The weak n-inner product space

TL;DR

This paper introduces the concept of weak n-inner products, generalizes the n-iterated 2-inner product, and explores their representations and applications in linear regression and Chebyshev functionals.

Contribution

It defines the weak n-inner product, relates it to standard k-inner products via Dodgson's identity, and applies it to statistical modeling and functional analysis.

Findings

Representation of n-iterated 2-inner product in terms of standard k-inner products

Characterization of linear regression models using weak n-inner products

Generalization of Chebyshev functional with n-iterated 2-inner product

Abstract

In this article we study a generalization of the n-inner product which we name weak n-inner product. As particular case we consider the n-iterated 2-inner product and we give its representation in terms of the standard k-inner products, k<= n, using the Dodgson's identity for determinants. Finally, we present several applications, including a brief characterization of a linear regression model for the random variables in discrete case and a generalization of the Chebyshev functional using the n-iterated 2-inner product.

Figures

Equations

$$|⟨v,w|v_2,...,v_n⟩|\le \sqrt{⟨v,v|v_2,...,v_n⟩}\sqrt{⟨w,w|v_2,...,v_n⟩},$$

$$⟨0,w|v_2,...,v_n⟩=⟨v,0|v_2,...,v_n⟩=⟨v,w|0,...,v_n⟩=0,$$

$$⟨v_2,w|v_2,...,v_n⟩=⟨v,v_2|v_2,...,v_n⟩=0,$$

$$\Vert v|v_2,...,v_n\Vert :=\sqrt{⟨v,v|v_2,...,v_n⟩},$$

$$\Vert v+w|v_2,...,v_n{\Vert }^2+\Vert v-w|v_2,...,v_n{\Vert }^2=2\Vert v|v_2,...,v_n{\Vert }^2+2\Vert w|v_2,...,v_n{\Vert }^2,$$

$$\Vert v+w|v_2,...,v_n{\Vert }^2-\Vert v-w|v_2,...,v_n{\Vert }^2=4⟨v,w|v_2,...,v_n⟩,$$

$$⟨v,w|v_2,...,v_n⟩:=\left|\begin{array}{cccc}⟨v,w⟩& ⟨v,v_2⟩& ...& ⟨v,v_n⟩\\ ⟨v_2,w⟩& ⟨v_2,v_2⟩& ...& ⟨v_2,v_n⟩\\ ⋮& ⋮& ⋮& ⋮\\ ⟨v_n,w⟩& ⟨v_n,v_2⟩& ...& ⟨v_n,v_n⟩\end{array}\right|,$$

$$(0,y|x_n,...,x_2)=(x,0|x_n,...,x_2)=(x,y|0,...,x_2)=0.$$

$$(x,y|x_n,\dots ,x_2)=\Theta (x,y|x_n\dots ,x_2)\cdot \Psi (x_{n-1},\dots ,x_2),$$

$$(x,y|x_n,\dots ,x_2)=0.$$

$$u=(y,y|x_n,\dots ,x_2)x-(x,y|x_n,\dots ,x_2)y.$$

$$(y,y|x_n,\dots ,x_2)[(x,x|x_n,\dots ,x_2)(y,y|x_n,\dots ,x_2)-(x,y|x_n,\dots ,x_2{)}^2]\ge 0.$$

$$-(y,y|x_n,\dots ,x_2)(x,y|x_n,\dots ,x_2{)}^2\ge 0.$$

$$(x,y|x_n,\dots ,x_2)=\frac{1}{2}[(x+y,x+y|x_n,\dots ,x_2)-(x,x|x_n,\dots ,x_2)-(y,y|x_n,\dots ,x_2)].$$

$$|(x,y|x_n,...,x_2)|\le \sqrt{(x,x|x_n,...,x_2)}\sqrt{(y,y|x_n,...,x_2)}.$$

$$|(x,y|x_n,...,x_2)|$$

$$=|\Psi (\hat{x},\hat{y})|\le \sqrt{\Psi (\hat{x},\hat{x})}\sqrt{\Psi (\hat{y},\hat{y})}=\sqrt{(x,x|x_n,...,x_2)}\sqrt{(y,y|x_n,...,x_2)}.$$

$$\Vert x|x_n,...,x_2\Vert :=\sqrt{(x,x|x_n,...,x_2)},\text{mbox}forallx,x_2,...,x_n\in X.$$

$$(x,y|z{)}_{\ast }:=⟨x,y|z⟩=\left|\begin{array}{cc}⟨x,y⟩& ⟨x,z⟩\\ ⟨z,y⟩& ⟨z,z⟩\end{array}\right|=⟨x,y⟩⟨z,z⟩-⟨x,z⟩⟨z,y⟩,x,y,z\in X.$$

$$(x,y|x_n,...,x_2{)}_{\ast }:=\left|\begin{array}{cc}(x,y|x_{n-1},...,x_2{)}_{\ast }& (x,x_n|x_{n-1},...,x_2{)}_{\ast }\\ (x_n,y|x_{n-1},...,x_2{)}_{\ast }& (x_n,x_n|x_{n-1},...,x_2{)}_{\ast }\end{array}\right|.$$

$$(x,y|x_{n+1},x_n,...,x_2{)}_{\ast }=\left|\begin{array}{cc}(x,y|x_n,...,x_2{)}_{\ast }& (x,x_{n+1}|x_n,...,x_2{)}_{\ast }\\ (x_{n+1},y|x_n,...,x_2{)}_{\ast }& (x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\end{array}\right|.$$

