# On Reverses of Some Inequalities inn-Inner Product Spaces

Authors: Renu Chugh; Sushma Lather
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We present some new reverses of Cauchy-Bunyakovsky-Schwarz inequality, and Triangle and Boas-Bellman Type inequalities inn-inner product spaces. The results obtained generalize the results of Dragomir(2003–2005)in n-inner product spaces. Also we provide some applications for determinantal integral inequalities.


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## Body

## 1. Introduction

In 1964, Gähler [1] introduced the concept of 2-norm and 2-inner product spaces as generalization of norm and inner product spaces. A systematic presentation of the results related to the theory of 2-inner product spaces can be found in the book in [2, 3] and in list of references in it. Generalization of 2-inner product space for n≥2 was developed by Misiak [4] in 1989. Gunawan and Mashadi [5] in 2001 introduced the concept of n-normedlinear spaces. Gunawan [6] obtained Cauchy-Bunyakovsky inequality in these spaces. Cho et al. [7] extended parallelogram law and proved Hlwaka’s type inequality in n-inner product spaces. In this paper by, using same ideas of [8–13] we establish some results regarding new reverses of Cauchy-Bunyakovsky-Schwarz inequality, and Triangle and Boas-Bellman Type inequalities in n-inner product spaces and we also provide some applications for determinantal integral inequalities.


## 2. Preliminaries

Definition 2.1 (see [4]).
Assume thatn is a positive integer and X is a vector space over the field K=R of real numbers or the field K=C of complex numbers, such that dim X≥n and 〈·,·∣·,…,·︸n-1〉 is a K valued function defined on X×X×⋯×X︸n+1 such that:(nI1)
〈x1,x1∣x2,…,xn〉≥0, for any x1,x2,…,xn∈X and 〈x1,x1∣x2,…,xn〉=0 if and only if x1,x2,…,xn are linearly dependent vectors;(nI2)
〈a,b∣x1,…,xn-1〉=〈a,b∣π(x1),…,π(xn-1)〉, for any a,b,x1,x2,…,xn-1∈X and for any bijections π:{x1,x2,…,xn-1}→{x1,x2,…,xn-1};(nI3)
Ifn>1, then 〈x1,x1∣x2,…,xn〉=〈x2,x2∣x1,x3,…,xn〉, for any x1,x2,…,xn∈X;(nI4)
〈a,b∣x1,…,xn-1〉=〈b,a∣x1,…,xn-1〉¯(nI5)
〈αa,b∣x1,…,xn-1〉=α〈a,b∣x1,…,xn-1〉, for any a,b,x1,…,xn-1∈X and any scalar α∈R;(nI6)
〈a+a1,b∣x1,…,xn-1〉=〈a,b∣x1,…,xn-1〉+〈a1,b∣x1,…,xn-1〉, for any a,b,a1,x1,…,xn-1∈X.Then 〈·,·∣·,…,·︸n-1〉 is called the n-inner product and (X,〈·,·∣·,…,·︸n-1〉) is called the n-prehilbert space. If n=1, then Definition 2.1 reduces to the ordinary inner product. In any given n-inner product space (X·〈·,·∣·,·,…,·〉), we can define a function ∥·,·,…,·∥ on Xn=X×X×⋯×X︸n-times as
(2.1)∥x1,x2,…,xn∥=〈x1,x1∣x2,…,xn〉
for any x1,x2,…,xn∈X. It is easy to see that this function satisfies the following conditions:(nN1)
∥x1,x2,…,xn∥≥0 and ∥x1,x2,…,xn∥=0 if and only if x1,x2,…,xn are linearly dependent,(nN2)
∥x1,x2,…,xn∥ is invariant under any permutation,(nN3)
∥x1,x2,…,axn∥=|a|∥x1,x2,…,xn∥, for any a∈K,(nN4)
∥x1,x2,…,xn-1,y+z∥≤∥x1,x2,…,xn-1,y∥+∥x1,x2,…,xn-1,z∥.
A function∥·,·,…,·∥ defined on Xn and satisfying the above conditions is called an n-norm on X and the pair (X,∥·,…,·∥) is called n-normed linear space. Whenever an n-inner product space (X,〈·,·∣·,·,…,·〉) is given, we consider it as an n-normed space (X,∥·,·,…,·∥) with the n-norm defined by (2.1).
Let(H;〈·,·〉) be an inner product space over the real or complex number fieldK.The following inequality is known as Cauchy-Schwarz’s inequality:
(2.2)|〈x,y〉|2≤∥x∥2∥y∥2,x,y∈H,
where ∥z∥2=〈z,z〉,z∈H. The equality occurs in (2.2) if and only if x and y are linearly dependent.
In [8], Dragomir obtained the following reverse of Cauchy-Schwarz’s inequality:(2.3)0≤∥x∥2∥y∥2-|〈x,y〉|2≤14|A-a|2∥y∥4,
provided x,y∈H and a,A∈K are so that either
(2.4)Re〈Ay-x,x-ay〉≥0,
or, equivalently
(2.5)∥x-a+A2,y∥≤12|A-a|∥y∥,
holds. The constant 1/4 is best possible in (2.3) in the sense that it cannot be replaced by a smaller quantity.
Ifx,y,A,a satisfy either (2.4) or (2.5), then the following reverse of Cauchy-Schwarz’s inequality also holds:(2.6)∥x∥∥y∥≤12·Re[A〈x,y〉¯+a¯〈x,y〉][Re(a¯A)]1/2≤12·|A|+|a|[Re(a¯A)]1/2|〈x,y〉|
provided that, the complex numbers a and A satisfy the condition Re(a¯A)>0. In both inequalities in (2.6), the constant 1/2 is best possible.
The following reverse of the triangle inequality in inner product space was also obtained by Dragomir [9]
Let(H;〈·,·〉) be an inner product space overK and x,y∈H,M≥m>0 such that either Re〈Ay-x,x-ay〉≥0 or, equivalently ∥x-(a+A)/2,y∥≤1/2|A-a|∥y∥, holds. (2.7)0≤∥x∥+∥y∥-∥x+y∥≤M-mmMRe〈x,y〉
holds.
Gunawan [6] generalized the Cauchy-Bunyakovsky-Schwarz inequality (shortly, the CBS inequality) for inner product space to n-inner product space and obtained the following:(2.8)|〈x,y∣z2,…,zn〉|2≤〈x,x∣z2,…,zn〉〈y,y∣z2,…,zn〉.
Moreover, the equality holds if and only if x,y,z2,…,zn are linearly dependent. In terms of the n-norms, the (CBS)-inequality (2.8) can be written as
(2.9)|〈x,y∣z2,…,zn〉|2≤∥x,z2,…,zn∥2∥y,z2,…,zn∥2.
The equality holds if and only if x,y,z2,…,zn are linearly dependent.


