# Homogeneity of isosceles orthogonality and related inequalities

- Cuixia Hao
^{1}Email author and - Senlin Wu
^{2}

**2011**:84

https://doi.org/10.1186/1029-242X-2011-84

© Hao and Wu; licensee Springer. 2011

**Received: **14 May 2011

**Accepted: **11 October 2011

**Published: **11 October 2011

## Abstract

We study the homogeneity of isosceles orthogonality, which is one of the most important orthogonality types in normed linear spaces, from two viewpoints. On the one hand, we study the relation between homogeneous direction of isosceles orthogonality and other notions including isometric reflection vectors and *L*_{2}-summand vectors and show that a Banach space *X* is a Hilbert space if and only if the relative interior of the set of homogeneous directions of isosceles orthogonality in the unit sphere of *X* is not empty. On the other hand, we introduce a geometric constant *NH*_{
X
}to measure the non-homogeneity of isosceles orthogonality. It is proved that 0 ≤ *NH*_{
X
}≤ 2, *NH*_{
X
}= 0 if and only if *X* is a Hilbert space, and *NH*_{
X
}= 2 if and only if *X* is not uniformly non-square.

**Mathematics Subject Classification (2010):**

46B20; 46C15

## Keywords

## 1 Introduction

We denote by *X* a real Banach space with *origin o* and *norm* ||·||, by *B*_{
X
}and *S*_{
X
}the *unit ball* and *unit sphere* of *X*, respectively. When the dimension of *X* is two, *B*_{
X
}and *S*_{
X
}are called the *unit disc* and *unit circle* of *X*, respectively. For two linearly independent points *x* and *y* in *X*, we denote by *X*_{
x,y
}the two-dimensional subspace of *X* spanned by *x* and *y*.

*x*in

*X*is said to be

*isosceles orthogonal to*a vector

*y*in

*X*if the equality ||

*x*+

*y*|| = ||

*x - y*|| holds (we write

*x*⊥

_{ I }

*y*for this situation). James [4] proved that

*X*is a Hilbert space if and only if isosceles orthogonality is homogeneous, i.e., if and only if the implication

*x*⊥

_{ I }

*y*⇒

*x*⊥

_{ I }

*αy*holds for each real number

*α*. For the situation of Birkhoff orthogonality (cf. [5] and [6]), where a vector

*x*is said to be

*Birkhoff orthogonal to y*(denoted by

*x*⊥

_{ B }

*y*) if the inequality ||

*x*+

*αy*|| ≥ ||

*x*|| holds for each real number

*α*, we know that this orthogonality is not symmetric, i.e.,

*x*⊥

_{ B }

*y*does not necessarily imply that

*y*⊥

_{ B }

*x*. We also need the notion of Roberts orthogonality. A vector

*x*is said to be

*Roberts orthogonal to*another vector

*y*(denoted by

*x*⊥

_{ R }

*y*) if the equality ||

*x*+

*αy*|| = ||

*x - αy*|| holds for each real number

*α*(cf. [7]). Roberts orthogonality implies both Birkhoff orthogonality and isosceles orthogonality. More precisely, the implications

*x*⊥

_{R}

*y*⇒

*x*⊥

_{ B }

*y*and

*x*⊥

_{ R }

*y*⇒

*x*⊥

_{ I }

*y*hold. Roberts orthogonality is both homogeneous and symmetric, but it does not have the

*existence*property (cf. Example 2.1 in [4]): there exists a Minkowski plane (i.e., a real two-dimensional Banach space) such that

*x*∈

*S*

_{ X }such that the implication

holds. Such a unit vector *x* is said to be a *homogeneous direction of isosceles orthogonality*. In the following, we denote by *H*_{
X
}the set of all homogeneous directions of isosceles orthogonality in *X*. In Section 2, we study the relation of homogeneous direction of isosceles orthogonality to other notions including isometric reflection vectors and *L*_{2}-summand vectors (see Section 2 for the definitions) and prove a new characterization of Hilbert spaces.

In the meantime, we provide a quantitative characterization of the non-homogeneity of isosceles orthogonality by introducing a new geometric constant *NH*_{
X
}. We show that *NH*_{
X
}= 0 if and only if isosceles orthogonality is homogeneous and *NH*_{
X
}= 2 if and only if the underlying space is not uniformly non-square.

## 2 Homogeneous directions of isosceles orthogonality

First, we study the relation of homogeneous directions of isosceles orthogonality to other notions.

