Journal of Inequalities and Applications

J Inequal Appl. 2017; 2017(1): 203.
Published online 2017 August 31.
PMCID: PMC5579225

# A new upper bound of geometric constant D(X)

## Abstract

A new constant WD(X) is introduced into any real 2n-dimensional symmetric normed space X. By virtue of this constant, an upper bound of the geometric constant D(X), which is used to measure the difference between Birkhoff orthogonality and isosceles orthogonality, is obtained and further extended to an arbitrary m-dimensional symmetric normed linear space (m ≥ 2). As an application, the result is used to prove a special case for the reverse Hölder inequality.

Keywords: Birkhoff orthogonality, isosceles orthogonality, symmetric normed linear spaces, geometric constant

## Introduction

The notion of orthogonality has many forms when the underlying space is transferred from inner product spaces to real normed spaces. For example, Birkhoff [1] introduced Birkhoff orthogonality in which X is assumed to be a real normed linear space. If  ∥ xλy ∥  ≥  ∥ x ∥ , λ ∈ ℝ, then x is said to be Birkhoff orthogonal to y. It can be written as xBy. James [2] defined isosceles orthogonality, that is, if  ∥ xy ∥  =  ∥ x − y ∥ , then x is said to be isosceles orthogonal to y. It is denoted by xIy. When X is an inner product space, these two types of orthogonality are equivalent to inner-product orthogonality.

However, these two types of orthogonality are different in general linear normed spaces. In order to quantify their difference in a real normed space X, Ji and Wu [3] introduced the geometric constant D(X)

$D(X)=inf{infλ∈R∥x+λy∥:x,y∈S(X),x⊥Iy},$

where S(X) is the unit sphere of X, and obtained the bounds $2(2−1)≤D(X)≤1$. In particular, they provided the value of D(X) in any two-dimensional symmetric Minkowski plane. Recently, the value of the constant D(X) in the normed plane whose unit circle is affine regular (4n)-gon was given in [4], and a new lower bound cB( ⋅ ) of D( ⋅ ) was obtained in [5]. Note that the constant D(X) is considered only in the unit sphere S(X). In reference [6], the author considered two constants BI(X) and IB(X) to measure the difference between Birkhoff orthogonality and isosceles orthogonality in the entire space X:

$BI(X)=sup{∥x+y∥−∥x−y∥∥x∥:x,y∈X,x,y≠0,x⊥By}$

and

$IB(X)=sup{infλ∈R∥x+λy∥∥x∥:x,y∈X,x,y≠0,x⊥Iy}.$

And the estimations 0 ≤ BI(X) ≤ 2 and ½ ≤ IB(X) ≤ 1 were also obtained. Other constants used to measure the difference between Birkhoff orthogonality and Robert orthogonality [7] were studied by [5] and [8]. For more conclusions about the difference between orthogonality types, please refer to literature of references [310] and so on.

In this study, by considering the constant D(X) in 2n-dimensional real symmetric normed linear spaces, we obtain an upper bound WD(X). As we discuss in Corollary 1, this bound can be extended to any m-dimensional symmetric normed linear space (m ≥ 2). This article is organized as follows. In Section 2, we present some notations and definitions. In Section 3, the constant WD(X) is introduced and discussed. In Section 4, we consider WD(X) for the space $lp2n$ and present a special case for the reverse Hölder inequality.

## Preliminaries

Let us fix some notations. Let X be an n-dimensional real linear normed space. By  ∥  ⋅  ∥  and ∥⋅∥, we denote the norm of X and the norm of a dual space X, respectively. The notation S(X) is the unit sphere of X. Let and denote the real field and a positive integer set, respectively.

### Definition

Let X be an n-dimensional real normed linear space. If there exist e1e2, …, en ∈ S(X) such that, for any ai ∈ ℝ, i = 1, 2, …, n, the following equality

$∥|a1|e1+|a2|e2+⋯+|an|en∥=∥a1e1+a2e2+⋯+anen∥=∥∑k=1n±aΦ(i)ei∥$

always holds, where Φ(i) ∈ {1, 2, …, n} and Φ(i) ≠ Φ(j) (if i ≠ j), then we call X a symmetric normed linear space and {e1e2, …, en} a group of symmetric axes of X. In particular, we call $∑k=1n±aΦ(i)ei$ a symmetric element of x.

