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# Generalized Dunkl-Williams inequality in 2-inner product spaces

Journal of Inequalities and Applications20132013:36

https://doi.org/10.1186/1029-242X-2013-36

• Received: 6 September 2012
• Accepted: 17 January 2013
• Published:

## Abstract

We consider the generalized Dunkl-Williams inequality in 2-normed spaces. Also, we give necessary and sufficient conditions for having the equality case in the strictly convex 2-normed space X.

## Keywords

• 2-normed spaces
• 2-inner product spaces
• strictly convex
• Dunkl-William inequality
• Maligranda inequality

## 1 Introductions and preliminaries

In 1964, Dunkl and Williams  proved that if x, y are non-zero vectors in a normed linear space X, then
$\parallel \frac{x}{\parallel x\parallel }-\frac{y}{\parallel y\parallel }\parallel ⩽\frac{4\parallel x-y\parallel }{\parallel x\parallel +\parallel y\parallel }.$
(1.1)

Also, after that it was proved that the equality holds if and only if $x=y$.

Maligranda  obtained a refinement of the Dunkl-Williams inequality. He proved that if x, y are non-zero vectors in a normed linear space X, then
Also, the Maligranda inequality and its reverse in normed linear spaces were proved by Mercer in . In  Kato et al. improved the triangle inequality and provided the reverse by showing that

for all non-zero elements ${x}_{1},\dots ,{x}_{n}$ of a normed linear space.

Pečarić and Rajić  sharpened inequalities (1.4) and (1.5) (when $n>2$) and generalized inequalities (1.2) and (1.3) by showing that

for all non-zero elements ${x}_{1},\dots ,{x}_{n}$ of a normed linear space.

Dragomir  replaced arbitrary scalars instead of $\frac{1}{\parallel {x}_{i}\parallel }$ for $i=1,\dots ,n$ in inequalities (1.6) and (1.7) and obtained a generalization of inequalities (1.6) and (1.7). This paper contains two sections. In the first section, we will work on a generalized case of Dunkl-Williams inequality in 2-normed spaces and also give necessary and sufficient conditions for having equality. In the second part, we want to introduce a refinement of the inequality $\parallel \frac{x}{\parallel x\parallel }-\frac{y}{\parallel y\parallel }\parallel ⩽\frac{2\parallel x-y\parallel }{\parallel x\parallel +\parallel y\parallel }$ in 2-inner product spaces, which was done by Mercer  in inner product spaces.

The concept of 2-normed spaces was introduced by Gähler  in 1963. After that, in 1973 and 1977, Diminnie, Gähler and White introduced the concept of 2-inner product spaces (see [8, 9]). We offer  to readers for more details.

Definition 1.1 Let $\mathbb{K}$ be the symbol of the field or and X be a linear space on $\mathbb{K}$. Define the $\mathbb{K}$-valued function $\left(\cdot ,\cdot |\cdot \right)$ on $X×X×X$ with the following properties:
1. (1)

$\left(x,x|y\right)⩾0$; $\left(x,x|y\right)=0$ if and only if x and y are linearly dependent,

2. (2)

$\left(x,x|y\right)=\left(y,y|x\right)$,

3. (3)

$\left(x,y|z\right)=\overline{\left(y,x|z\right)}$,

4. (4)

$\left(\alpha x,y|z\right)=\alpha \left(x,y|z\right)$ for any $\alpha \in \mathbb{K}$,

5. (5)

$\left(x+{x}^{\mathrm{\prime }},y|z\right)=\left(x,y|z\right)+\left({x}^{\mathrm{\prime }},y|z\right)$,

for all $x,{x}^{\mathrm{\prime }},y,z\in X$. $\left(\cdot ,\cdot |\cdot \right)$ is called a 2-inner product and $\left(X,\left(\cdot ,\cdot |\cdot \right)\right)$ is called a 2-inner product space.

Lemma 1.2 Let X be a 2-inner product space. Then
$|\left(x,y|z\right)|⩽\sqrt{\left(x,x|z\right)}\sqrt{\left(y,y|z\right)}$
(1.8)

for every $x,y,z\in X$.