## 3. Main Results

The aim of the present paper is to generalize the above mentioned results of Dragomir [8, 9], that is, reverses of CBS inequality, Triangle inequality and Boas-bellman type inequalities in inner product space to n-inner product spaces.


### 3.1. Reverses of the CBS Inequality

Theorem 3.1.
LetA,a∈K(K=C,R) and x,y,z2,…,zn∈X, where (X,〈·,·∣·,…,·〉) is n-inner product space over K. If
(3.1)Re(〈Ay-x,x-ay∣z2,…,zn〉)≥0.
or, equivalently,
(3.2)∥x-a+A2y,z2,…,zn∥≤12|A-a|∥y,z2,…,zn∥.
holds, then one has
(3.3)0≤∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤14|A-a|2∥y,z2,…,zn∥4.
The constant 1/4 in (3.3) cannot be replaced by a smaller constant.Proof.
Using (nI2)–(nI6), we get
(3.4)〈z2,z2∣x±y,…,zn〉=〈x±y,x±y∣z2,…,zn〉=〈x,x∣z2,…,zn〉+〈y,y∣z2,…,zn〉±2Re〈x,y∣z2,…,zn〉,(3.5)Re〈x,y∣z2,…,zn〉=14[〈z2,z2∣x+y,…,zn〉-〈z2,z2∣x-y,…,zn〉].
Considering the vectors x,u,U,z2,…,zn∈X and using (3.5) and (nN1–nN6), we have
(3.6)Re〈U-x,x-u∣z2,…,zn〉=14[〈z2,z2∣U-u,…,zn〉-〈z2,z2∣-2x+U+u,…,zn〉]=14[〈U-u,U-u∣z2,…,zn〉-4〈x-u+U2,x-u+U2∣z2,…,zn〉]=14∥U-u,z2,…,zn∥2-∥x-u+U2,z2,…,zn∥2.
Therefore, Re〈U-x,x-u∣z2,…,zn〉≥0 if and only if
(3.7)∥x-u+U2,z2,…,zn∥≤12∥U-u,z2,…,zn∥.
Applying this to the vectors U=Ay and u=ay, we obtain that the (3.1) and (3.2) are equivalent. If we consider the real numbers
(3.8)I1=Re[(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)(A∥y,z2,…,zn∥2-〈x,y∣z2,…,zn〉)×(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)],I2=∥y,z2,…,zn∥2Re〈Ay-x,x-ay∣z2,…,zn〉.
Then we obtain, by using properties of n-inner product space,
(3.9)I1=∥y,z2,…,zn∥2Re[A〈x,y∣z2,…,zn〉¯+a¯〈x,y∣z2,…,zn〉]-|〈x,y∣z2,…,zn〉|2-∥y,z2,…,zn∥4Re(Aa¯).I2=∥y,z2,…,zn∥2Re[A〈x,y∣z2,…,zn〉¯+a¯〈x,y∣z2,…,zn〉]-∥x,z2,…,zn∥2∥y,z2,…,zn∥2-∥y,z2,…,zn∥4Re(Aa¯)
which gives
(3.10)I1-I2=∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2,
for any x,y,z2,…,zn∈X and a,A∈K. If (3.1) holds, then I2≥0 and thus I1-I2≤I1(3.11)⇒∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤Re[(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)(A∥y,z2,…,zn∥2-〈x,y∣z2,…,zn〉)×(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)].
Now using the elementary inequality Re(αβ¯)≤1/4|α+β|2 for any α,β∈K(K=R,C), one yields
(3.12)Re[(A∥y,z2,…,zn∥2-〈x,y∣z2,…,zn〉)(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)]≤14|A-a|2∥y,z2,…,zn∥4.
If we combine (3.11) and (3.12), we get the required inequality.
To prove the sharpness of the constant1/4, assume that (3.3) holds with a constant C>0, that is,
(3.13)∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤C|A-a|2∥y,z2,…,zn∥4,
where x,y,z2,…,zn,A, and a satisfy the hypothesis of the theorem.
Considery∈X with ∥y,z2,…,zn∥=1,a≠A,m∈X with ∥m,z2,…,zn∥=1 and 〈y,m∣z2,…,zn〉=0 and define
(3.14)x=A+a2y+A-a2m
then we have
(3.15)Re〈Ay-x,x-ay∣z2,…,zn〉=|A-a|24Re〈y-m,y+m∣z2,…,zn〉=0
and then the condition (3.1) is fulfilled. From (3.13) we deduce
(3.16)∥A+a2y+A-a2m,z2,…,zn∥-|〈A+a2y+A-a2m,y∣z2,…,zn〉|2≤C|A-a|2
and, since
(3.17)∥A+a2y+A-a2m,z2,…,zn∥2=|A+a2|2+|A-a2|2,|〈A+a2y+A-a2m,y∣z2,…,zn〉|2=|A+a2|2.
By (3.16), we have
(3.18)|A-a2|2≤C|A-a|2,
for any A,a∈K with a≠A, which implies C≥1/4. This completes the proof.Theorem 3.2.
Assume thatx,y,z2,…,zn,a and A are the same as in above Theorem 3.1. If Re(a¯A)>0, then one has
(3.19)∥x,z2,…,zn∥∥y,z2,…,zn∥≤12Re[(A¯+a¯)〈x,y∣z2,…,zn〉][Re(a¯A)]1/2≤12|A+a|[Re(a¯A)]1/2|〈x,y∣z2,…,zn〉|.
The constant 1/2 is best possible in both inequalities in the sense that it cannot be replaced by a smaller constant.Proof.
Define(3.20)I=Re〈Ay-x,x-ay∣z2,…,zn〉=Re[A〈x,y∣z2,…,zn〉¯+a¯〈x,y∣z2,…,zn〉]-∥x,z2,…,zn∥2-∥y,z2,…,zn∥2Re(Aa¯).
We know that, for a complex number α∈C,Re(α)=Re(α¯) and thus
(3.21)Re[A〈x,y∣z2,…,zn〉¯]=Re[A¯〈x,y∣z2,…,zn〉]
which implies
(3.22)I=Re[(A¯+a¯)〈x,y∣z2,…,zn〉]-∥x,z2,…,zn∥2-∥y,z2,…,zn∥2Re(Aa¯).
Since x,y,z2,…,zn,a and A are assumed to satisfy the condition (3.1), by (3.22), we deduce
(3.23)∥x,z2,…,zn∥2+∥y,z2,…,zn∥2Re(Aa¯)≤Re[(A¯+a¯)〈x,y∣z2,…,zn〉],
which gives
(3.24)1[Re(a¯A)]1/2∥x,z2,…,zn∥2+[Re(a¯A)]1/2∥y,z2,…,zn∥2≤Re[(A¯+a¯)〈x,y∣z2,…,zn〉][Re(a¯A)]1/2
since Re(Aa¯)>0.
On the other hand, by the elementary inequality(3.25)αp2+1αq2≥2pq
for p,q≥0 and α>0, we have
(3.26)2∥x,z2,…,zn∥∥y,z2,…,zn∥≤1[Re(a¯A)]1/2∥x,z2,…,zn∥2+[Re(a¯A)]1/2∥y,z2,…,zn∥2.
Using (3.24) and (3.26), we deduce the first inequality in (3.22). The last part is obvious by the fact, for z∈C,|Re(z)|≤|z|.
To prove the sharpness of the constant1/2 in the first inequality in (3.22), we assume that (3.22) holds with a constant c>0, that is,(3.27)∥x,z2,…,zn∥∥y,z2,…,zn∥≤cRe[(A¯+a¯)〈x,y∣z2,…,zn〉][Re(a¯A)]1/2
provided x,y,z2,…,zn,a and A satisfy (2.1). If we take a=A=1,y=x≠0, then obviously (2.1) holds and from (3.27), we obtain
(3.28)∥x,z2,…,zn∥2≤2c∥x,z2,…,zn∥2
for any linearly independent vectors x,z2,…,zn∈X, which implies c≥1/2. This completes the proof.When the constants involved are assumed to be positive, then we may state the following result.Corollary 3.3.
LetM≥m>0 and assume that, for x,y,z2,…,zn∈X, one has
(3.29)Re〈My-x,x-my∣z2,…,zn〉≥0
or, equivalently,
(3.30)∥x-M+m2,z2,…,zn∥≤12(M-m)∥y,z2,…,zn∥.
Then one has the following reverse of CBS inequality:
(3.31)∥x,z2,…,zn∥∥y,z2,…,zn∥≤12M+mmMRe〈x,y∣z2,…,zn〉≤12M+mmM|〈x,y∣z2,…,zn〉|.
The constant 1/2 is sharp in (3.31).Corollary 3.4.
With the assumptions of the Theorem3.2, one has
(3.32)0<∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤14|A-a|2Re(a¯A)|〈x,y∣z2,…,zn〉|2.
The constant 1/4 is best possible in (3.32).Corollary 3.5.
With the assumption of Corollary3.3, one has
(3.33)0≤∥x,z2,…,zn∥∥y,z2,…,zn∥-|〈x,y∣z2,…,zn〉|≤∥x,z2,…,zn∥∥y,z2,…,zn∥-Re(〈x,y∣z2,…,zn〉)≤12(M-m)2mMRe(〈x,y∣z2,…,zn〉)≤12(M-m)2mM|〈x,y∣z2,…,zn〉|,(3.34)0≤∥x,z2,…,zn∥2∥y,z2,…,zn∥-|〈x,y∣z2,…,zn〉|2≤∥x,z2,…,zn∥2∥y,z2,…,zn∥2-[Re(〈x,y∣z2,…,zn〉)]2≤14(M-m)2mM[Re(〈x,y∣z2,…,zn〉)]2≤14(M-m)2mM|〈x,y∣z2,…,zn〉|2.
The constant 1/2 in (3.33) and the constant 1/4in (3.34) are best possible.