$$(x,x|x_{n+1},\dots ,x_2{)}_{\ast }$$

$$(\lambda x+x_{n+1},\lambda x+x_{n+1}|x_n,...,x_2{)}_{\ast }\ge 0,\text{mbox}forall\lambda \in \mathbb{R}.$$

$${\lambda }^2(x,x|x_n,...,x_2{)}_{\ast }+2\lambda (x,x_{n+1}|x_n,...,x_2{)}_{\ast }+(x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\ge 0,\forall \lambda .$$

$$(x,x|x_n,...,x_2{)}_{\ast }(x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }-(x,x_{n+1}|x_n,...,x_2{)}_{\ast }^2=0.$$

$$(x,x|x_{n+1},...,x_2{)}_{\ast }=\left|\begin{array}{cc}(x,x|x_n,...,x_2{)}_{\ast }& (x,x_{n+1}|x_n,...,x_2{)}_{\ast }\\ (x_{n+1},x|x_n,...,x_2{)}_{\ast }& (x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\end{array}\right|$$

$$=\left|\begin{array}{cc}(x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }& (x_{n+1},x|x_n,...,x_2{)}_{\ast }\\ (x,x_{n+1}|x_n,...,x_2{)}_{\ast }& (x,x|x_n,...,x_2{)}_{\ast }\end{array}\right|=(x_{n+1},x_{n+1}|x,x_n,...,x_2{)}_{\ast }.$$

$$(x,y|x_{n+1},x_n,...,x_2{)}_{\ast }=\left|\begin{array}{cc}(x,y|x_n,...,x_2{)}_{\ast }& (x,x_{n+1}|x_n,...,x_2{)}_{\ast }\\ (x_{n+1},y|x_n,...,x_2{)}_{\ast }& (x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\end{array}\right|$$

$$=\left|\begin{array}{cc}(y,x|x_n,...,x_2{)}_{\ast }& (x_{n+1},y|x_n,...,x_2{)}_{\ast }\\ (x,x_{n+1}|x_n,...,x_2{)}_{\ast }& (x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\end{array}\right|=(y,x|x_{n+1},x_n,...,x_2{)}_{\ast },$$

$$(\alpha x,y|x_{n+1},x_n,...,x_2{)}_{\ast }=\left|\begin{array}{cc}(\alpha x,y|x_n,...,x_2{)}_{\ast }& (\alpha x,x_{n+1}|x_n,...,x_2{)}_{\ast }\\ (x_{n+1},y|x_n,...,x_2{)}_{\ast }& (x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\end{array}\right|$$

$$=\left|\begin{array}{cc}\alpha (x,y|x_n,...,x_2{)}_{\ast }& \alpha (x,x_{n+1}|x_n,...,x_2{)}_{\ast }\\ (x_{n+1},y|x_n,...,x_2{)}_{\ast }& (x_{n+1},x_{n+1}|x_n,...,x_2{)}_{\ast }\end{array}\right|=\alpha (x,y|x_{n+1},x_n,...,x_2{)}_{\ast },$$

Full text

2000 Mathematics Subject Classification: 46C05, 26D15, 26D10

Key words: $n$-inner product space, $n$-pre-Hilbert space, Cauchy-Schwarz inequality

1 Introduction

The concept of linear 2-normed spaces and 2-metric spaces has been investigated by Gähler [13]. In [6] and [7], Diminnie, Gähler and White studied the 2-inner product spaces.

A classification of results related to the theory of 2-inner product spaces can be found in book [3]. Here, several properties of 2-inner product spaces are given. In [10] Dragomir et al. show the corresponding version of Boas-Bellman inequality in 2-inner product spaces. Others properties of a 2-inner product space can be found in [4].

Misiak [20] generalizes this concept of a 2-inner product space, in 1989, in the following way: let $n$ be a nonnegative integer $(n\geq 2)$ and $X$ be a vector space of dimension $\dim X=d\geq n$ ($d$ may be infinite) over the field of real numbers $\mathbb{R}$. An $\mathbb{R}$-valued function $\langle\cdot,\cdot\mid\cdot,...,\cdot\rangle$ on $X^{n+1}$ satisfying the following properties:

I1) $\langle v_{1},v_{1}|v_{2},...,v_{n}\rangle\geq 0;\langle v_{1},v_{1}|v_{2},...,v_{n}\rangle=0$ if and only if $v_{1},v_{2},...,v_{n}$ are linearly dependent;

I2) $\langle v_{1},v_{1}|v_{2},...,v_{n}\rangle=\langle v_{i_{1}},v_{i_{1}}|v_{i_{2}},v_{i_{3}},...,v_{i_{n}}\rangle$, for every permutation$(i_{1},i_{2},...,i_{n})$ of $(1,2,...,n)$;

I3) $\langle v,w|v_{2},...,v_{n}\rangle=\langle w,v|v_{2},...,v_{n}\rangle$;

I4) $\langle\alpha v,w|v_{2},...,v_{n}\rangle=\alpha\langle v,w|v_{2},...,v_{n}\rangle$, for every scalar $\alpha\in\mathbb{R}$.

I5) $\langle v+v^{\prime},w|v_{2},...,v_{n}\rangle=\langle v,w|v_{2},...,v_{n}\rangle+\langle v^{\prime},w|v_{2},...,v_{n}\rangle$;

is called an * n-inner product* on $X$, and the pair $(X,\langle\cdot,\cdot|\cdot,...,\cdot\rangle)$ is called an n-inner product space or n-pre-Hilbert space.