### 2.1 Relations to isometric reflection vectors and *L*_{2}-summand vectors

A *reflection* on *X* is an operator defined as follows: ${T}_{x,{x}^{*}}:z\to z-2{x}^{*}\left(z\right)\cdot x$, where *x* ∈ *X, x** ∈ *X**, and *x**(*x*) = 1. Let *x* be a point in *S*_{X}. If there exists a point ${x}^{*}\in {S}_{{X}^{*}}$ such that the reflection ${T}_{x,{x}^{*}}$ is an isometry then *x* is said to be an *isometric reflection vector* and *x** is said to be the corresponding *isometric reflection functional*. For any isometric reflection vector *x*, there is a unique isometric reflection functional *x** corresponding to it (cf. [8]). For the relation between isometric reflection vectors and Roberts orthogonality, Chan He et al. proved the following lemma.

**Lemma 1 (**[9]

**)**

*Let X be a real Banach space, x*∈

*S*

_{ X }, ${x}^{*}\in {S}_{{X}^{*}}$,

*and*${T}_{x,{x}^{*}}$

*be a reflection. Then*, ${T}_{x,{x}^{*}}$

*is an isometric reflection if and only if*

From Lemma 1, one can see that the notions of "homogeneous direction of isosceles orthogonality" and "isometric reflection vector" are closely connected. One may even expect that these two notions coincide. However, this is not true even when the underlying space is two-dimensional.

*Example 1* Let *X* = (ℝ^{2}, ||·||_{∞}), *x* = (0,1), *y* = (1,0), and *z* = (2,1). Then, it is not difficult to verify that *x* is an isometric reflection vector. However, one can observe that *x* ∉ *H*_{
X
}since *x* ⊥_{
I
}*z* but $x{\perp \u0338}_{I}\left(z\u22152\right)$.

In the meantime, we have the following proposition.

**Proposition 1** *Let X be a Minkowski plane and x* ∈ *H*_{
X
}*. Then, x is an isometric reflection vector*.

*Proof* Let *y* be a point in *S*_{
X
}such that *x* ⊥_{
I
}*y*. Then, *x* ⊥_{
R
}*y* since *x* ∈ *H*_{
X
}. Recall that Roberts orthogonality implies Birkhoff orthogonality. Hence *x* ⊥_{
B
}*y*. Thus, there exists a functional ${x}^{*}\in {S}_{{X}^{*}}$ such that *x**(*x*) = 1 and *x**(*y*) = 0 (cf. [6]). Then, ${T}_{x,{x}^{*}}$ is a reflection and the set *H* := {*z* ∈ *X* : *x**(*z*) = 0} is precisely the line passing through -*y* and *y*. Since Roberts orthogonality is homogeneous, we have *x* ⊥_{
R
}*H*. Then, it follows from Lemma 1 that ${T}_{x,{x}^{*}}$ is an isometry and *x* is an isometric reflection vector. □

Example 1 shows that the converse of Proposition 1 is not true in general, but it holds when the underlying Minkowski plane is strictly convex. More precisely, we have the following proposition.

**Proposition 2** *Let X be a Minkowski plane, x* ∈ *S*_{
X
}*be an isometric reflection vector. If there does not exist a nontrivial line segment contained in S*_{
X
}*and parallel to the line passing through -x and x then x* ∈ *H*_{
X
}.

*Proof* Since *x* is an isometric reflection vector, by Lemma 1, there exists a point *y* ∈ *S*_{
X
}such that *x* ⊥_{
R
}{*αy* : *α* ∈ ℝ}. On the other hand, by the assumption of the proposition and Theorem 2.3 in [10] (see also [11]), for each number *ρ* > 0, there exist precisely two points *p* and -*p* in *X* such that ||*p*|| = *ρ* and *x* ⊥_{
I
}*p* hold. Clearly, these two points have to be the points of intersection of the line {*αy* : *α* ∈ ℝ} and the sphere *ρS*_{
X
}. Thus, for each point *z* ∈ *X* satisfying *x* ⊥_{
I
}*z*, we have *z* = ||*z*|| *y* or *z* = -||*z*|| *y*. Since Roberts orthogonality is homogeneous, this means that *x* ⊥_{
R
}*z*. Thus, *x* ∈ *H*_{
X
}. □

Proposition 1 does not hold in higher dimensional cases. See the following example.

*Example 2* Let *X* = (ℝ^{3}, ||·||_{∞}) and *x* = (1,1,1). Then, *x* ∈ *H*_{
X
}, and it is not an isometric reflection vector.

