometric Reflection Vectors and Characterizations of Hilbert Spaces

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 S X of a Banach space X has nonempty interior in S X , then X 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.


Introduction
Throughout this paper  = (, ‖ ⋅ ‖) always denotes a real Banach space with origin , unit ball   , and unit sphere   .The dimension dim  of such a space is always assumed to be at least 2. When dim  = 2,  is called a Minkowski plane.For two linearly independent vectors  and  in , we denote by (, ) the Minkowski plane spanned by  and .
Following [1], 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., [2]) 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] For more information concerning geometric properties of isometric reflection vectors we refer to the recent papers [3,5,6]. Becerra Guerrero and Rodriguez Palacios [4] proved the following interesting characterization of Hilbert spaces (see [1] for a different proof).
Our aim is to improve Theorem 2. Before stating our main result, we still need some definitions and notation about arc lengths.
Let  and V be two unit vectors in a Minkowski plane .
is called the (minor) arc connecting  to V and denoted by arc(, V).The length of arc(, V) is denoted by where |  | is the circumference of   .We will also make use of directed arc length.Fix an orientation  on .Let  and V be two distinct points in   .The part of   connecting  to V in the positive orientation is called the directed arc connecting  to V, and its length is called the directed arc length from  to V and denoted 2 Journal of Function Spaces by  →   (, V).See [7, p. 112] or the survey paper [8] for the definitions of arc length and circumference.For an arbitrary number  ∈ (0, 1) we denote by   () and  −1  () the two points in   such that Moreover, we define  0  () :=  and, for an integer  ≥ 1, The following theorem is our main result.
Theorem 3. If there exists a vector  ∈    such that each twodimensional subspace  of  containing  contains a vector V ∈    such that   (, V)/| (,V) | 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.

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 (0, 1).
Based on the above lemma, Martini and Wu [10] proved the following lemma, which is our main tool.
Lemma 5 (cf.[10]).Let  be a Minkowski plane.Then, for any point  ∈   and any irrational number  ∈ (0, 1), the set 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 (V) ∈ (  ) and each number  ∈ R, 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.
(2) For each  ∈ , the midpoint of  and   () is contained in the one-dimensional space spanned by  and the difference between them lies in   .
(4) For each vector V 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   (, V) =   (, −V), from which (4) follows.
The following lemma is a direct corollary of (3.3  ) in [11], 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 V ∈  such that   (, V) is irrational and less than (1/4)|  |.Suppose that V =  +   , where  and  are two real numbers.Then  > 0. Replacing V by   (V), which is also an isometric reflection vector, we may, if necessary, also require that  > 0. Then V ∈ arc(,   ). Put Then, since V is the unique vector in   such that  →   (, V) = |  |, we have Since   maps arc(, V) to arc(,   (V)) and preserves arc length, we have Thus  0 ,  1 , and  −1 are all isometric reflection vectors.We continue by setting On the one hand, since  V maps arc(, V) to arc(V,  V ()), Similarly, (15) From Lemma 5 it follows that the set {  :  ∈ Z} is dense in   .Thus  lies in the closure of {  :  ∈ Z}.Since the set of isometric reflection vectors in  is closed, this implies that  itself is an isometric reflection vector.This completes the proof.
).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.