Let X be an n-dimensional symmetric normed linear space and e1, …, en be a group of symmetric axes. For x ∈ X, x is denoted by the coordinate representation of this group of symmetric axes, i.e., x = (x1, …, xn) = x1e1 +  ⋯  + xnen.

## Main results

Firstly, the following elementary results are presented. Throughout this paper, the symbol 〈 ⋅ ,  ⋅ 〉 denotes the natural inner product of two n-dimensional vectors. The first two lemmas are known, but we fail to find literature sources.

### Lemma 1

Let X be a normed space (ℝn,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, …, 0, 1) be a basis of X. Assume that B is a skew-symmetric matrix, i.e., BT = −B. Then xBx〉 = 0 for any x ∈ X.

### Proof

Given that xBx〉 = 〈BTxx〉 = −〈xBx, then xBx〉 = 0.

### Lemma 2

Let X be a normed space (ℝ2n,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, e2n = (0, 0, …, 0, 1) be a basis of X. There exist rank 2n matrices

B2n,1B2n,2, …, B2n,2n−1

such that

• (i) each square matrix B2n,i (i = 1, 2, …, 2n − 1) is a skew-symmetric orthogonal matrix, i.e., $B2n,iT=−B2n,i$, $B2n,iTB2n,i=Id2n$;
• (ii) each row and column in B2n,i (i = 1, 2, …, 2n − 1) has one and only one non-zero element, and this element is 1 or −1;
• (iii) matrices $B2n,iTB2n,j$, i ≠ j, i, j = 1, 2, …, 2n − 1, satisfy the preceding two properties, i.e., (i) and (ii).

### Proof

The result is proven by induction. For n = 1, the only matrix

$B2,1=[0−110]$

satisfies conditions (i)-(iii).

Assume that this lemma holds for nk, namely, the matrices B2k,1, B2k,2, …, B2k,2k−1 exist, satisfying the conditions (i) to (iii).

Now, we shall prove this lemma holds for nk + 1. First, we need to introduce three rank 2 square matrices σ1, σ2, σ and zero matrix 0, which are defined as

$σ1=[100−1],σ2=[0110],σ=[0−110],0=[0000],$

respectively. Next, we claim that the set {B2k+1,1B2k+1,2, …, B2k+1,2k+1−1} can be written as the following set:

{B2k,i(σ1), B2k,i(σ2), Id2k(σ):i = 1, 2, …, 2k − 1}.
1

So we only need to prove that the matrices in (1) satisfy the three properties in this lemma, where B2k,i(σj) denotes the matrix in which the entries 1, −1, 0 in matrix B2k,i are replaced by the matrices σj, σj, 0 respectively; and Id2k(σ) denotes the matrix in which the entries 1, 0 in the unit matrix Id2k are replaced by the matrices σ, 0, respectively.

Let

$[B2k,i(σj)]TB2k,i(σj)=B2k,iT(σjT)B2k,i(σj)=Id2k(σjTσj)=Id2k+1,$

where i = 1, …, 2k − 1, j = 1, 2, and

[Id2k(σ)]TId2k(σ) = Id2k(σT)Id2k(σ) = Id2k(σTσ) = Id2k+1.

Then the matrices in (1) are proven to be orthogonal.

By induction, $B2k,iTB2k,l$ (where i ≠ l, j = 1, 2) is a skew-symmetric orthogonal matrix and

$[B2k,i(σj)]TB2k,l(σj)=B2k,iT(σjT)B2k,l(σj)=B2k,iTB2k,l(σjTσj)=B2k,iTB2k,l(Id2).$

We obtain that [B2k,i(σj)]TB2k,l(σj) is also a skew-symmetric orthogonal matrix that satisfies condition (ii). Similarly, [B2k,i(σj)]TB2k,l(σm), i ≠ l, j ≠ m and [B2k,i(σj)]TId2k(σ) satisfy conditions (i) and (ii). Thus, the matrices in set (1) satisfy condition (iii).