### 3.2. Reverse of the Triangle Inequality

Corollary 3.6.
Assumex,y,z2,…,zn,m,M are the same as in Corollary 3.3 and Re〈x,y∣z2,…,zn〉≥0. Then one has the following reverse of the triangle inequality:
(3.35)0≤∥x,z2,…,zn∥+∥y,z2,…,zn∥-∥x+y,z2,…,zn∥≤M-m(mM)1/4Re〈x,y∣z2,…,zn〉.Proof.
It is easy to see that(3.36)0≤(∥x,z2,…,zn∥+∥y,z2,…,zn∥)2-∥x+y,z2,…,zn∥2=2[∥x,z2,…,zn∥∥y,z2,…,zn∥-Re(〈x,y∣z2,…,zn〉)]
for any x,y,z2,…,zn∈X. If the assumption of Corollary 3.3 holds, then (3.33) is valid and, by (3.36), we deduce
(3.37)0≤(∥x,z2,…,zn∥+∥y,z2,…,zn∥)2-∥x+y,z2,…,zn∥2≤(M-m)2mMRe(〈x,y∣z2,…,zn〉)
which gives
(3.38)(∥x,z2,…,zn∥+∥y,z2,…,zn∥)2≤∥x+y,z2,…,zn∥2+(M-m)2mMRe(〈x,y∣z2,…,zn〉).
Taking square root in (3.38), we have
(3.39)∥x,z2,…,zn∥+∥y,z2,…,zn∥≤∥x+y,z2,…,zn∥2+(M-m)2mMRe(〈x,y∣z2,…,zn〉)≤∥x+y,z2,…,zn∥+M-m(mM)1/4Re〈x,y∣z2,…,zn〉.
From where we deduce the desire inequality (3.35). This completes the proof.Let(X,〈·,·∣·,…〉) be a n-inner product space over real or complex number field K. If (ei)1<i<m are linearly independent vector in the n-inner product space X, and, for given z2,…,zn∈X,〈ei,ej∣z2,…,zn〉=δij for all i,j∈{1,2,…,m} where δij is the Kronecker delta, then the following inequality is the Bessel’s inequality for z2,…,zn orthonormal family (ei)1<i<m in n-inner product space (X,〈·,·∣·,…〉):
(3.40)∑i=1m|〈x,ei∣z2,…,zn〉|2≤∥x,z2,…,zn∥2foranyx∈X.Theorem 3.7.
Letx1,x2,…,xm,z2,…,zn∈X and α1,…,αm∈K. then one has
(3.41)∥∑i=1mαixi,z2,…,zn∥2≤{max1≤i≤m|αi|2∑i=1m∥xi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥xi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1,∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2+{max1≤i≠j≤m{|αiαj|}∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|,[(∑i=1m|αi|c)2-(∑i=1m|αi|2c)]1/c×(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,[(∑i=1m|αi|)2-∑i=1m|αi|2]max1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|.Proof.
We observe that(3.42)∥∑i=1mαixi,z2,…,zn∥2=〈∑i=1mαixi,∑j=1mαjxj,∣z2,…,zn〉=∑i=1m∑j=1mαiαj¯〈xi,xj∣z2,…,zn〉=|∑i=1m∑j=1mαiαj¯〈xi,xj∣z2,…,zn〉|≤∑i=1m∑j=1m|αi||αj¯||〈xi,xj∣z2,…,zn〉|=∑i=1m|αi|2∥xi,z2,…,zn∥2+∑1≤i≠j≤m|αi||αj||〈xi,xj∣z2,…,zn〉|.
Using Holder’s inequality, we may write that
(3.43)∑i=1m|αi|2∥xi,z2,…,zn∥2≤{max1≤i≤m|αi|2∑i=1m∥xi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥xi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1,∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2.
By Holder’s inequality for double sum, we also have
(3.44)∑1≤i≠j≤m|αi||αj||〈xi,xj∣z2,…,zn〉|≤{max1≤i≠j≤m|αiαj|∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|,(∑1≤i≠j≤m|αi|c|αj|c)1/c×(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,∑1≤i≠j≤m|αi||αj|max1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|={max1≤i≠j≤m{|αiαj|}∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|,[(∑i=1m|αi|c)2-(∑i=1m|αi|2c)]1/c×(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,[(∑i=1m|αi|)2-∑i=1m|αi|2]max1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|.
Using (3.43) and (3.44) in (3.42), we may deduce the desired result.Corollary 3.8.
With the assumption in above Theorem3.7, one has
(3.45)∥∑i=1mαixi,z2,…,zn∥2≤∑i=1n|αi|2{(∑1≤i≠j≤n|(xi,xj∣z2,…,zn)|2)2max1≤i≤n∥xi,z2,…,zn∥2+[(∑i=1n|αi|2)2-∑i=1n|αi|4]1/2∑i=1n|αi|2×(∑1≤i≠j≤n|(xi,xj∣z2,…,zn)|2)1/2[(∑i=1n|αi|2)2-∑i=1n|αi|4]12∑i=1n|αi|2}≤∑i=1m|αi|2{max1≤i≤m∥xi,z2,…,zn∥2+(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|2)1/2}.
The first inequality follows by taking the third branch in the first curly bracket with the second branch in the second curly bracket for c=d=2. The second inequality in (3.45) follows by the fact that
(3.46)[(∑i=1m|αi|2)2-∑i=1m|αi|4]1/2≤∑i=1m|αi|2.
By applying the following Cauchy-Bunyakovsky-Schwarz inequality:
(3.47)(∑i=1mai)2≤m∑i=1mai2,ai∈R+,1≤i≤m.
one may write that
(3.48)(∑i=1m|αi|c)2-∑i=1m|αi|2c≤(m-1)∑i=1m|αi|2c,(m≥1),(3.49)(∑i=1m|αi|)2-∑i=1m|αi|2≤(m-1)∑i=1m|αi|2,(m≥1).
It is obvious that
(3.50)max1≤i≠j≤m{|αiαj|}≤max1≤i≤m|αi|2.Corollary 3.9.
With the assumption in above Theorem3.7, one has
(3.51)∥∑i=1mαixi,z2,…,zn∥2≤{max1≤i≤m|αi|2∑i=1m∥xi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥xi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2+{max1≤i≤m|αi|2∑1≤i≤m|〈xi,xi∣z2,…,zn〉|,(m-1)1/c(∑i=1m|αi|2c)1/c(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,(m-1)∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2.