It is easy to see that the $n$-inner product is a linear function of its two first arguments. Several results related to the theory of the $n$-inner product spaces can be found in [15], [21]: $\langle v,w|\alpha v_{2},...,v_{n}\rangle=\alpha^{2}\langle v,w|v_{2},...,v_{n}\rangle$, for every real number $\alpha$ and for $v,w,v_{2},..,v_{n}\in X$; $\langle v,w|v_{2}+v_{2}^{\prime},v_{3},...,x_{n}\rangle-\langle v,w|v_{2}-v_{2}^{\prime},v_{3},...,v_{n}\rangle=\langle v_{2},v_{2}^{\prime}|v+w,v_{3},...,v_{n}\rangle-\langle v_{2},v_{2}^{\prime}|v-w,v_{3},...,v_{n}\rangle$, for all $v,w,v_{2},v_{3},...,v_{n},v_{2}^{\prime}\in X$ and an extension of the Cauchy-Schwarz inequality to arbitrary n:

[TABLE]

for all $v,w,v_{2},...,v_{n}\in X$. The equality holds in (1) if and only if $v,w,v_{2},...,v_{n}$ are linearly dependent.

Other consequences from the above properties can be inferred very easily:

[TABLE]

[TABLE]

for all $v,w,v_{2},...,v_{n}\in X$.

Let $(X,\langle\cdot,\cdot|\cdot,...,\cdot\rangle)$ be an $n$-inner product space, $n\geq 2$. We can define a function $\|\cdot,...,\cdot\|$ on $X\times X\times...\times X=X^{n}$ by

[TABLE]

for all $v,v_{2},...,v_{n}\in X$, which in [20] is shown that satisfies the following conditions:

N1) $\|v|v_{2},...,v_{n}\|\geq 0$ and $\|v|v_{2},...,v_{n}\|=0$ if and only if $v,v_{2},...,v_{n}$ are linearly dependent;

N2) $\|v|v_{2},...,v_{n}\|$ is invariant under permutation;

N3) $\|\alpha v|v_{2},...,v_{n}\|=|\alpha|\|v|v_{2},...,v_{n}\|$, for any scalar $\alpha\in\mathbb{R}.$

N4) $\|v+w|v_{2},...,v_{n}\|\leq\|v|v_{2},...,v_{n}\|+\|w|v_{2},...,v_{n}\|$,

for all $v,w,v_{2},...,v_{n}\in X.$

A function $\|\cdot|\cdot,...,\cdot\|$ defined on $X^{n}$ and satisfying the above conditions is called an n-norm on X and $(X,\|\cdot|\cdot,...,\cdot\|)$ is called a linear n-normed space.

It is easy to see that if $(X,\langle\cdot,\cdot|\cdot,...,\cdot\rangle)$ is an n-inner product space over the field of real numbers $\mathbb{R}$, then $(X,\|\cdot|\cdot,...,\cdot\|)$ is a linear n-normed space and the n-norm $\|\cdot|\cdot,...,\cdot\|$ is generated by the n-inner product $\langle\cdot,\cdot|\cdot,...,\cdot\rangle$.

Furthermore, we have the parallelogram law [3],

[TABLE]

for all $v,w,v_{2},..,v_{n}\in X$ and the polarization identity (see e.g. [3] and [4]),

[TABLE]

for all $v,w,v_{2},...,v_{n}\in X.$

The standard $n$-inner product on an inner product space $X=(X,\langle\cdot.\cdot\rangle)$ is given by:

[TABLE]

which generates $n$-norm $\|v|v_{2},...,v_{n}\|:=\sqrt{\langle v,v|v_{2},...,v_{n}\rangle},$ representing the volume of the $n$-dimensional parallelepiped spanned by $v,v_{2},...,v_{n}.$

Various type of applications of $n$-inner products and $n$-norms can be found in recent papers [2], [16], [17], [18], [22], [23], [25], [26].

Remark 1**.**

The standard $n$-inner product satisfies also the following additional condition:

I6) If $v,v_{2},\ldots,v_{n}$ are linearly dependent, then $\langle v,w|v_{2},\ldots,v_{n}\rangle=0$,

for $v,w,v_{2},\ldots,v_{n}\in X$.**

The motivation of this article is to study another type of $n$-inner product built based on the properties of the $n$-inner product, except property I2. We will define the weak $n$-inner product and the $n$-iterated $2$-inner product and we will give its representation in terms of the standard $k$-inner products, $k\leq n$, using the Dodgson’s identity for determinants. We also present a brief characterization of a linear regression model for the random variables in discrete case. Finally, we generalize the Chebyshev functional using the $n$-iterated $2$-inner product.

2 The weak $n$-inner product

Let $X$ be a real vector space.

Definition 1**.**

An $\mathbb{R}$-valued function $(\cdot,\cdot\mid\cdot,...,\cdot)$ on $X^{n+1}$, $n\geq 2$, satisfying the following properties:

P1)* Positivity: $(x,x|x_{n},...,x_{2})\geq 0\;\mbox{and}\;(x,x|x_{n},...,x_{2})=0$ if and only if $x,x_{2},x_{3},...,x_{n}$ are linearly dependent;*

P2)* Interchangeability: $(x,x|x_{n},...,x_{2})=(x_{n},x_{n}|x,x_{n-1},...,x_{2})$;*

P3)* Symmetry: $(x,y|x_{n},...,x_{2})=(y,x|x_{n},...,x_{2})$;*

P4)* Homogeneity: $(\alpha x,y|x_{n},...,x_{2})=\alpha(x,y|x_{n},...,x_{2})$, for every scalar $\alpha\in\mathbb{R}$.*

P5)* Additivity: $(x+x^{\prime},y|x_{n},...,x_{2})=(x,y|x_{n},...,x_{2})+(x^{\prime},y|x_{n},...,x_{2})$;

is called a weak $n$-inner product on $X$, and the pair $(X,(\cdot,\cdot|\cdot,...,\cdot))$ is called a weak $n$-inner product space or weak n-pre-Hilbert space.*

Remark 2**.**

It is easy to see that:

[TABLE]

Remark 3**.**

Obviously an $n$-inner product is a weak $n$-inner product, so an $n$-inner product space is a weak $n$-inner product space, but the reciprocal is not true. This fact will be shown in Remark 4.