*Proof* Let *y* be an arbitrary point in *S*_{
X
}such that *x* ⊥_{
I
}*y*. Then, it is clear that *x* and *y* are linearly independent. Next, we show that *x* ⊥_{
R
}*y*.

Assume that *y* = (*α*, *β*, *γ*). Then, since *y* ∈ *S*_{
X
}, max{|*α*|, |*β*|, |*γ*|} = 1. We only deal with the case when |*α*| = 1, and the other two cases can be proved in a similar way.

*y*with -

*y*if it is necessary, we may assume that

*α*= 1. Then,

*x*⊥

_{ I }

*y*it follows that

We only need to consider the subcase when *β* = - 1, and the other subcase when *γ* = -1 can be proved similarly.

*μ*, we have the following equations:

which implies that *x* ⊥_{
R
}*y*.

In the rest of the proof, we show that *x* is not an isometric reflection vector.

*H*passing through the origin

*o*such that

*x*⊥

_{ R }

*H*. Let

*z*

_{1}= (-1, -1, 1) and

*z*

_{2}= (-1, 1, -1). Then, it is clear that

*x*⊥

_{ R }

*z*

_{1}and

*x*⊥

_{ R }

*z*

_{2}. Since

*z*

_{1}is the unique (except for the sign) point in ${S}_{{X}_{x,{z}_{1}}}$ such that

*x*⊥

_{ R }

*z*

_{1}, and

*H*intersects ${X}_{x,{z}_{1}}$, we have that

*z*

_{1}∈

*H*. Similarly,

*z*

_{2}∈

*H*. However, for the point

*w*= (

*z*

_{1}+

*z*

_{2})/2, we have

which imply that $x{\perp \u0338}_{R}w$. This is a contradiction to the fact that *H* is a hyperplane in *X*. □

Nevertheless, we have the following lemma.

**Lemma 2** *If x* ∈ *H*_{
X
}*is a smooth point of S*_{
X
}*then x is an isometric reflection vector, and therefore, x is Roberts orthogonal to a hyperplane*.

*Proof* By Lemma 1, it suffices to show that *x* is Roberts orthogonal to a hyperplane. Since *x* is a smooth point, there exists a unique hyperplane *H* such that *x* ⊥_{
B
}*H*. In the following, we show that *x* ⊥_{
R
}*H*.

For each vector *z* ∈ *H*\{*o*}, there exists a unit vector *z'* ∈ *X*_{
x,z
}such that *x* ⊥_{
I
}*z'*. From the relation *x* ∈ *H*_{
X
}, it follows that *x* ⊥_{
R
}*z'*, which implies *x* ⊥_{
B
}*z'*. Since *x* is a smooth point, either *z*/||*z*|| = *z'* or *z*/||*z*|| = *-z'* holds. Thus, *x* ⊥_{
R
}*z*. The case when *z* = *o* is trivial. □

*M*be a closed subspace of

*X*. If there exists another closed subspace

*N*of

*X*such that

*X*=

*M*⊕

*N*and that, for each pair of points

*m*∈

*M*and

*n*∈

*N*, the equality

holds, then *M* is said to be an *L*_{2}*-summand subspace* (cf. [12]). Note that, when *M* is an *L*_{2}-summand subspace, *N* is uniquely determined. Let *x* be a point in *X*. If the subspace spanned by *x* is an *L*_{2}-summand subspace then *x* is said to be an *L*_{2}*-summand vector*.

**Theorem 1** *Let x* ∈ *S*_{
X
}*be an L*_{2}*-summand vector. Then, x* ∈ *H*_{
X
}.

*Proof*We denote by

*M*the one-dimensional subspace spanned by

*x*, by

*N*the closed subspace of

*X*such that

*X*=

*M*⊕

*N*and that the equality

*m*∈

*M*and

*n*∈

*N*. Let

*y*be an arbitrary point in

*X*such that

*x*⊥

_{ I }

*y*,

*y*

_{ M }∈

*M*, and

*y*

_{ N }∈

*N be*the two points such that

*y*=

*y*

_{ M }+

*y*

_{ N }. Then,

*y*

_{ M }=

*o*. Thus, we have that

*y*=

*y*

_{ N }∈

*N*. Next, we show that, for each point

*z*∈

*N, x*⊥

_{ I }

*z*. Actually, this is an easy consequence of the equations

Since *N* is a linear subspace of *X*, it follows that *x* ⊥_{
R
}*z* holds for each point *z* ∈ *N*. We have shown that, for each point *y* ∈ *X, x* ⊥_{
I
}*y* implies that *x* ⊥_{
R
}*y*, i.e., *x* ∈ *H*_{
X
}. □

### 2.2 A characterization of Hilbert spaces

**Theorem 2** *Let X be a Banach space with* dim*X* ≥ 2. *Then, X is a Hilbert space if and only if the relative interior of H*_{
X
}*in S*_{
X
}*is not empty*.