The proof is obvious by Theorem 3.7 on applying (3.48)–(3.50).Theorem 3.10.
Letx,y1,…,ym,z2,…,zn be vectors of an n-inner product space (X,〈·,·∣·,…〉) and α1,α2,…,αn∈K(K=R,C). Then
(3.52)|∑i=1mαi〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2{max1≤i≤m|αi|2∑i=1m∥yi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥yi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1,∑i=1m|αi|2max1≤i≤m∥yi,z2,…,zn∥2+∥x,z2,…,zn∥2{max1≤i≠j≤m{|αiαj|}∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|,[(∑i=1m|αi|c)2-(∑i=1m|αi|2c)]1/c×(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|d)1/d,[(∑i=1m|αi|)2-(∑i=1m|αi|2)]∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2.Proof.
Since(3.53)∑i=1mαi〈x,yi∣z2,…,zn〉=〈x,∑αi¯yi∣z2,…,zn〉
and by Schwarz’s inequality in n-inner product spaces,
(3.54)|∑i=1mαi〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2∥∑i=1mαi¯yi,z2,…,zn∥2.
Using αi=αi¯,zi=yi(i=1,2,…,n) in the Theorem 3.7, we get the desired inequality (3.52).Corollary 3.11.
With the assumptions in Theorem3.10, the following holds:
(3.55)|∑i=1mαi〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2×{∑i=1m|αi|2{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2},max1≤i≤m|αi|2{∑i=1m∥yi,z2,…,zn∥2+∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|},(∑i=1m|αi|2p)1/p×{(∑i=1m∥yi,z2,…,zn∥2q)1/q+(m-1)(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|q)1/q},∑i=1m|αi|2{max1≤i≤m∥yi,z2,…,zn∥2+(m-1)max1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}.


### 3.3. Some Boas-Bellman Type Inequalities inn-Inner Product Spaces

If we putαi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), in the first inequality of (3.55)(3.56)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2∑i=1m|〈x,yi∣z2,…,zn〉|2×{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2}which is equivalent to the following Boass-Bellman type inequality forn-inner products:(3.57)∑i=1m|〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2}.
Now, if we take αi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), in second inequality of (3.55), we have(3.58)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2max1≤i≤m|〈x,yi∣z2,…,zn〉|2×{∑i=1m∥yi,z2,…,zn∥2+∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}.
By taking the square root in (3.59)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥max1≤i≤m|〈x,yi∣z2,…,zn〉|×{∑i=1m∥yi,z2,…,zn∥2+∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}1/2
for any x,y1,…,ym,z2,…,zn be vectors of an n-inner product space (X,〈·,·∣·,…〉). If we assume that (ei)1≤i≤m is an orthonormal family in X with respect to the vector z2,…,zn,(ei,ej∣z2,…,zn)=δij for all i,j∈{1,…,m} then by (3.57) we deduce Bessel’s inequality ∑i=1m|〈x,ei∣z2,…,zn〉|2≤∥x,z2,…,zn∥2, and (3.59) implies(3.60)∑i=1m|〈x,ei∣z2,…,zn〉|2≤m∥x,z2,…,zn∥max1≤i≤m|〈x,ei∣z2,…,zn〉|2.
For the third inequality in (3.55) αi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), we have(3.61)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2(∑i=1m|〈x,yi∣z2,…,zn〉|2p)1/p×{(∑i=1m∥yi,z2,…,zn∥2q)1/q+(m-1)(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|q)1/q}
for p>1,(1/p)+(1/q)=1. Taking the square root in this inequality, we get (3.62)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥(∑i=1m|〈x,yi∣z2,…,zn〉|2p)1/2p×{(∑i=1m∥yi,z2,…,zn∥2q)1/q+(m-1)(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|q)1/q}1/2.
For any x,y1,…,ym,z2,…,zn∈X, and p>1, with (1/p)+(1/q)=1, then the above inequality (3.62) becomes, for an orthonormal family with respect to the vector z2,…,zn,(3.63)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤m1/q∥x,z2,…,zn∥(∑i=1m|〈x,ei∣z2,…,zn〉|2p)1/2p.
We take αi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), in the last inequality of (3.55)(3.64)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2∑i=1m|〈x,yi∣z2,…,zn〉|2×{max1≤i≤m∥yi,z2,…,zn∥2+(m-1)max1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}
which implies
(3.65)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥2{max1<i<m∥yi,z2,…,zn∥2+(m-1)max1<i≠j<m|〈yi,yj∣z2,…,zn〉|}
for any x,y1,…,ym,z2,…,zn∈X.