For $n=2$ a weak $n$-inner product is also an $n$-inner product. For $n\geq 3$ a weak $n$-inner product can be build, for instance, by formula

[TABLE]

where $\Theta(x,y|x_{n}\ldots,x_{2})$ is a $n$-inner product* and $\Psi:X^{n-2}\to\mathbb{R}$ is a function with properties $\Psi(x_{n-1},\ldots,x_{2})\geq 0.\;\forall(x_{n-1},\ldots,x_{2})$ and $\Psi(x_{n-1},\ldots,x_{2})=0$ iff $x_{n-1},\ldots,x_{2}$ are linearly dependent (in the case $n=3$, this means $x_{2}=0$).*

In the next lemma we generalize a property that exists in the case of $2$-inner products. The method of the proof is based on the method used in [4].

Lemma 1**.**

Let $,x_{2},\ldots,x_{n},x,y\in X$. If $x_{2},\ldots,x_{n},x$ are linearly dependent, then

[TABLE]

Proof We consider two cases.

Case 1. $x_{2},\ldots,x_{n},y$ are linearly independent. Consider the vector

[TABLE]

Then from P1) we hve $(u,u|x_{n},\ldots,x_{2})\geq 0$. This inequality is equivalent to

[TABLE]

Since $x_{2},\ldots,x_{n},x$ are linearly dependent, from P1) we obtain $(x,x|x_{n},\ldots,x_{2})=0$ and hence

[TABLE]

Since $x_{2},\ldots,x_{n},y$ are linearly independent it follows that $(y,y|x_{n},\ldots,x_{2})>0$. Consequently one obtains (5).

Case 2. $x_{2},\ldots,x_{n},y$ are linearly dependent. Then also $x_{2},\ldots,x_{n},x+y$ are are linearly dependent. We have

[TABLE]

Because $(x,x|x_{n},\ldots,x_{2})=0$, $(y,y|x_{n},\ldots,x_{2})=0$, $(x+y,x+y|x_{n},\ldots,x_{2})=0$ relation (5) follows.

Theorem 1**.**

Suppose that $(X,(\cdot,\cdot|\cdot,...,\cdot))$ is a weak n-inner product space over the field of real numbers $\mathbb{R}$. Let $x_{2},\ldots,x_{n}\in X$, $n\geq 2$ be fixed. Denote $Y=span\{x_{2},\ldots,x_{n}\}$. Define the quotient space $X/Y=\{\hat{x}|\;x\in X\}$, where $\hat{x}=\{u\in X|u-x\in Y\}$, $x\in X$. Then function $\psi:(X/Y)^{2}\to\mathbb{R}$, $\psi(\hat{x},\hat{y}):=(x,y|x_{n},...,x_{2})$, $\hat{x},\hat{y}\in X/Y$ is well defined and is a semi-inner product on $X/Y$. Moreover, if $x_{2},\ldots,x_{n}$ are linearly independent, then $\psi$ is an inner product.

Proof Let $x,x^{\prime},y,y^{\prime}\in X$, such that $x^{\prime}-x\in Y$ and $y^{\prime}-y\in Y$. Using Lemma 1 we get $\psi(\widehat{x^{\prime}},\hat{y^{\prime}})=(x^{\prime},y^{\prime}|x_{n},...,x_{2})=(x,y|x_{n},...,x_{2})+(x^{\prime}-x,y|x_{n},...,x_{2})+(x,y^{\prime}-y|x_{n},...,x_{2})+(x^{\prime}-x,y^{\prime}-y|x_{n},...,x_{2})=(x,y|x_{n},...,x_{2})=\psi(\widehat{x},\hat{y})$. This means that $\psi$ is well defined.

From P1) we have $\psi(\hat{x},\hat{x})=(x,x|x_{n},...,x_{2})\geq 0$. Moreover, if $\psi(\hat{x},\hat{x})=0$, then $(x,x|x_{2},\ldots,x_{n})=0$, which implies that $x,x_{2},\ldots,x_{n}$ are linearly dependent. If $x_{2},\ldots,x_{n}$ are linearly independent it follows that $x\in Y$. Then $\hat{x}=\hat{0}$.

The other properties of the inner product follow in a simple manner from conditions P3), P4) and P5).

Theorem 2**.**

( Schwarz type inequality)* Let $(X,(\cdot,\cdot|\cdot,...,\cdot))$ be a weak $n$-inner product space. For any $x,y,x_{2},...,x_{n}\in X$ we have*

[TABLE]

In the case when $x_{2},\ldots,x_{n}$ are linearly independent, then the equality holds in (6) if and only if there exist $\mu\in\mathbb{R}_{+}$ and $u\in Y:=span\{x_{2},\ldots,x_{n}\}$ such that $y=\mu x+u$.

Proof By taking into account Theorem 1 and the notations given there, we have

[TABLE]

If $x_{2},\ldots,x_{n}$ are linearly independent, then the equality holds in (6) iff there is $\mu\geq 0$, such that $\hat{y}=\mu\hat{x}$, i.e. exists $u\in Y$ for which $y=\mu x+u$. .