Definition 1.3  Let X be a linear space of dimension greater than 1 on the field $\mathbb{K}$ and let $\parallel \cdot ,\cdot \parallel :X×X\to \left[0,+\mathrm{\infty }\right)$ be a function satisfying the following conditions:
1. (1)

$\parallel x,y\parallel =0$ if and only if x and y are linearly dependent,

2. (2)

$\parallel x,y\parallel =\parallel y,x\parallel$,

3. (3)

$\parallel \alpha x,y\parallel =|\alpha |\parallel x,y\parallel$ for all $\alpha \in \mathbb{K}$,

4. (4)

$\parallel x+y,z\parallel ⩽\parallel x,z\parallel +\parallel y,z\parallel$,

for all $x,y,z\in X$. $\parallel \cdot ,\cdot \parallel$ is called a 2-norm and $\left(X,\parallel \cdot ,\cdot \parallel \right)$ is called a linear 2-normed space.

It follows from (4) that
$|\parallel x,z\parallel -\parallel y,z\parallel |⩽\parallel x-y,z\parallel$

for all $x,y,z\in X$.

Let X be a 2-inner product space of dimension greater than 1 on the field . If we define $\parallel x,y\parallel =\sqrt{\left(x,x|y\right)}$ for all $x,y\in X$, then $\parallel \cdot ,\cdot \parallel$ is a 2-norm on X and
(see Theorem 3.1.9 of ). In this case, (1.8) means
$|\left(x,y|z\right)|⩽\parallel x,z\parallel \parallel y,z\parallel .$

A linear 2-normed space $\left(X,\parallel \cdot ,\cdot \parallel \right)$ is said to be strictly convex if $\parallel x+y,z\parallel =\parallel x,z\parallel +\parallel y,z\parallel$, $\parallel x,z\parallel =\parallel y,z\parallel =1$ and $z\notin span\left\{x,y\right\}$, then $x=y$.

## 2 Main results

In this section, we establish a generalization of Dunkl-Williams inequality and its reverse in 2-normed spaces.

Theorem 2.1 Let X be a 2-normed space on the field $\mathbb{K}$. For ${x}_{1},\dots ,{x}_{n},z\in X$ and ${a}_{1},\dots ,{a}_{n}\in \mathbb{K}$, we have
Proof For a fixed $1⩽i⩽n$, we have
$\begin{array}{rl}\parallel \sum _{j=1}^{n}{a}_{j}{x}_{j},z\parallel & =\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j}+\sum _{j=1}^{n}\left({a}_{j}-{a}_{i}\right){x}_{j},z\parallel \\ ⩽\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j},z\parallel +\parallel \sum _{j=1}^{n}\left({a}_{j}-{a}_{i}\right){x}_{j},z\parallel \\ ⩽\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j},z\parallel +\sum _{j=1}^{n}\parallel \left({a}_{j}-{a}_{i}\right){x}_{j},z\parallel \\ =|{a}_{i}|\parallel \sum _{j=1}^{n}{x}_{j},z\parallel +\sum _{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel .\end{array}$

By taking minimum over $i=1,\dots ,n$, we obtain (2.1).

Now, we have
$\begin{array}{rl}\parallel \sum _{j=1}^{n}{a}_{j}{x}_{j},z\parallel & =\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j}-\sum _{j=1}^{n}\left({a}_{i}-{a}_{j}\right){x}_{j},z\parallel \\ ⩾|\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j},z\parallel -\parallel \sum _{j=1}^{n}\left({a}_{i}-{a}_{j}\right){x}_{j},z\parallel |\\ ⩾\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j},z\parallel -\sum _{j=1}^{n}\parallel \left({a}_{i}-{a}_{j}\right){x}_{j},z\parallel \\ =|{a}_{i}|\parallel \sum _{j=1}^{n}{x}_{j},z\parallel -\sum _{j=1}^{n}|{a}_{i}-{a}_{j}|\parallel {x}_{j},z\parallel .\end{array}$

By taking maximum over $1⩽i⩽n$, we obtain (2.2). □

Theorem 2.2 

Let X be a real linear 2-normed space. The following statements are equivalent:
1. (1)

$\left(X,\parallel \cdot ,\cdot \parallel \right)$ is strictly convex.

