# A Remark on the Homogeneity of Isosceles Orthogonality

Authors: Chan He; Dan Wang
Journal: Journal of Function Spaces
 (2014)

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## Abstract

Inspired by the definition of homogeneous direction of isosceles orthogonality, we introduce the notion of almost homogeneous direction of isosceles orthogonality and show that, surprisingly, these two notions coincide. Several known characterizations of inner products are improved.


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## Body

## 1. Introduction

Throughout this paper,X=(X,∥·∥) always denotes a real normed linear space with origino, unit ballBX, and unit sphereSX. The dimension of such a space is always assumed to be at least 2.Isosceles orthogonality, an extension of the orthogonality in inner product spaces to normed linear spaces, was introduced by James in [1]. Two vectors (or points) x and y in X are said to be isosceles orthogonal (denoted by x⊥Iy) if and only if
(1)∥x+y∥=∥x-y∥.In the same paper James proved that the implication(2)x,y∈X,α∈ℝ,x⊥Iy⟹x⊥Iαy
characterizes inner product spaces. In other words, isosceles orthogonality is homogeneous if and only if the underlying space is an inner product space. Although implication (2) does not hold in a normed linear space which is not an inner product space, it might hold “locally.” For example, let X be the normed linear space on ℝ2 with the maximum norm and x=(1,1). Then
(3)y∈X,α∈ℝ,x⊥Iy⟹x⊥Iαy.
In fact, in this example one can easily verify that x⊥Iy if and only if y is a multiple of (-1,1). The existence of vectors satisfying implication (3) motivated the authors of [2] to introduce the following definition.Definition 1.
A unit vectorx in X such that implication (3) holds is called a homogeneous direction of isosceles orthogonality.We denote byHX the set of all homogeneous directions of isosceles orthogonality. Note that if X is an inner product space then HX=SX and that HX might be an empty set for some normed linear space X (cf. Example 2.1 in [1]).In the book [3], Amir mentioned the following implication similar to (2) which also characterizes inner product spaces (cf. [3, (4.12)] and [4, I1)]):
(4)thereexistsanumberα∈(0,1)suchthatx,y∈X,hhhhhhhhhhhhhhhhhhhhhhhhhx⊥Iy⟹x⊥Iαy.
A related characterization can be found in [3, (10.13)] (or [5]). By modifying (4) a little bit, we can introduce the notion of almost homogeneous direction of isosceles orthogonality. Note that, in the following definition, we do not need the number α to be chosen uniformly.Definition 2.
Letx be a unit vector in a normed linear space X. If for each y∈X, which is isosceles orthogonal to x, there exists a number α:=α(x,y)∈(0,1) such that x⊥Iαy, then x is called an almost homogeneous direction of isosceles orthogonality.We denote byHX′ the set of all almost homogeneous directions of isosceles orthogonality. It is clear that
(5)x∈HX⟹x∈HX′.
We will show that, surprisingly, the converse of (5) is also true.