Definition 2**.**

Let $(X,(\cdot,\cdot|\cdot,...,\cdot))$ be a weak $n$-inner product space, $n\geq 2$. We can define a function $\|\cdot,...,\cdot\|$ on $X\times X\times...\times X=X^{n}$ by

[TABLE]

Proposition 1**.**

If $(X,(\cdot,\cdot|\cdot,...,\cdot))$ is a weak $n$-inner product space, then function $\|\cdot,...,\cdot\|$ defined in (7) satisfies the following conditions:

• C1)

$\|x|x_{n},...,x_{2}\|\geq 0$* and $\|x|x_{n},...,x_{2}\|=0$ if and only if $x,x_{2},...,x_{n}$ are linearly dependent;*

• C2)

$\|x|x_{n},x_{n-1},...,x_{2}\|=\|x_{n}|x,x_{n-1},...,x_{2}\|$;

• C3)

$\|\alpha x|x_{n},...,x_{2}\|=|\alpha|\|x|x_{n},...,x_{2}\|$, for any scalar $\alpha\in\mathbb{R};$

• C4)

$\|x+y|x_{n},...,x_{2}\|\leq\|x|x_{n},...,x_{2}\|+\|y|x_{n},...,x_{2}\|$**

for all $x,y,x_{2},...,x_{n}\in X.$

Proof Conditions C1)-C4) follow immediately from conditions P1)-P5) and Definition 2.

Definition 3**.**

Let $X$ be a real vector space. A real function $\|\cdot|\cdot,...,\cdot\|$ defined on $X^{n}$ and satisfying conditions C1)-C4) is called a weak $n$-norm on $X$ and $(X,\|\cdot|\cdot,...,\cdot\|)$ is called a linear weak $n$-normed space.

It follows that if $(X,(\cdot,\cdot|\cdot,...,\cdot))$ is a weak n-inner product space over the field of real numbers $\mathbb{R}$, then $(X,\|\cdot|\cdot,...,\cdot\|)$ is a linear weak n-normed space and the weak n-norm $\|\cdot|\cdot,...,\cdot\|$ is generated by the weak n-inner product $(\cdot,\cdot|\cdot,...,\cdot)$.

Theorem 3**.**

In conditions of Theorem 1, function $\varphi:X/Y\to\mathbb{R}_{+}$, $\varphi(\hat{x}):=\|x|x_{n},...,x_{2}\|$, $\hat{x}\in X/Y$ is well defined and is a semi-norm on $X/Y$. Moreover, if $x_{2},\ldots,x_{n}$ are linearly independent, then $\varphi$ is a norm.

Proof It follows immediately from Theorem 1, since $\varphi(\hat{x})=\sqrt{\psi(\hat{x},\hat{x})}$, $\hat{x}\in X/Y$, where function $\psi$ was defined in this theorem.

In an inner product space, a special weak $n$-inner product can be defined by recurrence starting from the $2$-inner product. Recall that the $2$-inner product was studied in [3], [4].

Definition 4**.**

Let $(X,\langle\cdot,\cdot\rangle)$ be a real pre-Hilbert space. The $n$-iterated $2$-inner product, or standard weak $n$-inner product $(\cdot,\cdot|\cdot,\ldots,\cdot)_{*}:X^{n+1}\to\mathbb{R}$ is defined for $n\geq 2$ as follows. For $n=2$, $(\cdot,\cdot|\cdot)_{*}$ coincides with the standard $2$-inner product, i.e.

[TABLE]

Then, if $n\geq 3$ and $x,y,x_{2},...,x_{n}\in X$, define:

[TABLE]

Theorem 4**.**

If $(X,\langle\cdot,\rangle)$ is a real pre-Hilbert space, then for any $n\geq 2$ function $(\cdot,\cdot|\cdot,\ldots,\cdot)_{*}:X^{n+1}\to\mathbb{R}$ given in Definition 4 is a weak $n$-inner product.

Proof Consider proposition $S(n)$: the $n$-iterated $2$-inner product satisfies conditions P1)-P6). We prove this proposition by mathematical induction, for $n\geq 2$.

For $n=2$, $S(n)$ is true since we know from [3], [4], that the standard $2$-inner product, $(x,y|z)_{*}=\det\begin{pmatrix}\langle x,y\rangle&\langle x,z\rangle\\ \langle z,y\rangle&\langle z,z\rangle\end{pmatrix}$, satisfies conditions $P1)-P6)$.

Suppose $S(n)$ is true and prove that proposition $S(n+1)$ is true. The $(n+1)$-iterated $2$-inner product is given by

[TABLE]

Let us prove P1) for $n+1$. First we prove that $(x,x|x_{n+1},...,x_{2})_{*}\geq 0$, for $x,x_{2},\ldots x_{n+1}\in X$.

Case 1: $(x,x|x_{n},...,x_{2})_{*}=0$. Then, from property P1) for $n$, it results that $x,x_{2},...,x_{n}$ are linearly dependent. From the hypothesis of induction and from Lemma 1 it follows that $(x,x_{n+1}|x_{n},...,x_{2})_{*}=0$. Then

[TABLE]

Case 2: $(x,x|x_{n},...,x_{2})_{*}>0$. From P1) for $n$ we have $(z,z|x_{n},...,x_{2})_{*}\geq 0$, for all $z\in X$, then

[TABLE]

We obtain the following relation:

[TABLE]

Since $(x,x|x_{n},...,x_{2})_{*}>0$, the discriminant $\Delta_{\lambda}$ of this polynomial in variable $\lambda$ is not strictly positive. Hence $(x,x|x_{n+1},x_{n},...,x_{2})_{*}=-\frac{1}{4}\Delta_{\lambda}\geq 0.$ So, in both cases we obtain $(x,x|x_{n+1},...,x_{2})_{*}\geq 0$.