*Proof* Clearly, if *X* is a Hilbert space then isosceles orthogonality coincides with Roberts orthogonality, which implies that *H*_{
X
}= *S*_{
X
}.

Now assume that the relative interior of *H*_{
X
}in *S*_{
X
}, which is denoted by *P*, is not empty. By Theorem 2.2 in [8], it suffices to show that each point *x* in *P* is an isometric reflection vector. By Lemma 2, we only need to show that each point *x* in *P* is a smooth point.

*x*be an arbitrary point in

*P*. Suppose to the contrary that

*x*is not a smooth point. Then, there exists a two-dimensional subspace

*Y*containing

*x*such that

*x*is not a smooth point of

*S*

_{ Y }. Let

*w*be a point in

*S*

_{ Y }such that

*x*⊥

_{ B }

*w*. Since

*x*is a relative interior point of

*P*, it is also a relative interior point of

*P*∩

*S*

_{ Y }. Thus, there exist two points

*u*and

*v*in

*S*

_{ Y }such that

*x*is a relative interior point (with respect to

*S*

_{ Y }) of the set

*v*∈ arc(

*x*,

*w*) and that each point of arc(

*u*,

*v*)\{

*x*} is a smooth point. The points

*u*and

*v*are also chosen in a way such that there exist two numbers

*α*

_{0}≥ 0,

*β*

_{0}≤ 0 and that the relations

*u*⊥

_{ B }(

*α*

_{0}

*x*+

*w*) and

*v*⊥

_{ B }(

*β*

_{0}

*x*+

*w*) hold. I.e., we assume that the supporting lines of

*B*

_{ Y }at

*u*and

*v*both intersect the line passing through

*w*and parallel to 〈-

*x*,

*x*〉. Let {

*u*

_{ n }} and {

*v*

_{ n }} be two sequences such that

*α*

_{ n }} and {

*β*

_{ n }} such that

*A*and

*B*such that

*S*

_{ Y }is a closed convex curve,

*B*≥

*A*. Now, we have that

*x*is not a smooth point,

*B*>

*A*. Recall that Roberts orthogonality implies Birkhoff orthogonality. Thus,

*x*∈

*H*

_{ X }, we have

Due to the uniqueness property of isosceles orthogonality on the unit sphere, this is impossible. It follows that *x* is a smooth point. □

## 3 A measure of non-homogeneity of isosceles orthogonality

*x*∈

*S*

_{ X }

Now, the following result follows from Theorem 2 and the observation that *x* ∈ *H*_{
X
}if and only if *NH*_{
X
}(*x*) = 0.

**Theorem 3** *Let X be a Banach space with* dim*X* ≥ 2. *If the relative interior of* {*x* ∈ *S*_{
X
}: *NH*_{
X
}(*x*) = 0} *in S*_{
X
}*is not empty then X is a Hilbert space*.

*X*, we have $\sqrt{2}\le J\left(X\right)\le 2$. It is well known that a Banach space

*X*is

*uniformly non-square*if and only if

*J*(

*X*) < 2. Some preliminaries about ultrapower are also necessary. Let $\mathcal{U}$ be an ultrafilter on ℕ. A sequence {

*x*

_{ n }} in

*X converges to x with respect to*$\mathcal{U}$, denoted by $\underset{\mathcal{U}}{lim}{x}_{n}=x$, if, for each neighborhood

*U*of

*x*, $\left\{i\in \mathbb{N}:{x}_{i}\in U\right\}\in \mathcal{U}$. The

*ultrapower*of

*X*, which is denoted by $\tilde{X}$, is the quotient space ${l}_{\infty}\left(X\right)\u2215{N}_{\mathcal{U}}\left(X\right)$ equipped with the quotient norm, where

and $\parallel \tilde{x}\parallel =\underset{\mathcal{U}}{lim}\parallel {x}_{n}\parallel $ for $\tilde{x}={\left({x}_{n}\right)}_{\mathcal{U}}\in \tilde{X}$. For more information about ultra-techniques in Banach space theory, we refer to [16] and [17].

First, we prove the following inequality between *NH*_{
X
}and *J*(*X*).

**Lemma 3**

*Let X be a Banach space with*dim

*X*≥ 2.