2. (2)

If $\parallel x+y,z\parallel =\parallel x,z\parallel +\parallel y,z\parallel$ and $z\notin span\left\{x,y\right\}$, then $x=\lambda y$ for some $\lambda >0$.

Let X be a real linear 2-normed space. If $z\notin span\left\{x,y\right\}$ and $|\left(x,y|z\right)|=\parallel x,z\parallel \parallel y,z\parallel$, then $x=\lambda y$ for some real λ.

Lemma 2.3 Let X be a linear strictly convex 2-normed space with respect to a 2-inner product on the field . For non-zero elements ${x}_{1},\dots ,{x}_{n}\in X$ satisfying ${\sum }_{j=1}^{n}{x}_{j}\ne 0$ , $z\notin span\left\{{x}_{1},\dots ,{x}_{n}\right\}$ and non-zero elements ${a}_{1},\dots ,{a}_{n}$ of such that ${a}_{i}\ne {a}_{j}$ for some i, j, the following statements are equivalent:
1. (1)

$\parallel {\sum }_{j=1}^{n}{a}_{j}{x}_{j},z\parallel =|{a}_{i}|\parallel {\sum }_{j=1}^{n}{x}_{j},z\parallel +{\sum }_{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel$;

2. (2)

${\sum }_{j=1}^{n}\left({x}_{j},{x}_{k}|z\right){a}_{i}\left({a}_{k}-{a}_{i}\right)=\parallel {\sum }_{j=1}^{n}{x}_{j},z\parallel |{a}_{i}||{a}_{k}-{a}_{i}|\parallel {x}_{k},z\parallel$.

Proof Let (1) hold. For a fixed $1⩽i⩽n$, we have
$\begin{array}{rcl}\parallel \sum _{j=1}^{n}{a}_{j}{x}_{j},z\parallel & =& \parallel \sum _{j=1}^{n}{a}_{i}{x}_{j}+\sum _{j=1}^{n}\left({a}_{j}-{a}_{i}\right){x}_{j},z\parallel \\ =& |{a}_{i}|\parallel \sum _{j=1}^{n}{x}_{j},z\parallel +\sum _{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel .\end{array}$
(2.3)
From the assumption, there exists a non-empty maximal subset $\left\{{j}_{1},\dots ,{j}_{m}\right\}$ of $\left\{1,\dots ,n\right\}$ for some $1⩽m⩽n$ such that ${a}_{{j}_{k}}\ne {a}_{i}$ for all $1⩽k⩽m$. Hence, (2.3) holds if and only if
$\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j}+\sum _{k=1}^{m}\left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}},z\parallel =|{a}_{i}|\parallel \sum _{j=1}^{n}{x}_{j},z\parallel +\sum _{k=1}^{m}|{a}_{{j}_{k}}-{a}_{i}|\parallel {x}_{{j}_{k}},z\parallel .$
Using Theorem 2.2, we deduce that there exists ${\beta }_{{j}_{k}}>0$ such that ${\sum }_{j=1}^{n}{a}_{i}{x}_{j}={\beta }_{{j}_{k}}\left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}}$ which is equivalent to
$\left(\sum _{j=1}^{n}{a}_{i}{x}_{j},\left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}}|z\right)=\parallel \sum _{j=1}^{n}{a}_{i}{x}_{j},z\parallel \parallel \left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}},z\parallel .$

So, (1) and (2) are equivalent. □

Lemma 2.4 Let X be a linear strictly convex 2-normed space with respect to a 2-inner product on . For non-zero elements ${x}_{1},\dots ,{x}_{n}\in X$ satisfying ${\sum }_{j=1}^{n}{x}_{j}=0$, $z\notin span\left\{{x}_{1},\dots ,{x}_{n}\right\}$ and non-zero elements ${a}_{1},\dots ,{a}_{n}$ of such that ${a}_{i}\ne {a}_{j}$ for some i, j, the following statements are equivalent:
1. (1)

$\parallel {\sum }_{j=1}^{n}{a}_{j}{x}_{j},z\parallel ={\sum }_{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel$.