On the other hand, let us suppose that $(x,x|x_{n+1},x_{n},...,x_{2})_{*}=0$, which means that

[TABLE]

If $(x_{n+1},x_{n+1}|x_{n},...,x_{2})_{*}\not=0$, the expression above is equal to $-\frac{1}{4}\Delta_{\lambda}$, where $\Delta_{\lambda}$ is the discriminat of the polynomial equation of degree 2 in $\lambda$: $Q(\lambda)=0$, where $Q(\lambda)=(x+\lambda x_{n+1},x+\lambda x_{n+1}|x_{n},...,x_{2})_{*}$. Since the discriminat is [math], then there exists $\lambda_{0}\in\mathbb{R}$, for which $Q(\lambda_{0})=0$. From condition P1) for $n$ it follows that $x+\lambda_{0}x_{n+1},x_{n},...,x_{2}$ are linearly dependent. Then, there are the numbers $\alpha,\alpha_{i}\in\mathbb{R}$, not all null, such that $\alpha(x+\lambda_{0}x_{n+1})+\sum_{i=2}^{n}\alpha_{i}x_{i}=0$. Therefore, $x,x_{2},...,x_{n+1}$ are linearly dependent. If $(x_{n+1},x_{n+1}|x_{n},...,x_{2})_{*}=0$, then $x_{2},...,x_{n+1}$ are linearly dependent from P1) for $n$. Then $x,x_{2},...,x_{n+1}$ are linearly dependent. Condition P1) is completely proved for $n+1$.

We prove condition P2) for $n+1$:

[TABLE]

[TABLE]

Consequently, condition P2) is true for $n+1$.

We pass to the verification of condition P3). We have

[TABLE]

[TABLE]

because $\det A=\det A^{T}$, for any square matrix $A$ and $(x,y|x_{n},...,x_{2})_{*}=(y,x|x_{n},..,x_{2})_{*}$. So, the $(n+1)$-iterated $2$-inner product satisfies condition P3) for $n+1$.

We pass now to condition P4). Since we have

[TABLE]

[TABLE]

it follows that condition P4) is proved for $n+1$.

For condition P5) for $n+1$, we take into account that $(x+x^{\prime},y|x_{n+1},x_{n},...,x_{2})_{*}$ can be expressed by a determinat of second order, having on the first line the elements $(x+x^{\prime},y|x_{n},...,x_{2})_{*}$ and $(x+x^{\prime},x_{n+1}|x_{n},...,x_{2})_{*}$, respectively, and on the second line, elements which do not depend on $x$ and $x^{\prime}$. Then, using by induction the additivity in the first argument of the products above, and then the additivity of the determinant with regard to the first line, it follows immediately that

[TABLE]

Proposition 2**.**

Let $(X,\langle\cdot,\cdot\rangle)$ be a real pre-Hilbert space. For $x,y,x_{2},\ldots,x_{n}\in X$, $n\geq 2$ and $t\in\mathbb{R}$:

[TABLE]

Proof For $n=2$, $(tx,ty|tx_{2})_{*}=t^{4}(x,y|x_{2})_{*}.$ Then, it follows by mathematical induction.

Remark 4**.**

Theorem 4 allows us to furnish an example of weak $n$-inner product which is not a $n$-inner product. For this, let $X=\mathbb{R}^{3}$ endowed with the usual inner product. Then, from Theorem 4, $3$-iterated $2$-inner product $(\cdot,\cdot|\cdot,\cdot)_{*}:X^{4}\to\mathbb{R}$ is a weak $3$-inner product, but it is not a $3$-inner product. Indeed, if axiom I2) would be true for $3$-iterated $2$-inner product then we must have:

[TABLE]

But, if we choose $x=(1,0,0)$, $u=(1,1,1)$, $v=(2,1,2)$ we have

[TABLE]

and on other hand

[TABLE]

Hence relation (11) is not true. Consequently axiom I2) is not satisfied. Therefore $3$-iterated $2$-inner product is not a $3$-inner product.

3 Representation of the $n$-iterated $2$-inner product in terms of the standard $k$-inner products, $(k\leq n)$

In this section we obtain a representation of the $n$-iterated $2$-inner product, given in Definition 4 in terms of the standard $k$-inner products $k\leq n$. For this we use Dodgson’s identity for determinants, [8], [9]. Historical notes about this identity, in connection with Chiò’s formula can be found in [1]. To express this identity we adopt the compact notation used by Eves [11]. If $A=(a_{i,j})_{1\leq i,j\leq n}$ is a square matrix, denote the determinant of $A$ by $|a_{1,1}\;\ldots a_{n,n}|$ and the sub-determinant involving rows $i_{1},\ldots,i_{s}$ and columns $j_{1},\ldots,j_{s}$ by $|a_{i_{1},j_{1}}\;\ldots a_{i_{s},j_{s}}|$. In [11] - Theorem 3.6.3, the following Dodgson type identity $(n\geq 3)$ is proved:

[TABLE]

For us it is more convenient to use the following identity $(n\geq 3)$:

[TABLE]

For $n=3$ one has:

[TABLE]

Formula (13) can be easily obtained applying formula (12). Indeed, first note that

[TABLE]

Then, applying rule (12) for our new matrix we find

[TABLE]

Note that, conversely, from relation (13) one can deduce relation (12).

