# Isometric Reflection Vectors and Characterizations of Hilbert Spaces

Authors: Donghai Ji; Senlin Wu
Journal: Journal of Function Spaces
 (2014)

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## 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 sphereS
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.


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## Body

## 1. Introduction

Throughout this paperX
=
(
X
,
∥
·
∥
) always denotes a real Banach space withorigin
o,unit ball
B
X, andunit sphere
S
X. The dimension dim
⁡
X of such a space is always assumed to be at least 2. When dim
⁡
X
=
2, X is called aMinkowski plane. For two linearly independent vectors x and y in X, we denote by L
(
x
,
y
) the Minkowski plane spanned by x and y.Following [1], a closed linear subspace M of X is said to be anisometric reflection subspace of X if there exists a linear subspace N of X such that X
=
M
⊕
N and that ∥
x
+
y
∥
=
∥
x
-
y
∥ holds (i.e., x isisosceles orthogonal to y; see, e.g., [2]) for every x
∈
M and every y
∈
N. In this situation the linear map R
M
:
X
↦
X defined by R
M
(
x
+
y
)
=
x
-
y, where x
∈
M and y
∈
N, is a surjective linear isometry and is called theisometric reflection of M in X. In particular, if an isometric reflection subspace M is spanned by a unit vector u, then such a unit vector is called anisometric reflection vector. In this case the isometric reflection R
M is simply denoted by R
u. The set of isometric reflection vectors in S
X is denoted by I
X
r. The notion of isometric reflection vector is closely related to another type of orthogonality in normed linear spaces: a vector x
∈
X is said to beRoberts orthogonal to another vector y (denoted by x
⊥
R
y) if ∥
x
+
α
y
∥
=
∥
x
-
α
y
∥ holds for each real number α. It is not difficult to verify the following lemma.Lemma 1 (cf. [3, 4]).
A unit vectoru lies in I
X
r if and only if there exists a hyperplane H
u passing through the origin o such that u
⊥
R
H
u. Moreover, if u lies in I
X
r, then the hyperplane H
u 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 [4] proved the following interesting characterization of Hilbert spaces (see [1] for a different proof).Theorem 2 (cf. [1, 4]).
IfI
X
r has nonempty relative interior in S
X, then X is a Hilbert space.Our aim is to improve Theorem2. Before stating our main result, we still need some definitions and notation about arc lengths.Letu and v be two unit vectors in a Minkowski plane X. If u
≠
-
v, then the set
(1)
{
α
u
+
β
v
:
α
,
β
≥
0
}
∩
S
X
is called the(minor) arc connecting
u
to
v and denoted by arc
(
u
,
v
). Thelength of arc
(
u
,
v
) is denoted by δ
X
(
u
,
v
). If u
=
-
v then δ
X
(
u
,
v
) is set to be |
S
X
|
/
2, where |
S
X
| is the circumference of S
X. We will also make use ofdirected arc length. Fix an orientation ω on X. Let u and v be two distinct points in S
X. The part of S
X connecting u to v in the positive orientation is called thedirected arc connecting
u
to
v, and its length is called thedirected arc length from u to v and denoted by δ
X
→
(
u
,
v
). See [7, p. 112] or the survey paper [8] for the definitions of arc length and circumference. For an arbitrary number α
∈
(
0,1
) we denote by A
α
(
u
) and A
α
-
1
(
u
) the two points in S
X such that
(2)
δ
X
→
(
A
α
-
1
(
u
)
,
u
)
=
δ
X
→
(
u
,
A
α
(
u
)
)
=
α
|
S
X
|
.
Moreover, we define A
α
0
(
u
)
∶
=
u and, for an integer n
≥
1,
(3)
A
α
n
(
u
)
∶
=
A
α
(
A
α
n
-
1
(
u
)
)
,
A
α
-
n
(
u
)
∶
=
A
α
-
1
(
A
α
-
n
+
1
(
u
)
)
.The following theorem is our main result.Theorem 3.
If there exists a vectoru
∈
I
X
r such that each two-dimensional subspace L of X containing u contains a vector v
∈
I
X
r such that δ
X
(
u
,
v
)
/
|
S
L
(
u
,
v
)
| is irrational and less than 1/4, then X is a Hilbert space.Compared with Theorem2, the cardinality of the set of isometric reflection vectors involved in the hypothesis of Theorem 3 is substantially reduced.