2. (2)

There exist $1⩽i,l⩽n$ such that ${a}_{l}\ne {a}_{i}$ and $\left({x}_{k},{x}_{l}|z\right)\left({a}_{k}-{a}_{i}\right)\left({a}_{l}-{a}_{i}\right)=\parallel {x}_{k},z\parallel \parallel {x}_{l},z\parallel |{a}_{k}-{a}_{i}||{a}_{l}-{a}_{i}|$ for all k.

Proof Let i ($1⩽i⩽n$) be fixed. By ${\sum }_{j=1}^{n}{x}_{j}=0$, we get
$\sum _{j=1}^{n}{a}_{j}{x}_{j}=\sum _{j=1}^{n}\left({a}_{j}-{a}_{i}\right){x}_{j}.$
Now, by the above equality and (1), we get
$\parallel \sum _{j=1}^{n}\left({a}_{j}-{a}_{i}\right){x}_{j},z\parallel =\parallel \sum _{j=1}^{n}{a}_{j}{x}_{j},z\parallel =\sum _{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel .$
Let $\left\{{j}_{1},\dots ,{j}_{m}\right\}$ be as in the proof of Lemma 2.3. So,
$\parallel \sum _{k=1}^{m}\left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}},z\parallel =\sum _{k=1}^{m}|{a}_{{j}_{k}}-{a}_{i}|\parallel {x}_{{j}_{k}},z\parallel .$
Now, let $1⩽l⩽m$. Then we get
$\parallel \underset{k\ne l}{\overset{m}{\sum _{k=1}}}\left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}}+\left({a}_{{j}_{l}}-{a}_{i}\right){x}_{{j}_{l}},z\parallel =\underset{k\ne l}{\overset{m}{\sum _{k=1}}}|{a}_{{j}_{k}}-{a}_{i}|\parallel {x}_{{j}_{k}},z\parallel +|{a}_{{j}_{l}}-{a}_{i}|\parallel {x}_{{j}_{l}},z\parallel .$
Hence, for some ${\beta }_{jk}>0$, $\left({a}_{{j}_{k}}-{a}_{i}\right){x}_{{j}_{k}}={\beta }_{jk}\left({a}_{{j}_{l}}-{a}_{i}\right){x}_{{j}_{l}}$. This is equivalent to
$\left({x}_{{j}_{k}},{x}_{{j}_{l}}|z\right)\left({a}_{{j}_{k}}-{a}_{i}\right)\left({a}_{{j}_{l}}-{a}_{i}\right)=\parallel {x}_{{j}_{k}},z\parallel \parallel {x}_{{j}_{l}},z\parallel |{a}_{{j}_{k}}-{a}_{i}||{a}_{{j}_{l}}-{a}_{i}|,$

as desired. □

As an application of Lemmas 2.3 and 2.4, we offer the following theorem.

Theorem 2.5 Let X be a linear strictly convex 2-normed space with respect to a 2-inner product on . Also, let ${x}_{1},\dots ,{x}_{n}\in X$ be non-zero elements, $z\notin span\left\{{x}_{1},\dots ,{x}_{n}\right\}$ and ${a}_{1},\dots ,{a}_{n}\in \mathbb{R}$ be non-zero elements such that ${a}_{i}\ne {a}_{j}$ for some i, j.
1. (1)
If ${\sum }_{j=1}^{n}{x}_{j}\ne 0$, then
$\parallel \sum _{j=1}^{n}{a}_{j}{x}_{j},z\parallel =\underset{1⩽k⩽n}{min}\left\{|{a}_{i}|\parallel \sum _{j=1}^{n}{x}_{j},z\parallel +\sum _{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel \right\}$

if and only if there exists $1⩽i⩽n$ such that
$\sum _{j=1}^{n}\left({x}_{j},{x}_{k}|z\right){a}_{i}\left({a}_{k}-{a}_{i}\right)=\parallel \sum _{j=1}^{n}{x}_{j},z\parallel |{a}_{i}||{a}_{k}-{a}_{i}|\parallel {x}_{k},z\parallel .$
1. (2)
If ${\sum }_{j=1}^{n}{x}_{j}=0$, then
$\parallel \sum _{j=1}^{n}{a}_{j}{x}_{j},z\parallel =\underset{1⩽k⩽n}{min}\sum _{j=1}^{n}|{a}_{j}-{a}_{i}|\parallel {x}_{j},z\parallel$

if and only if there exist $1⩽i,l⩽n$ such that ${a}_{l}\ne {a}_{i}$ and
$\left({x}_{k},{x}_{l}|z\right)\left({a}_{k}-{a}_{i}\right)\left({a}_{l}-{a}_{i}\right)=\parallel {x}_{k},z\parallel \parallel {x}_{l},z\parallel |{a}_{k}-{a}_{i}||{a}_{l}-{a}_{i}|$

for all k.