Let $(X,\langle\cdot,\cdot\rangle)$ be an inner product space. For $x,y,z,v\in X$, from (13), for $n=3$ we deduce

[TABLE]

Hence, we obtained

[TABLE]

Also, using formula (13), for $n=4$ and then formula (15) we obtain :

[TABLE]

[TABLE]

Hence

[TABLE]

The results obtained in (15) and (16) can be generalized as it is shown in the next theorem. We extend the definition of the standard weak $n$-inner product, for $n=1$, by the convention $\langle x,y|x_{1},\ldots,x_{2}\rangle=\langle x,y\rangle$.

Theorem 5**.**

Let $(X,\langle\cdot,\cdot\rangle)$ be an inner product space. For $n\geq 2$, consider the vectors $x,y,x_{2},\ldots,x_{n}\in X.$. Then

[TABLE]

where $E_{2}=1$ and

[TABLE]

Proof For $n=2$ the theorem is immediate, since $(x,y|x_{2})_{*}=\langle x,y|x_{2}\rangle$ and $E_{2}=1$. For $n=3$ the theorem follows from relation (15), for the choice $v=x_{3}$ and $z=x_{2}$. Then $E_{3}=\langle z,z\rangle$. For $n\geq 4$ we prove by induction. Suppose the theorem true for $n\geq 3$ and let us prove it for $n+1$. Using the hypothesis of induction we get

[TABLE]

We transform all the four elements from the above determinant. Each of them is a determinant of order $n$. First, in the following determinant, changing the order of the last $n-1$ lines and then changing the order of the last $n-1$ columns we obtain successively

[TABLE]

Next, for the second determinant, we change the order of all the $n$ columns and then we change the order of the last $n-1$ lines we obtain:

[TABLE]

since $\frac{(n-1)n}{2}+\frac{(n-2)(n-1)}{2}=(n-1)^{2}.$

Similar operations there can be made for the third determinant. We change the order of all the $n$ lines and then we change the order of the last $n-1$ columns and we get:

[TABLE]

Finally, applying formula I2) we have

[TABLE]

Consider the matrix

[TABLE]

Denote the elements of $A$ by $a_{i,j}$, $1\leq i,j\leq n+1$. Using the notation given in the beginning of the section we can write $|A|=|a_{1,1}\;a_{2,2}\;\ldots\;a_{n+1,n+1}|$.

From formula (20) we obtain $\langle x,y|x_{n},\ldots,x_{2}\rangle=|a_{1,1}\;a_{2,2}\;\ldots\;a_{n,n}|$.

From formula (21) we obtain $\langle x,x_{n+1}|x_{n},\ldots,x_{2}\rangle=(-1)^{(n-1)^{2}}|a_{1,2}\;a_{2,3}\;\ldots\;a_{n,n+1}|$.

From formula (22) we obtain $\langle x_{n+1},y|x_{n},\ldots,x_{2}\rangle=(-1)^{(n-1)^{2}}|a_{2,1}\;a_{3,2}\;\ldots\;a_{n+1,n}|$.

From formula (23) we obtain $\langle x_{n+1},x_{n+1}|x_{n},\ldots,x_{2}\rangle=|a_{2,2}\;a_{3,3}\;\ldots\;a_{n+1,n+1}|$.

Then applying formula (13) for $n+1$ instead of $n$ we arrive to

[TABLE]

If we change the order of the last $n$ lines and of the last $n$ columns in $|A|$ the determinant does not change, i.e.

[TABLE]

But

[TABLE]

Therefore,

[TABLE]

Also, if we change the order of the lines and columns in determinat $|a_{2,2}\;a_{3,3}\;\ldots\;a_{n,n}|$, the value does not change. Hence

[TABLE]

But

[TABLE]

Therefore,

[TABLE]

From relations (24), (25) and (26) we conclude that

[TABLE]

Replacing in (19) we obtain

[TABLE]

Since $(E_{n})^{2}\langle x_{n},x_{n}|x_{n-1},\ldots,x_{2}\rangle=E_{n+1}$ it results, finally, that

[TABLE]

4 Several applications of the $n$ iterated $2$-inner product

1. Let $X=(X,\langle\cdot,\cdot\rangle)$ be an inner product space. Let $x,w,z\in X$. From Definition 4 we deduce

[TABLE]

Relation (28) can be written as

[TABLE]

Since $(x,x|w,z)_{*}\geq 0$, then we obtain the inequality from Lupu and Schwarz [19] given by the following:

[TABLE]

2. Formula (15) can be written in the form

[TABLE]

Therefore, for $\|z\|\neq 1$, we have $(x,y|w,z)_{*}$ $\neq\langle x,y|w,z\rangle.$ Also, since in the case $x=y$, the determinant in (31) is the Gram’s determinant $\Gamma(x,w,z)$, from relation (31) we can deduce:

[TABLE]

Since, also $(x,x|z,w)_{*}=\Gamma(x,z,w)\cdot\|w\|^{2}$ and $\Gamma(x,w,z)=\Gamma(x,z,w)$ it results

[TABLE]

3. From Theorem 5 for $n=4$ we find that the $4$ iterated $2$-inner product can be given in the following way:

[TABLE]

From relation (34), for $x=y$ we deduce:

[TABLE]

where $\Gamma(x,v,w,z)$ is the Gram’s determinant.