In order to present an upper bound of D(X), a new constant WD(X) for any real normed linear space X = (ℝ2n,  ∥  ⋅  ∥ ) is introduced.

### Definition 1

Let X be a normed space (ℝ2n,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, e2n = (0, 0, …, 0, 1) be a basis of X. The geometric constant WD(X) is defined as

$WD(X)=inf{infλ∈R∥x+λy∥:x∈S(X),y=Bx,B∈{B2n,1,B2n,2,…,B2n,2n−1}},$

where B2n,1, B2n,2, …, B2n,2n−1 are given as in Lemma 2.

### Proposition 1

Let X be a normed space (ℝ2n,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, e2n = (0, 0, …, 0, 1) be a basis of X. And the normed space X is such that, for any x ∈ S(X) and any yB2n,ix, B2n,i ∈ {B2n,1, …, B2n,2n−1}, there exists λ0 ∈ ℝ such that xλ0yBH, where H = span{yB2n,1x, …, B2n,i−1xB2n,i+1x, …, B2n,2n−1x}. Then

$WD(X)=min{〈x,x〉∥x∥∗:x∈S(X)}=min{〈x,x〉∥x∥∥x∥∗:x≠0,x∈X}.$
2

### Proof

Assume that x ∈ S(X), yBx, where B ∈ {B2n,1B2n,2, …, B2n,2n−1}. Without losing generality, let yB2n,1x. Then there exists λ0 ∈ ℝ such that  ∥ xλ0y ∥  = minλ∈ℝ ∥ xλy ∥ , i.e., xλ0yBy. Based on Corollary 4.2 in [11], we have the following equalities:

f1(y) = 0,  f1(xλ0y) =  ∥ xλ0y ∥ ,  ∥f1 = 1,

for some f1 ∈ X. Let My = {f ∈ X:f(y) = 0}. Then dim(My) = 2n − 1. Based on Lemma 2, the matrices B2n,i and $B2n,iTB2n,j$ (i ≠ j) are skew-symmetric. Then, by Lemma 1, the following can be obtained:

xB2n,ix〉 = 0,  〈B2n,ixB2n,jx〉 = 0,  i ≠ j

which imply that all of the vectors x, B2n,2x, …, B2n,2n−1x are in My and linearly independent. Thus,

My = span{xB2n,2x, …, B2n,2n−1x}( ⊂ X).

Given that f1 ∈ My, we may assume that f1α1xα2B2n,2x +  ⋯  + α2n−1B2n,2n−1x, where α1, …, α2n−1 ∈ ℝ. We get f1(xλ0y) = 〈α1xx〉 = 〈α1xxλ0y〉 =  ∥ xλ0y ∥ . This leads to f1|span{x,y}α1x|span{x,y} and α1x|span{x,y} ≥ 1. Hence, we obtain that 1 ≤ ∥f1|span{x,y} ≤ ∥f1 = 1. Then we have α1x|span{x,y} = 1.

Since xλ0yBH, where H = span{yB2n,2x, …, B2n,2n−1x} ⊂ X, and x ∉ H, then X = span{xλ0y} + H. Thus, for any z ∈ H and any real number a, b, the following inequality

∥ a(xλ0y) ∥  ≤  ∥ a(xλ0y) + bz ∥

holds. Thus, the inequalities |〈α1xa(xλ0y) + bz〉| ≤ ∥α1x|span{x,y} ∥ a(xλ0y) ∥  ≤  ∥ a(xλ0y) + bz ∥  indicate that α1x ≤ 1, and then α1x = 1, namely, $|α1|=1∥x∥∗$ is independent of y and  ∥ xλ0y ∥  = 〈α1xx. So Eq. (2) is obtained, and thereby we complete the proof.

### Lemma 3

Let X be a real symmetric linear normed space (ℝ2n,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, e2n = (0, 0, …, 0, 1) be a basis of X.