## 3.1. Reverses of the CBS Inequality

Theorem 3.1.
LetA,a∈K(K=C,R) and x,y,z2,…,zn∈X, where (X,〈·,·∣·,…,·〉) is n-inner product space over K. If
(3.1)Re(〈Ay-x,x-ay∣z2,…,zn〉)≥0.
or, equivalently,
(3.2)∥x-a+A2y,z2,…,zn∥≤12|A-a|∥y,z2,…,zn∥.
holds, then one has
(3.3)0≤∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤14|A-a|2∥y,z2,…,zn∥4.
The constant 1/4 in (3.3) cannot be replaced by a smaller constant.Proof.
Using (nI2)–(nI6), we get
(3.4)〈z2,z2∣x±y,…,zn〉=〈x±y,x±y∣z2,…,zn〉=〈x,x∣z2,…,zn〉+〈y,y∣z2,…,zn〉±2Re〈x,y∣z2,…,zn〉,(3.5)Re〈x,y∣z2,…,zn〉=14[〈z2,z2∣x+y,…,zn〉-〈z2,z2∣x-y,…,zn〉].
Considering the vectors x,u,U,z2,…,zn∈X and using (3.5) and (nN1–nN6), we have
(3.6)Re〈U-x,x-u∣z2,…,zn〉=14[〈z2,z2∣U-u,…,zn〉-〈z2,z2∣-2x+U+u,…,zn〉]=14[〈U-u,U-u∣z2,…,zn〉-4〈x-u+U2,x-u+U2∣z2,…,zn〉]=14∥U-u,z2,…,zn∥2-∥x-u+U2,z2,…,zn∥2.
Therefore, Re〈U-x,x-u∣z2,…,zn〉≥0 if and only if
(3.7)∥x-u+U2,z2,…,zn∥≤12∥U-u,z2,…,zn∥.
Applying this to the vectors U=Ay and u=ay, we obtain that the (3.1) and (3.2) are equivalent. If we consider the real numbers
(3.8)I1=Re[(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)(A∥y,z2,…,zn∥2-〈x,y∣z2,…,zn〉)×(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)],I2=∥y,z2,…,zn∥2Re〈Ay-x,x-ay∣z2,…,zn〉.
Then we obtain, by using properties of n-inner product space,
(3.9)I1=∥y,z2,…,zn∥2Re[A〈x,y∣z2,…,zn〉¯+a¯〈x,y∣z2,…,zn〉]-|〈x,y∣z2,…,zn〉|2-∥y,z2,…,zn∥4Re(Aa¯).I2=∥y,z2,…,zn∥2Re[A〈x,y∣z2,…,zn〉¯+a¯〈x,y∣z2,…,zn〉]-∥x,z2,…,zn∥2∥y,z2,…,zn∥2-∥y,z2,…,zn∥4Re(Aa¯)
which gives
(3.10)I1-I2=∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2,
for any x,y,z2,…,zn∈X and a,A∈K. If (3.1) holds, then I2≥0 and thus I1-I2≤I1(3.11)⇒∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤Re[(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)(A∥y,z2,…,zn∥2-〈x,y∣z2,…,zn〉)×(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)].
Now using the elementary inequality Re(αβ¯)≤1/4|α+β|2 for any α,β∈K(K=R,C), one yields
(3.12)Re[(A∥y,z2,…,zn∥2-〈x,y∣z2,…,zn〉)(〈x,y∣z2,…,zn〉¯-a¯∥y,z2,…,zn∥2)]≤14|A-a|2∥y,z2,…,zn∥4.
If we combine (3.11) and (3.12), we get the required inequality.
To prove the sharpness of the constant1/4, assume that (3.3) holds with a constant C>0, that is,
(3.13)∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤C|A-a|2∥y,z2,…,zn∥4,
where x,y,z2,…,zn,A, and a satisfy the hypothesis of the theorem.
Considery∈X with ∥y,z2,…,zn∥=1,a≠A,m∈X with ∥m,z2,…,zn∥=1 and 〈y,m∣z2,…,zn〉=0 and define
(3.14)x=A+a2y+A-a2m
then we have
(3.15)Re〈Ay-x,x-ay∣z2,…,zn〉=|A-a|24Re〈y-m,y+m∣z2,…,zn〉=0
and then the condition (3.1) is fulfilled. From (3.13) we deduce
(3.16)∥A+a2y+A-a2m,z2,…,zn∥-|〈A+a2y+A-a2m,y∣z2,…,zn〉|2≤C|A-a|2
and, since
(3.17)∥A+a2y+A-a2m,z2,…,zn∥2=|A+a2|2+|A-a2|2,|〈A+a2y+A-a2m,y∣z2,…,zn〉|2=|A+a2|2.
By (3.16), we have
(3.18)|A-a2|2≤C|A-a|2,
for any A,a∈K with a≠A, which implies C≥1/4. This completes the proof.Theorem 3.2.
Assume thatx,y,z2,…,zn,a and A are the same as in above Theorem 3.1. If Re(a¯A)>0, then one has
(3.19)∥x,z2,…,zn∥∥y,z2,…,zn∥≤12Re[(A¯+a¯)〈x,y∣z2,…,zn〉][Re(a¯A)]1/2≤12|A+a|[Re(a¯A)]1/2|〈x,y∣z2,…,zn〉|.
The constant 1/2 is best possible in both inequalities in the sense that it cannot be replaced by a smaller constant.Proof.
Define(3.20)I=Re〈Ay-x,x-ay∣z2,…,zn〉=Re[A〈x,y∣z2,…,zn〉¯+a¯〈x,y∣z2,…,zn〉]-∥x,z2,…,zn∥2-∥y,z2,…,zn∥2Re(Aa¯).
We know that, for a complex number α∈C,Re(α)=Re(α¯) and thus
(3.21)Re[A〈x,y∣z2,…,zn〉¯]=Re[A¯〈x,y∣z2,…,zn〉]
which implies
(3.22)I=Re[(A¯+a¯)〈x,y∣z2,…,zn〉]-∥x,z2,…,zn∥2-∥y,z2,…,zn∥2Re(Aa¯).
Since x,y,z2,…,zn,a and A are assumed to satisfy the condition (3.1), by (3.22), we deduce
(3.23)∥x,z2,…,zn∥2+∥y,z2,…,zn∥2Re(Aa¯)≤Re[(A¯+a¯)〈x,y∣z2,…,zn〉],
which gives
(3.24)1[Re(a¯A)]1/2∥x,z2,…,zn∥2+[Re(a¯A)]1/2∥y,z2,…,zn∥2≤Re[(A¯+a¯)〈x,y∣z2,…,zn〉][Re(a¯A)]1/2
since Re(Aa¯)>0.
On the other hand, by the elementary inequality(3.25)αp2+1αq2≥2pq
for p,q≥0 and α>0, we have
(3.26)2∥x,z2,…,zn∥∥y,z2,…,zn∥≤1[Re(a¯A)]1/2∥x,z2,…,zn∥2+[Re(a¯A)]1/2∥y,z2,…,zn∥2.
Using (3.24) and (3.26), we deduce the first inequality in (3.22). The last part is obvious by the fact, for z∈C,|Re(z)|≤|z|.
To prove the sharpness of the constant1/2 in the first inequality in (3.22), we assume that (3.22) holds with a constant c>0, that is,(3.27)∥x,z2,…,zn∥∥y,z2,…,zn∥≤cRe[(A¯+a¯)〈x,y∣z2,…,zn〉][Re(a¯A)]1/2
provided x,y,z2,…,zn,a and A satisfy (2.1). If we take a=A=1,y=x≠0, then obviously (2.1) holds and from (3.27), we obtain
(3.28)∥x,z2,…,zn∥2≤2c∥x,z2,…,zn∥2
for any linearly independent vectors x,z2,…,zn∈X, which implies c≥1/2. This completes the proof.When the constants involved are assumed to be positive, then we may state the following result.Corollary 3.3.
LetM≥m>0 and assume that, for x,y,z2,…,zn∈X, one has
(3.29)Re〈My-x,x-my∣z2,…,zn〉≥0
or, equivalently,
(3.30)∥x-M+m2,z2,…,zn∥≤12(M-m)∥y,z2,…,zn∥.
Then one has the following reverse of CBS inequality:
(3.31)∥x,z2,…,zn∥∥y,z2,…,zn∥≤12M+mmMRe〈x,y∣z2,…,zn〉≤12M+mmM|〈x,y∣z2,…,zn〉|.
The constant 1/2 is sharp in (3.31).Corollary 3.4.
With the assumptions of the Theorem3.2, one has
(3.32)0<∥x,z2,…,zn∥2∥y,z2,…,zn∥2-|〈x,y∣z2,…,zn〉|2≤14|A-a|2Re(a¯A)|〈x,y∣z2,…,zn〉|2.
The constant 1/4 is best possible in (3.32).Corollary 3.5.
With the assumption of Corollary3.3, one has
(3.33)0≤∥x,z2,…,zn∥∥y,z2,…,zn∥-|〈x,y∣z2,…,zn〉|≤∥x,z2,…,zn∥∥y,z2,…,zn∥-Re(〈x,y∣z2,…,zn〉)≤12(M-m)2mMRe(〈x,y∣z2,…,zn〉)≤12(M-m)2mM|〈x,y∣z2,…,zn〉|,(3.34)0≤∥x,z2,…,zn∥2∥y,z2,…,zn∥-|〈x,y∣z2,…,zn〉|2≤∥x,z2,…,zn∥2∥y,z2,…,zn∥2-[Re(〈x,y∣z2,…,zn〉)]2≤14(M-m)2mM[Re(〈x,y∣z2,…,zn〉)]2≤14(M-m)2mM|〈x,y∣z2,…,zn〉|2.
The constant 1/2 in (3.33) and the constant 1/4in (3.34) are best possible.