*Then*,

*Proof* Let *x* and *y* be two arbitrary unit vectors that are isosceles orthogonal to each other and *α* be an arbitrary positive real number. Without loss of generality, we may assume that ||*x* + *αy*|| ≥ ||*x* - *αy*||. In the following, we distinguish two cases.

**Case 1**: 0 <

*α*≤ 1. It follows from the convexity of

*f*(

*α*) = ||

*x*+

*αy*|| that

**Case 2**:

*α*> 1. By the triangle inequality, we have

The desired inequality now follows directly from the definitions of *NH*_{
X
}and *J*(*X*). □

**Lemma 4** *Let X be a Banach space with* dim*X* ≥ 2. *If NH*_{
X
}= 2 *then J*(*X*) = 2.

*Proof* If *NH*_{
X
}= 2 then 2 = *NH*_{
X
}≤ 2(*J*(*X*) - 1) ≤ 2. Thus *J*(*X*) = 2. □

For the lower and upper bounds of *NH*_{
X
}and *NH*_{
X
}(*x*), we have the following theorem.

**Theorem 4**

*Let X be a Banach space with*dim

*X*≥ 2.

*Then*,

*NH*_{
X
} = 0 *if and only if X is a Hilbert space and NH*_{
X
} = 2 *if and only if X is not uniformly non-square*.

*Proof*To prove the inequalities 0 ≤

*NH*

_{ X }≤ 2, it suffices to show that 0 ≤

*NH*

_{ X }(

*x*) ≤ 2 hold for each

*x*∈

*S*

_{ X }, which follow from the following observation: for each number

*α*> 0,

By Theorem 3, it is clear that if *NH*_{
X
}= 0 then *X* is a Hilbert space. Conversely, if *X* is Hilbert space then isosceles orthogonality coincides with Roberts orthogonality, which implies that *NH*_{
X
}= 0. In the following, we prove that *NH*_{
X
}= 2 if and only if *J*(*X*) = *2*. By Lemma 4, we only need to show the implication *J*(*X*) = 2 ⇒ *NH*_{
X
}= 2.

*J*(

*X*) = 2 holds. Then, there exist {

*x*

_{ n }}, {

*y*

_{ n }} ⊂

*S*

_{ X }such that

*x*

_{ n }⊥

_{ I }

*y*

_{ n }holds for each

*n*and

*α*,

*β*≥ 0 satisfying

*α*+

*β*

*>*0, we have

For each sufficiently large *k* ∈ ℕ and each *n* ∈ ℕ, let *v*_{
n,k
}be a point in the unit circle of ${X}_{{x}_{n},{y}_{n}}$ such that $\mid {v}_{n,k}-{x}_{n}\parallel =\frac{1}{k}$ and that *x*_{
n
}∈ arc(*v*_{
n,k
},*y*_{
n
}); let *u*_{
n,k
}be a point in arc(*x*_{
n
}, *y*_{
n
}) such that *u*_{
n
}_{,}_{
k
}⊥_{
I
}*v*_{
n,k
}. Then, there exist {*α*_{
n,k
}}, {*β*_{
n
}, *k*}, {*γ*_{
n,k
}}, and {*η*_{
n
}, *k*} ⊂ (0, +∞) such that *x*_{
n
}= α_{
n,k
}*v*_{
n,k
}*+ β*_{
n,k
}*y*_{
n
}and *u*_{
n,k
}*= γ*_{
n,k
}*x*_{
n
}+ *η*_{
n,k
}*y*_{
n
}. By extracting subsequences if it is necessary, we may assume that {*u*_{
n,k
}}, {*v*_{
n,k
}}, {*α*_{
n,k
}}, {*β*_{
n,k
}}, {*γ*_{
n,k
}}, and {*η*_{
n,k
}} all converge as *k* tends to infinity.

*k*and $\alpha \le \frac{1}{2k}$, we have

Since 1/*k* tends to 0 when *k* tends to infinity, we have *NH*_{
X
}*= 2*.

## Declarations

### Acknowledgements

The authors are grateful for useful advice from the anonymous referee which leads to a improvement of the proof of Theorem 4. The research of the first named author is supported by a grant from Ministry of Education of Heilongjiang Province supporting overseas returned scholars. The research of the second named author is supported by National Nature Science Foundation of China (grant number 11001068), a foundation from the Ministry of Education of Heilongjiang Province (grant number 11541069), a foundation from Harbin University of Science and Technology (grant number 2009YF028), and by the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry.

## Authors’ Affiliations

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