1. Assume that each row and each column in the 2n × 2n matrix B has only one non-zero element, which takes the value of 1 or −1. Then matrix B is an isometric operator on X.
2. Assume that
• (i)
B is a skew-symmetric and orthogonal matrix, i.e., BT = −B, BTBId;
• (ii)
Each row and each column in matrix B has only one non-zero element, which takes the value of 1 or −1.
Then  ∥ Bx ∥  =  ∥ x ∥  and xIBx for any x ∈ X.

### Proof

(1) The equality yBx indicates that y is merely the vector in which the elements are a rearrangement of the corresponding elements of x; some items in the elements change their sign. Thus, based on the definition of a real symmetric linear normed space, we have  ∥ Bx ∥  =  ∥ x ∥ .

(2) Let yBx, by Lemma 3(1), B is an isometric operator. Thus,  ∥ Bx ∥  =  ∥ x ∥  = 1. Meanwhile,  ∥ xy ∥  =  ∥ (IdB)x ∥  =  ∥ B(Id − B)x ∥  =  ∥ (Id − B)x ∥  =  ∥ x − y ∥ , and xIy are obtained.

Then the main theorem can be obtained.

### Theorem 1

Let X be a real symmetric linear normed space (ℝ2n,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, e2n = (0, 0, …, 0, 1) be a basis of X. And the normed space X is such that, for any x ∈ S(X) and any yB2n,ix, B2n,i ∈ {B2n,1, …, B2n,2n−1}, there exists λ0 ∈ ℝ such that xλ0yBH, where H = span{yB2n,1x, …, B2n,i−1xB2n,i+1x, …, B2n,2n−1x}. Then $2(2−1)≤D(X)≤WD(X)≤1$.

### Proof

The first inequality has been proven in Theorem 1 of [3]; thus, the last inequality can be easily obtained. The second inequality can be proven as follows by assuming that x ∈ S(X). Given that X is a symmetric normed linear space and B2n,i, i = 1, …, 2n − 1, satisfies properties (i) and (ii) in Lemma 2. By Lemma 3(2), B2n,ix ∈ S(X) and xIB2n,ix (i = 1, …, 2n − 1) can be obtained. Hence, we get D(X) ≤ WD(X).

It is easy to extend the above result to any m-dimensional real symmetric normed linear space.

### Corollary 1

Let X be a real symmetric linear normed space (ℝm,  ∥  ⋅  ∥ ) and e1 = (1, 0, …, 0), e2 = (0, 1, 0, …, 0), …, em = (0, 0, …, 0, 1) be a basis of X. And the normed space X is such that there exists a subspace Y ⊂ X with dimY = 2n and, for any x ∈ S(Y) and any yB2n,ix, B2n,i ∈ {B2n,1, …, B2n,2n−1}, there exists λ0 ∈ ℝ such that xλ0yBH, where H = span{yB2n,1x, …, B2n,i−1xB2n,i+1x, …, B2n,2n−1x}. Then $2(2−1)≤D(X)≤WD(Y)≤1$.

It is worth mentioning that the upper bound WD(X) of the geometric constant D(X), which is given in Theorem 1, has several advantages. Firstly, it is defined unrelated to isosceles orthogonality compared to D(X). Secondly, due to (2), WD(X) has a simple expression, which makes calculation feasible. Finally, it is less than one in general. For example, we consider WD(X) for the space $lp2n$ in the next section.

## The case of $lp2n$

The space $lp2n$ is used to show that the aforementioned upper bound WD(X) is optimal for D(X).