In [4], Cho, Matić and Pec̆arić, used Gram’s determinant $\Gamma(x_{1},x_{2},...,x_{k}|z)$ of the vectors $x_{1},x_{2},...,x_{k}$ with respect to the vector $z$ by:

[TABLE]

We consider the following determinant, which can be rewritten using formula (14):

[TABLE]

From relations (34) and (37) we find the following identity:

[TABLE]

which implies the relation:

[TABLE]

4. Let $x,y,e,w$ be vectors in the inner product space $X,$ over the field of real numbers and the vectors $\{e,x,y\}$ being linearly independent, such that

[TABLE]

where $a,b,c\in\mathbb{R}.$

We want to study the problem of determining the scalars $a,b,c$. Using the inner product and its properties, we deduce

[TABLE]

Therefore, we have to solve this system with three equations and three unknowns $a,b,c\in\mathbb{R}.$ The matrix of the system is

[TABLE]

From formula (31) we find:

[TABLE]

From P1) $(x,x|y,e)_{*}$ is zero if and only if the vectors $x,y,e$ are linearly dependent. But, the vectors $\{e,x,y\}$ are linearly independent, therefore, we have $(x,x|y,e)_{*}>0$. Using the Cramer method, we find that

[TABLE]

In the particular case when $\|e\|=1$ we obtain:

[TABLE]

5. Next, we will make a correlation of the previous calculations with the coefficients that appear in the case of a multiple linear regression model.

A process is called multiple linear regression, when we have more than one independent variable [12]. For a general linear model for two independent variables $V$ and $W$ and a dependent variable $Z$, $Z=aV+bW+c$, where $V=\begin{pmatrix}x_{i}\\ \frac{1}{n}\end{pmatrix}_{1\leq i\leq n}$; $W=\begin{pmatrix}y_{i}\\ \frac{1}{n}\end{pmatrix}_{1\leq i\leq n}$; $Z=\begin{pmatrix}z_{i}\\ \frac{1}{n}\end{pmatrix}_{1\leq i\leq n}$ with probabilities $P(V=x_{i})=\dfrac{1}{n}$, $P(W=y_{i})=\dfrac{1}{n}$, $P(Z=z_{i})=\dfrac{1}{n},$ for any $i=\overline{1,n}.$

We can describe the underlying relationship between $z_{i}$ and $x_{i},y_{i}$ involving error term $\epsilon_{i}$ by $\epsilon_{i}=z_{i}-ax_{i}-by_{i}-c$.

If we take $S(a,b,c)=\sum_{i=1}^{n}\epsilon_{i}^{2}=\sum_{i=1}^{n}(z_{i}-ax_{i}-by_{i}-c)^{2}$, then we have to find $\underset{a,b,c\in\mathbb{R}}{min}S(a,b,c).$ Using the Lagrange method, we obtain

[TABLE]

By simple calculations, we deduce

[TABLE]

Now, we take the vector space $(X=\mathbb{R}^{n},\langle\cdot,\cdot\rangle).$ For $x=(x_{1},x_{2},...,x_{n}),\ y=(y_{1},y_{2},...,y_{n}),\ z=(z_{1},z_{2},...,z_{n}),$ we have

[TABLE]

[TABLE]

and

[TABLE]

If $e=\frac{u}{\|u\|}$, where $u=(1,1,...,1)\in\mathbb{R}^{n}$, then the average of vector $x$ is \mu_{x}=\bigg{\langle}\frac{x}{\|u\|},e\bigg{\rangle}=\frac{1}{n}\sum_{i=1}^{n}x_{i}, and we have

[TABLE]

Therefore, in $(\mathbb{R}^{n},\langle\cdot,\cdot\rangle)$, we define the variance of a vector $x$ by var(x):=\bigg{\|}\frac{x}{\|u\|}|e\bigg{\|}^{2}.

The standard deviation $\sigma(x)$ of $x\in\mathbb{R}^{n}$ is defined by $\sigma(x):=\sqrt{var(x)}$, so we deduce that \sigma(x)=\bigg{\|}\frac{x}{\|u\|}|e\bigg{\|}. Since, using the standard $2$-inner product, we have

[TABLE]

it is easy to define the covariance of two vectors $x$ and $y$ by

[TABLE]

It is easy to see that, we obtain

[TABLE]

We observe that by the vector method we obtain the same coefficients as by the Lagrange method.

6. In [24], the Chebyshev functional is defined by

[TABLE]

for all $x,y\in X$, where $z\in X$ is a given nonzero vector.

It is easy to see that if the standard $2$-inner product $(\cdot,\cdot|\cdot)$ is defined by the inner product $\langle\cdot,\cdot\rangle,$ then we have $T_{z}(x,y)=(x,y|z)_{*}=(x,y|z)$.

Therefore, we generalize this Chebyshev functional to the following functional:

[TABLE]

which we will call n-Chebyshev functional, so

[TABLE]

for all $x,y\in X$, where $x_{2},...,x_{n}\in X$ are given nonzero vectors.

In a particular case, when $n=3$, we have

[TABLE]

so, we have

[TABLE]

Therefore, the Cauchy-Schwarz inequality in terms of the $n$-Chebyshev functional becomes:

[TABLE]

5 Conclusions

In this paper we exemplified the weak $n$-inner product only by the weak $n$ iterated $2$-inner product. This particular case of weak $n$-inner product does not exhaust all the possibilities of particular cases. The weak $n$-inner product is clearly more general then the $n$-inner product and consequently it offers more possibilities. An important connection is between the vector method and the Lagrange method given above. In the future, we will determine a formula for multiple regression for $n$ independent variables.

Acknowledgment. The authors would like to thank to the reviewers for the pertinent remarks, which led to an improvement of the paper.