## 2. Proof of Theorem3

The following result is well known in number theory.Lemma 4 (Kronecker’s Theorem, cf. [9, Theorem 439, p. 376]).
Ifα is an irrational number, then the set of all numbers of the form
(4)
{
n
α
}
∶
=
n
α
-
[
n
α
]
,
where [
n
α
] is the largest integer which does not exceed n
α, 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]).
LetX be a Minkowski plane. Then, for any point u
∈
S
X and any irrational number α
∈
(
0,1
), the set {
A
α
n
(
u
)
}
n
=
-
∞
+
∞ is dense in S
X.Lemma 6.
Letu be a unit vector in I
X
r and T be a linear isometry on X. Then T
(
u
)
∈
I
X
r.Proof.
Sinceu
∈
I
X
r, there exists a hyperplane H
u passing through the origin and satisfying u
⊥
R
H
u. It is clear that T
(
H
u
) is also a hyperplane passing through the origin. For each vector T
(
v
)
∈
T
(
H
u
) and each number α
∈
ℝ, we have the following equalities:
(5)
∥
T
(
u
)
+
α
T
(
v
)
∥
=
∥
u
+
α
v
∥
=
∥
u
-
α
v
∥
=
∥
T
(
u
)
-
α
T
(
v
)
∥
.
This implies that T
(
u
)
⊥
R
T
(
H
u
). Thus T
(
u
)
∈
I
X
r.Before proving Theorem3 we list some elementary properties of the isometric reflection.Lemma 7.
Letu
∈
S
X be an isometric reflection vector. Then the following statements hold true.(1)
R
u
(
R
u
(
x
)
)
=
x holds for each x
∈
X.
(2)
For eachx
∈
X, the midpoint of x and R
u
(
x
) is contained in the one-dimensional space spanned by u and the difference between them lies in H
u.
(3)
For two unit vectorsx
,
y satisfying x
≠
-
y, R
u maps arc
(
x
,
y
) to arc
(
R
u
(
x
)
,
R
u
(
y
)
) preserving arc length.
(4)
For each vectorv in H
u,
(6)
δ
X
(
u
,
v
)
=
1
4
|
S
L
(
u
,
v
)
|
.
(5)
The image ofI
X
r under R
u is contained in I
X
r.Proof.
By Lemma1, there exists a unique hyperplane H
u passing through the origin o such that u
⊥
R
H
u. Then, for each point x
∈
X, there exist a unique number α
(
x
) and a unique point x
H
∈
H
u such that
(7)
x
=
α
(
x
)
u
+
x
H
.
It is clear that R
u
(
x
)
=
α
(
x
)
u
-
x
H. Thus (1) and (2) are both true. (3) follows directly from the fact that R
u is a linear isometry. (5) is a direct corollary of Lemma 6. It remains to prove (4).
Clearly,(8)
S
L
(
u
,
v
)
=
arc
(
u
,
v
)
∪
arc
(
u
,
-
v
)
∪
arc
(
-
u
,
-
v
)
∪
arc
(
-
u
,
v
)
.
Moreover, (3) implies that δ
X
(
u
,
v
)
=
δ
X
(
u
,
-
v
), from which (4) follows.The following lemma is a direct corollary of (3.3′) in [11], and we omit the proof.Lemma 8.
LetX be a Banach space. If I
X
r
=
S
X, then X is a Hilbert space.Proof of Theorem3.
We only need to show that, under the hypothesis of the theorem, each unit vectorx is an isometric reflection vector.
We may assume thatx and u are linearly independent, since otherwise there is nothing to prove. Let L
=
L
(
u
,
x
) and fix an orientation ω on L. Denote by u
H the point in S
L
∩
H
u such that the orientation from u to u
H is ω. By the hypothesis of the theorem, there exists an isometric reflection vector v
∈
L such that δ
X
(
u
,
v
) is irrational and less than (
1
/
4
)
|
S
L
|. Suppose that v
=
γ
u
+
η
u
H, where γ and η are two real numbers. Then γ
>
0. Replacing v by R
u
(
v
), which is also an isometric reflection vector, we may, if necessary, also require that η
>
0. Then v
∈
arc
(
u
,
u
H
).
Put(9)
α
=
δ
X
(
u
,
v
)
|
S
L
|
,
x
0
∶
=
A
α
0
(
u
)
=
u
.
Then, since v is the unique vector in S
L such that δ
X
→
(
u
,
v
)
=
α
|
S
L
|, we have
(10)
x
1
∶
=
A
α
(
u
)
=
v
.
Since R
u maps arc
(
u
,
v
) to arc
(
u
,
R
u
(
v
)
) and preserves arc length, we have
(11)
x
-
1
∶
=
A
α
-
1
(
u
)
=
R
u
(
v
)
=
R
x
0
(
x
1
)
.
Thus x
0, x
1, and x
-
1 are all isometric reflection vectors. We continue by setting
(12)
x
2
∶
=
A
α
(
v
)
=
A
α
2
(
u
)
,
x
-
2
∶
=
A
α
-
1
(
x
-
1
)
=
A
α
-
2
(
u
)
.
On the one hand, since R
v maps arc
(
u
,
v
) to arc
(
v
,
R
v
(
u
)
),
(13)
x
2
=
R
v
(
u
)
=
R
x
1
(
x
0
)
.
Similarly,
(14)
x
-
2
=
R
x
-
1
(
u
)
=
R
x
-
1
(
x
0
)
.
Now x
2 and x
-
2 are both isometric reflection vectors of X. By repeating such constructions, we can obtain a set of isometric reflection vectors {
x
n
:
n
∈
ℤ
} of X in L such that, for each n
≥
2,
(15)
x
n
=
A
α
n
(
u
)
=
R
x
n
-
1
(
x
n
-
2
)
,
x
-
n
=
A
α
-
n
(
u
)
=
R
x
-
n
+
1
(
x
-
n
+
2
)
.
From Lemma 5 it follows that the set {
x
n
:
n
∈
ℤ
} is dense in S
L. Thus x lies in the closure of {
x
n
:
n
∈
ℤ
}. Since the set of isometric reflection vectors in X is closed, this implies that x itself is an isometric reflection vector. This completes the proof.



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*Source: 634082-2014-03-18.xml*