## 3.2. Reverse of the Triangle Inequality

Corollary 3.6.
Assumex,y,z2,…,zn,m,M are the same as in Corollary 3.3 and Re〈x,y∣z2,…,zn〉≥0. Then one has the following reverse of the triangle inequality:
(3.35)0≤∥x,z2,…,zn∥+∥y,z2,…,zn∥-∥x+y,z2,…,zn∥≤M-m(mM)1/4Re〈x,y∣z2,…,zn〉.Proof.
It is easy to see that(3.36)0≤(∥x,z2,…,zn∥+∥y,z2,…,zn∥)2-∥x+y,z2,…,zn∥2=2[∥x,z2,…,zn∥∥y,z2,…,zn∥-Re(〈x,y∣z2,…,zn〉)]
for any x,y,z2,…,zn∈X. If the assumption of Corollary 3.3 holds, then (3.33) is valid and, by (3.36), we deduce
(3.37)0≤(∥x,z2,…,zn∥+∥y,z2,…,zn∥)2-∥x+y,z2,…,zn∥2≤(M-m)2mMRe(〈x,y∣z2,…,zn〉)
which gives
(3.38)(∥x,z2,…,zn∥+∥y,z2,…,zn∥)2≤∥x+y,z2,…,zn∥2+(M-m)2mMRe(〈x,y∣z2,…,zn〉).
Taking square root in (3.38), we have
(3.39)∥x,z2,…,zn∥+∥y,z2,…,zn∥≤∥x+y,z2,…,zn∥2+(M-m)2mMRe(〈x,y∣z2,…,zn〉)≤∥x+y,z2,…,zn∥+M-m(mM)1/4Re〈x,y∣z2,…,zn〉.
From where we deduce the desire inequality (3.35). This completes the proof.Let(X,〈·,·∣·,…〉) be a n-inner product space over real or complex number field K. If (ei)1<i<m are linearly independent vector in the n-inner product space X, and, for given z2,…,zn∈X,〈ei,ej∣z2,…,zn〉=δij for all i,j∈{1,2,…,m} where δij is the Kronecker delta, then the following inequality is the Bessel’s inequality for z2,…,zn orthonormal family (ei)1<i<m in n-inner product space (X,〈·,·∣·,…〉):
(3.40)∑i=1m|〈x,ei∣z2,…,zn〉|2≤∥x,z2,…,zn∥2foranyx∈X.Theorem 3.7.
Letx1,x2,…,xm,z2,…,zn∈X and α1,…,αm∈K. then one has
(3.41)∥∑i=1mαixi,z2,…,zn∥2≤{max1≤i≤m|αi|2∑i=1m∥xi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥xi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1,∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2+{max1≤i≠j≤m{|αiαj|}∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|,[(∑i=1m|αi|c)2-(∑i=1m|αi|2c)]1/c×(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,[(∑i=1m|αi|)2-∑i=1m|αi|2]max1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|.Proof.
We observe that(3.42)∥∑i=1mαixi,z2,…,zn∥2=〈∑i=1mαixi,∑j=1mαjxj,∣z2,…,zn〉=∑i=1m∑j=1mαiαj¯〈xi,xj∣z2,…,zn〉=|∑i=1m∑j=1mαiαj¯〈xi,xj∣z2,…,zn〉|≤∑i=1m∑j=1m|αi||αj¯||〈xi,xj∣z2,…,zn〉|=∑i=1m|αi|2∥xi,z2,…,zn∥2+∑1≤i≠j≤m|αi||αj||〈xi,xj∣z2,…,zn〉|.
Using Holder’s inequality, we may write that
(3.43)∑i=1m|αi|2∥xi,z2,…,zn∥2≤{max1≤i≤m|αi|2∑i=1m∥xi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥xi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1,∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2.
By Holder’s inequality for double sum, we also have
(3.44)∑1≤i≠j≤m|αi||αj||〈xi,xj∣z2,…,zn〉|≤{max1≤i≠j≤m|αiαj|∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|,(∑1≤i≠j≤m|αi|c|αj|c)1/c×(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,∑1≤i≠j≤m|αi||αj|max1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|={max1≤i≠j≤m{|αiαj|}∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|,[(∑i=1m|αi|c)2-(∑i=1m|αi|2c)]1/c×(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,[(∑i=1m|αi|)2-∑i=1m|αi|2]max1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|.
Using (3.43) and (3.44) in (3.42), we may deduce the desired result.Corollary 3.8.
With the assumption in above Theorem3.7, one has
(3.45)∥∑i=1mαixi,z2,…,zn∥2≤∑i=1n|αi|2{(∑1≤i≠j≤n|(xi,xj∣z2,…,zn)|2)2max1≤i≤n∥xi,z2,…,zn∥2+[(∑i=1n|αi|2)2-∑i=1n|αi|4]1/2∑i=1n|αi|2×(∑1≤i≠j≤n|(xi,xj∣z2,…,zn)|2)1/2[(∑i=1n|αi|2)2-∑i=1n|αi|4]12∑i=1n|αi|2}≤∑i=1m|αi|2{max1≤i≤m∥xi,z2,…,zn∥2+(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|2)1/2}.
The first inequality follows by taking the third branch in the first curly bracket with the second branch in the second curly bracket for c=d=2. The second inequality in (3.45) follows by the fact that
(3.46)[(∑i=1m|αi|2)2-∑i=1m|αi|4]1/2≤∑i=1m|αi|2.
By applying the following Cauchy-Bunyakovsky-Schwarz inequality:
(3.47)(∑i=1mai)2≤m∑i=1mai2,ai∈R+,1≤i≤m.
one may write that
(3.48)(∑i=1m|αi|c)2-∑i=1m|αi|2c≤(m-1)∑i=1m|αi|2c,(m≥1),(3.49)(∑i=1m|αi|)2-∑i=1m|αi|2≤(m-1)∑i=1m|αi|2,(m≥1).
It is obvious that
(3.50)max1≤i≠j≤m{|αiαj|}≤max1≤i≤m|αi|2.Corollary 3.9.
With the assumption in above Theorem3.7, one has