## 2. Main Result

We will use also the notion of Birkhoff orthogonality. A vectorx is said to be Birkhoff orthogonal (see [6, 7]) to another vector y if the inequality ∥x+αy∥≥∥x∥ holds for each real number α. For relations between isosceles orthogonality and Birkhoff orthogonality and related results, we refer to the survey [8].Lemma 3 (cf. Section 3 in [7] or Theorems 4.8 and 4.9 in [8]).
Letx≠o and y be two vectors in a normed linear space. Then x⊥Bγx+y if and only if γ∈[α(x,y),β(x,y)], where
(6)α(x,y):=-1∥x∥lim⁡η→0+∥x+ηy∥-∥x∥η,β(x,y):=-1∥x∥lim⁡η→0-∥x+ηy∥-∥x∥η.Therefore, a pointz∈y+〈-x,x〉 satisfies x⊥Bz if and only if
(7)z∈[α(x,y)x+y,β(x,y)x+y].
Clearly, the midpoint of the segment [α(x,y)x+y,β(x,y)x+y] is the unique point z∈y+〈-x,x〉 satisfying α(x,z)+β(x,z)=0.One can easily verify the following lemma.Lemma 4.
Letx and y be two linearly independent vectors in X and λ a positive number. Then
(8)α(x,λy)=λα(x,y),β(x,λy)=λβ(x,y).Lemma 5.
Ifu∈HX′, then, for each point z that is isosceles orthogonal to u, α(u,z)+β(u,z)=0.Proof.
We only consider the nontrivial case whenz≠o. Set
(9)y=12(α(u,z)+β(u,z))u+z.
Then one can verify that
(10)α(u,y)+β(u,y)=0.
In the following we show that y=z, which will complete the proof.
By the construction ofy there exists a real number α0 such that z=y+α0u. Since u∈HX′, there exists a sequence {γn}n=1∞ of positive numbers, which is strictly decreasing and converges to 0, such that u⊥Iγnz for each integer n≥1. Thus we have the equality
(11)∥u+γn(α0u+y)∥=∥u-γn(α0u+y)∥n=1,2,…
or, equivalently,
(12)∥1γnu+(α0u+y)∥=∥1γnu-(α0u+y)∥n=1,2,….
When γn is sufficiently close to 0, one can verify the following:
(13)∥1γnu+(α0u+y)∥-∥1γnu+y∥=∥1γnu+1/γn1/γn+α0y+α0u+α01/γn+α0y∥-∥1γnu+y∥=∥1γnu+1/γn1/γn+α0y∥+α0∥u+11/γn+α0y∥-∥1γnu+y∥.
By combining (13) with the triangle inequalities we obtain
(14)∥1γnu+y∥-|α01/γn+α0|∥y∥+α0∥u+11/γn+α0y∥-∥1γnu+y∥≤∥1γnu+(α0u+y)∥-∥1γnu+y∥≤|1/γn1/γn+α0-1|∥y∥+α0∥u+11/γn+α0y∥,
from which it follows that
(15)lim⁡n→∞(∥1γnu+(α0u+y)∥-∥1γnu+y∥)=α0∥u∥.
Similarly, whenγn is sufficiently close to 0, we have the equality
(16)∥1γnu-(α0u+y)∥-∥1γnu-y∥=∥1γnu-11-α0γny-α0u+α0γn1-α0γny∥-∥1γnu-y∥=∥1γnu-11-α0γny∥-α0∥u-γn1-α0γny∥-∥1γnu-y∥.
From (16) and the triangle inequalities it follows that
(17)∥1γnu-y∥-∥α0γn1-α0γny∥-α0∥u-γn1-α0γny∥-∥1γnu-y∥≤∥1γnu-(α0u+y)∥-∥1γnu-y∥≤∥α0γn1-α0γny∥-α0∥u-γn1-α0γny∥.
From (17) it follows that
(18)-lim⁡n→∞(∥1γnu-(α0u+y)∥-∥1γnu-y∥)=α0∥u∥.
By adding (15) and (18), we obtain
(19)2α0∥u∥=lim⁡n→∞(∥1γnu+(α0u+y)∥-∥1γnu+y∥-∥1γnu-(α0u+y)∥+∥1γnu-y∥)=lim⁡n→∞(∥1γnu-y∥-∥1γnu+y∥)=-lim⁡n→∞(∥u-γny∥-∥u∥-γn+∥u+γny∥-∥u∥γn).
Then it follows from (10) that α0=0. Therefore z=y, as claimed.Our main result is the following theorem.Theorem 6.
Ifu∈HX′, then u∈HX.Proof.
Letz be an arbitrary point that is isosceles orthogonal to u and λ an arbitrary number. We only need to consider the case λ>0. By Lemma 5,
(20)α(u,z)+β(u,z)=0.
Thus
(21)α(u,λz)+β(u,λz)=0.
Moreover, it is clear that λz is the unique point p in λz+〈-u,u〉 such that α(u,p)+β(u,p)=0. Lemma 5 and the existence of a point in λz+〈-u,u〉 that is isosceles orthogonal to u show that u⊥Iλz, and the proof is complete.By Theorems 6 and 2 in [2] we obtain the following corollary, which improves (4.12) and (10.13) in [3]. Corollary 7 is formally stronger than Theorem 2 in [2].Corollary 7.
LetX be a Banach space whose dimension is at least two. If the relative interior of HX′ in SX is not empty, then X is a Hilbert space.



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*Source: 876015-2014-03-02.xml*