### Proposition 2

Let pq ≥ 1 such that $1p+1q=1$. Then

$WD(lp2n)=inf{1+t2(1+tq)1q(1+tp)1p:t∈[0,1]}.$

### Proof

Assume that $x=α1e1+α2e2+⋯+α2ne2n∈S(lp2n)$ and yB2n,ixB2n,i ∈ {B2n,1, …, B2n,2n−1}. For simplicity, we may take αi ≥ 0, i = 1, 2, …, 2n, and

$y=∑k=12n−2α2n−1−k+1ek+∑k=12n−2(−1)α2n−2−k+1e2n−2+k+∑k=12n−2α2n−k+1e2n−1+k+∑k=12n−2(−1)α2n−1+2n−2−k+1e2n−1+2n−2+k.$

Let f(λ) = ∥x+λyp and

$f1(λ)=∑k=12n−2(αk+λα2n−1−k+1)p+∑k=12n−2(α2n−2+k−λα2n−2−k+1)p+∑k=12n−2(α2n−1+k+λα2n−k+1)p+∑k=12n−2(α2n−1+2n−2+k−λα2n−1+2n−2−k+1)p.$

Then the equality f1(λ) = f(λ) holds on the interval [ξη], where

$ξ=max{maxk=12n−2{−αkα2n−1−k+1},maxk=12n−2{−α2n−1+kα2n−k+1}}$

and

$η=min{mink=12n−2{α2n−2+kα2n−2−k+1},mink=12n−2{α2n−1+2n−2+kα2n−1+2n−2−k+1}}.$

Since f(λ) is a convex function on [ξη], then f1(λ) is also a convex function on [ξη]. There exists one case in which the following equalities about λ hold:

$α2n−1(α1+λα2n−1)p−1=α1(α2n−2+2n−2−λα1)p−1,α2n−1−1(α2+λα2n−1−1)p−1=α2(α2n−1−1−λα2)p−1,…,α2n−2+1(α2n−2+λα2n−2+1)p−1=α2n−2(α2n−2+1−λα2n−2)p−1,…,α2n(α2n−1+1+λα2n)p−1=α2n−1+1(α2n−λα2n−1+1)p−1,…,α2n−2n−2+1(α2n−1+2n−2+λα2n−2n−2+1)p−1=α2n−1+2n−2(α2n−1+2n−2+1−λα2n−1+2n−2)p−1.$

In this case, $f1′(λ)=0$. If we let $γk=(α2n−1−k+1αk)1p−1$, 1 ≤ k ≤ 2n−2, then

$λ=α2n−1−k+1−γkαkαk+γkα2n−1−k+1.$

If we let $γ2n−2+k=(α2n−k+1α2n−1+k)1p−1$, 1 ≤ k ≤ 2n−2, then

$λ=α2n−k+1−γkα2n−1+kα2n−1+k+γkα2n−k+1.$

On the one hand, we have

$λ=α2n−1−k+1−γkαkαk+γkα2n−1−k+1≤α2n−1−k+1αk,1≤k≤2n−2$

and

$λ=α2n−k+1−γkα2n−1+kα2n−1+k+γkα2n−k+1≤α2n−k+1α2n−1+k,1≤k≤2n−2.$

On the other hand, we have

$λ−(−αkα2n−1−k+1)=αk2+α2n−1−k+12α2n−1−k+1(αk+γkα2n−1−k+1)≥0,1≤k≤2n−2,$

and

$λ−(−α2n−1+kα2n−k+1)=α2n−1+k2+α2n−k+12α2n−k+1(α2n−1+k+γkα2n−k+1)≥0,1≤k≤2n−2.$

These inequalities show that λ ∈ [ξη]. Hence, if all γk and γ2n−2+k (1 ≤ k ≤ 2n−2) are equal, then the preceding equalities about λ are equal. If f1(λ) has a minimum value on the interval [ξη], then f(λ) also has a minimum value on the interval [ξη].