(3.51)∥∑i=1mαixi,z2,…,zn∥2≤{max1≤i≤m|αi|2∑i=1m∥xi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥xi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2+{max1≤i≤m|αi|2∑1≤i≤m|〈xi,xi∣z2,…,zn〉|,(m-1)1/c(∑i=1m|αi|2c)1/c(∑1≤i≠j≤m|〈xi,xj∣z2,…,zn〉|d)1/d,wherec>1,1c+1d=1,(m-1)∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2.
The proof is obvious by Theorem 3.7 on applying (3.48)–(3.50).Theorem 3.10.
Letx,y1,…,ym,z2,…,zn be vectors of an n-inner product space (X,〈·,·∣·,…〉) and α1,α2,…,αn∈K(K=R,C). Then
(3.52)|∑i=1mαi〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2{max1≤i≤m|αi|2∑i=1m∥yi,z2,…,zn∥2,(∑i=1m|αi|2a)1/a(∑i=1m∥yi,z2,…,zn∥2b)1/b,wherea>1,1a+1b=1,∑i=1m|αi|2max1≤i≤m∥yi,z2,…,zn∥2+∥x,z2,…,zn∥2{max1≤i≠j≤m{|αiαj|}∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|,[(∑i=1m|αi|c)2-(∑i=1m|αi|2c)]1/c×(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|d)1/d,[(∑i=1m|αi|)2-(∑i=1m|αi|2)]∑i=1m|αi|2max1≤i≤m∥xi,z2,…,zn∥2.Proof.
Since(3.53)∑i=1mαi〈x,yi∣z2,…,zn〉=〈x,∑αi¯yi∣z2,…,zn〉
and by Schwarz’s inequality in n-inner product spaces,
(3.54)|∑i=1mαi〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2∥∑i=1mαi¯yi,z2,…,zn∥2.
Using αi=αi¯,zi=yi(i=1,2,…,n) in the Theorem 3.7, we get the desired inequality (3.52).Corollary 3.11.
With the assumptions in Theorem3.10, the following holds:
(3.55)|∑i=1mαi〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2×{∑i=1m|αi|2{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2},max1≤i≤m|αi|2{∑i=1m∥yi,z2,…,zn∥2+∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|},(∑i=1m|αi|2p)1/p×{(∑i=1m∥yi,z2,…,zn∥2q)1/q+(m-1)(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|q)1/q},∑i=1m|αi|2{max1≤i≤m∥yi,z2,…,zn∥2+(m-1)max1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}.


## 3.3. Some Boas-Bellman Type Inequalities inn-Inner Product Spaces

If we putαi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), in the first inequality of (3.55)(3.56)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2∑i=1m|〈x,yi∣z2,…,zn〉|2×{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2}which is equivalent to the following Boass-Bellman type inequality forn-inner products:(3.57)∑i=1m|〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2}.
Now, if we take αi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), in second inequality of (3.55), we have(3.58)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2max1≤i≤m|〈x,yi∣z2,…,zn〉|2×{∑i=1m∥yi,z2,…,zn∥2+∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}.
By taking the square root in (3.59)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥max1≤i≤m|〈x,yi∣z2,…,zn〉|×{∑i=1m∥yi,z2,…,zn∥2+∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}1/2
for any x,y1,…,ym,z2,…,zn be vectors of an n-inner product space (X,〈·,·∣·,…〉). If we assume that (ei)1≤i≤m is an orthonormal family in X with respect to the vector z2,…,zn,(ei,ej∣z2,…,zn)=δij for all i,j∈{1,…,m} then by (3.57) we deduce Bessel’s inequality ∑i=1m|〈x,ei∣z2,…,zn〉|2≤∥x,z2,…,zn∥2, and (3.59) implies(3.60)∑i=1m|〈x,ei∣z2,…,zn〉|2≤m∥x,z2,…,zn∥max1≤i≤m|〈x,ei∣z2,…,zn〉|2.
For the third inequality in (3.55) αi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), we have(3.61)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2(∑i=1m|〈x,yi∣z2,…,zn〉|2p)1/p×{(∑i=1m∥yi,z2,…,zn∥2q)1/q+(m-1)(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|q)1/q}
for p>1,(1/p)+(1/q)=1. Taking the square root in this inequality, we get (3.62)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥(∑i=1m|〈x,yi∣z2,…,zn〉|2p)1/2p×{(∑i=1m∥yi,z2,…,zn∥2q)1/q+(m-1)(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|q)1/q}1/2.
For any x,y1,…,ym,z2,…,zn∈X, and p>1, with (1/p)+(1/q)=1, then the above inequality (3.62) becomes, for an orthonormal family with respect to the vector z2,…,zn,(3.63)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤m1/q∥x,z2,…,zn∥(∑i=1m|〈x,ei∣z2,…,zn〉|2p)1/2p.
We take αi=〈x,yi∣z2,…,zn〉¯(i=1,2,…,m), in the last inequality of (3.55)(3.64)(∑i=1m|〈x,yi∣z2,…,zn〉|2)2≤∥x,z2,…,zn∥2∑i=1m|〈x,yi∣z2,…,zn〉|2×{max1≤i≤m∥yi,z2,…,zn∥2+(m-1)max1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|}
which implies
(3.65)(∑i=1m|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥2{max1<i<m∥yi,z2,…,zn∥2+(m-1)max1<i≠j<m|〈yi,yj∣z2,…,zn〉|}
for any x,y1,…,ym,z2,…,zn∈X.