Let $tk=α2n−1−k+1αk$ and $t2n−2+k=α2n−k+1α2n−1+k$ (1 ≤ k ≤ 2n−2). If α1 =  ⋯  = α2n−2α2n−1+1 =  ⋯  = α2n−1+2n−2 and α2n−2+1 =  ⋯  = α2n−1α2n−2n−2+1 =  ⋯  = α2n, then t1t2 =  ⋯  = t2n−1 are obtained. We may assume that t1t2 =  ⋯  = t2n−1t, and take $λ0=t−tq−11+tq$, where q is any positive number such that $1p+1q=1$, then $f1′(λ0)=0$ and

$f1(λ0)=α1p[(1+λ0t)p+(t−λ0)p]+α2p[(1+λ0t)p+(t−λ0)p]+⋯+α2n−2p[(1+λ0t)p+(t−λ0)p]+α2n−1+1p[(1+λ0t)p+(t−λ0)p]+⋯+α2n−1+2n−2p[(1+λ0t)p+(t−λ0)p]=(1+t2)p(1+tq)p−1(1+tp).$

Moreover, we may assume that α1 ≥ α2n−1, then 0 ≤ t ≤ 1. If α1 ≤ α2n−1, then we can take $t=tk=αkα2n−1−k+1$. The sufficient and necessary condition for the extreme points of a derived convex function is that it must be the stagnation point. Since f1(λ) is a strictly convex function, then λ0 is unique. Hence, we have

$WD(lp2n)=inf{inf{1+t12(1+t1q)1q(1+t1p)1p:t1∈[0,1]},…,inf{1+t22(1+t2q)1q(1+t2p)1p:t2∈[0,1]},inf{1+t2n−12(1+t2n−1q)1q(1+t2n−1p)1p:t2n−1∈[0,1]}}=inf{1+t2(1+tq)1q(1+tp)1p:t∈[0,1]}.$

### Remark 1

According to the above proof, $WD(lp2n)$ is independent of the selection of B. Thus, it may verify the existence of the space X satisfying the condition of Proposition 1.

### Corollary 2

Let m, n be any positive integer such that 2n ≤ m. Then

$2(2−1)≤D(lpm)≤D(lp2n)≤WD(lp2n)≤1.$

### Corollary 3

$limp→∞D(lpm)=2(2−1)$. Specially, $WD(l12n)=2(2−1)$.

### Corollary 4

Let pq ≥ 1 such that $1p+1q=1$. Assume that $lp2n$ has the property that, for any $x∈S(lp2n)$ and any yB2n,ix, B2n,i ∈ {B2n,1, …, B2n,2n−1}, there exists λ0 ∈ ℝ such that xλ0yBH, where

H = span{yB2n,1x, …, B2n,i−1xB2n,i+1x, …, B2n,2n−1x}.

Then, for any $x∈lp2n$ and x ≠ 0, there exists a positive constant $b=WD(lp2n)$ such that

$2(2−1)∥x∥p∥x∥q≤b∥x∥p∥x∥q≤〈x,x〉≤∥x∥p∥x∥q.$
3

The third inequality in (3) is the classical Hölder inequality

$∑k=1nakbk≤(∑k=1nakp)1p(∑k=1nbkq)1q,$

where ak ≥ 0, bk ≥ 0 (k = 1, 2, …, n), p > 1, q > 1 and $1p+1q=1$. This classical inequality plays a very important role in many areas of pure and applied mathematics. Various generalizations and improvements of this classical inequality have been studied by many mathematicians. The second inequality in (3) is a special case of the reverse version of Hölder inequality, which differs from other known results (e.g., see [12]). One of its values is that the constant b is always greater than or equal to $2(2−1)$.

## Conclusion

In this paper, by studying the geometric constant D(X) of any real 2n-dimensional symmetric normed space X = (ℝ2n,  ∥  ⋅  ∥ ), we obtained an upper bound WD(X), which is not greater than 1. And using the special properties of a finite dimensional normed space (ℝ2n,  ∥  ⋅  ∥ ) and the constraints on (ℝ2n,  ∥  ⋅  ∥ ), we also give a simple formula for WD(X). In particular, when $X=lp2n$, this formula is used to give a special form of the reverse Hölder inequality.

## Acknowledgements

The second author is supported by the National Natural Science Foundation of China (Grant No. 11401451). The third author is supported by the National Natural Science Foundation of China (Grant No. 61373174).

## Footnotes

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors of the manuscript have read and agreed to its content and are accountable for all aspects of the accuracy and integrity of the manuscript.

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