## 4. Applications for Integral Inequalities

Let(Ω,Σ,μ) be a measure space consisting of a set Ω, a σ-algebra Σ of subsets of Ω and a countably additive measure μ on Σ with values in R∪{∞}. Denote by L2ρ(Ω) the Hilbert space of all real-valued functions f defined on Ω that are n-ρ-integrable on Ω, that is, ∫Ωρ(s)|f(s)|ndμ(s)<∞ where ρ:Ω→[0,∞) is a measurable function on Ω.We can introduce the followingn-inner product on L2ρ(Ω):(4.1)〈f,g∣h2,…,hn〉ρ=1n!∫Ω∫Ω⋯∫Ω︸n-timesρ(s1)ρ(s2)⋯ρ(sn)×|f(s1)f(s2)…f(sn)h2(s1)h2(s2)…h2(sn)⋮…⋮hn(s1)hn(s2)…hn(sn)|×|g(s1)g(s2)…g(sn)h2(s1)h2(s2)…h2(sn)⋮…⋮hn(s1)hn(s2)…hn(sn)|dμ(s1)dμ(s2)⋯dμ(sn),
where by
(4.2)|f(s1)f(s2)…f(sn)h2(s1)h2(s2)…h2(sn)⋮hn(s1)…hn(s2)…⋮hn(sn)|
we denote the determinant of matrix
(4.3)[f(s1)f(s2)⋯⋯f(sn)h2(s1)h2(s2)⋯⋯h2(sn)⋮⋮⋯⋯⋮hn(s1)hn(s2)⋯⋯hn(sn)].Generating then-norm on L2ρ(Ω) is expressed by(4.4)∥f,h2,…,hn∥p=(1n!∫Ω∫Ω⋯∫Ω︸n-timesρ(s1)ρ(s2)⋯ρ(sn)×|f(s1)f(s2)⋯⋯f(sn)h2(s1)h2(s2)⋯⋯h2(sn)⋮⋮⋯⋯⋮hn(s1)hn(s2)⋯⋯hn(sn)|2dμ(s1)dμ(s2)⋯dμ(sn)|f(s1)f(s2)⋯⋯f(sn)h2(s1)h2(s2)⋯⋯h2(sn)⋮⋮⋯⋯⋮hn(s1)hn(s2)⋯⋯hn(sn)|2)1/2.
A simple calculation with integrals shows that(4.5)〈f,g∣h2,…,hn〉ρ=|∫Ωρfgdμ∫Ωρfh2dμ⋯⋯∫Ωρfhndμ∫Ωρfh2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρghndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|,(4.6)∥f,h2,…,hn∥p=|∫Ωρf2dμ∫Ωρfh2dμ⋯⋯∫Ωρfhndμ∫Ωρfh2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρghndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|,
where, for simplicity, instead of ∫Ωρ(s)f(s)g(s)dμ(s), we have written ∫Ωρfgdμ.Proposition 4.1.
Letf,g1,…,gm,h2,…,hn∈L2ρ(Ω), where ρ:Ω→[0,∞) a measure function is on Ω. Then one has
(4.7)∑i=1m|∫Ωρfgidμ∫Ωρfh2dμ⋯⋯∫Ωρfhndμ∫Ωρgih2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρgihndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|2≤|∫Ωρf2dμ∫Ωρfh2dμ⋯⋯∫Ωρfhndμ∫Ωρfh2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρfhndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|×{max1≤i≤m|∫Ωρgi2dμ∫Ωρgih2dμ⋯⋯∫Ωρgihndμ∫Ωρgih2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρgihndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|+(∑1≤i≠j≤m|∫Ωρgjgidμ∫Ωρgjh2dμ⋯⋯∫Ωρgjhndμ∫Ωρgih2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρgihndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|2)1/2.}Proof.
Applying then-inner product and n-norm defined in (4.1) and (4.4), and using (4.5) and (4.6) in (3.57)
(4.8)∑i=1m|〈x,yi∣z2,…,zn〉|2≤∥x,z2,…,zn∥2×{max1≤i≤m∥yi,z2,…,zn∥2+(∑1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|2)1/2},
we get the required proof of the proposition.Proposition 4.2.
Letf,g1,…,gm,h2,…,hn∈L2ρ(Ω), where ρ:Ω→[0,∞) a measure function is on Ω. Then one has
(4.9)∑i=1m|∫Ωρfgidμ∫Ωρfh2dμ⋯⋯∫Ωρfhndμ∫Ωρgih2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρgihndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|2≤|∫Ωρf2dμ∫Ωρfh2dμ⋯⋯∫Ωρfhndμ∫Ωρfh2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρfhndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|×{max1≤i≤m|∫Ωρgi2dμ∫Ωρgih2dμ⋯⋯∫Ωρgihndμ∫Ωρgih2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρgihndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|+(m-1)max1≤i≠j≤m|∫Ωρgjgidμ∫Ωρgjh2dμ⋯⋯∫Ωρgjhndμ∫Ωρgih2dμ∫Ωρh22dμ⋯⋯∫Ωρh2hndμ⋮⋮⋯⋯⋮∫Ωρgihndμ∫Ωρh2hndμ⋯⋯∫Ωρhn2dμ|}.Proof.
By (3.65),
(4.10)(∑|〈x,yi∣z2,…,zn〉|2)≤∥x,z2,…,zn∥2×{max1≤i≤m∥yi,z2,…,zn∥2+(m-1)max1≤i≠j≤m|〈yi,yj∣z2,…,zn〉|},
and using (4.1) and (4.4), we get the proof of the proposition.



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*Source: 385824-2010-09-28.xml*