## 3 Improvement of Dunkl-Williams inequality with two elements

In , Dunkl and Williams proved that in inequality (1.1) the constant 4 is the best choice in normed linear spaces. Moreover, they proved that in an inner product space, the constant 4 can be replaced by 2; that is,
$\parallel \frac{x}{\parallel x\parallel }-\frac{y}{\parallel y\parallel }\parallel ⩽\frac{2\parallel x-y\parallel }{\parallel x\parallel +\parallel y\parallel }.$
(3.1)

In addition, the equality in (3.1) holds if and only if $\parallel x\parallel =\parallel y\parallel$.

In 2-inner product spaces, we have the following theorem.

Theorem 3.1 For non-zero vectors x, y, z in a 2-inner product space X with $z\notin span\left\{x,y\right\}$,
$\parallel \frac{x}{\parallel x,z\parallel }-\frac{y}{\parallel y,z\parallel },z\parallel ⩽\frac{2\parallel x-y,z\parallel }{\parallel x,z\parallel +\parallel y,z\parallel }.$
(3.2)
If X is a linear 2-normed space in Theorem 3.1, then we have the following inequality:
$\parallel \frac{x}{\parallel x,z\parallel }-\frac{y}{\parallel y,z\parallel },z\parallel ⩽\frac{4\parallel x-y,z\parallel }{\parallel x,z\parallel +\parallel y,z\parallel }.$

A refinement of (3.1) has been obtained by Mercer . Now, we use Mercer’s inequality and give a refinement of (3.2).

Theorem 3.2 Let x, y, z be non-zero vectors in a 2-inner product space X with $z\notin span\left\{x,y\right\}$. Then we have
Proof Let $\alpha =\parallel \frac{x}{\parallel x,z\parallel }-\frac{y}{\parallel y,z\parallel },z\parallel$. By using (2.2) for two elements x, −y with constants $\frac{1}{\parallel x,z\parallel }$, $\frac{1}{\parallel y,z\parallel }$ respectively, we obtain
$\alpha ⩾\frac{\parallel x-y,z\parallel -|\parallel x,z\parallel -\parallel y,z\parallel |}{min\left(\parallel x,z\parallel ,\parallel y,z\parallel \right)}.$
So,
$\parallel y,z\parallel +\parallel x,z\parallel -\parallel x-y,z\parallel ⩾\left(2-\alpha \right)min\left(\parallel x,z\parallel ,\parallel y,z\parallel \right).$
A simple computation shows that
$\frac{Re\left(x,y|z\right)}{\parallel x,z\parallel \parallel y,z\parallel }=1-\frac{1}{2}{\alpha }^{2},\phantom{\rule{1em}{0ex}}{\alpha }^{2}=\frac{{\parallel x-y,z\parallel }^{2}-{\left(\parallel y,z\parallel -\parallel x,z\parallel \right)}^{2}}{\parallel x,z\parallel \parallel y,z\parallel }.$
Hence,
${\parallel x-y,z\parallel }^{2}-{\left(\frac{\parallel y,z\parallel +\parallel x,z\parallel }{2}\right)}^{2}{\alpha }^{2}⩾\frac{{\left(\parallel y,z\parallel -\parallel x,z\parallel \right)}^{2}}{2}\left(2-\alpha \right).$
Therefore,
${\alpha }^{2}-2\alpha {\left(\frac{\parallel y,z\parallel -\parallel x,z\parallel }{\parallel y,z\parallel +\parallel x,z\parallel }\right)}^{2}+4\frac{{\left(\parallel y,z\parallel -\parallel x,z\parallel \right)}^{2}-{\parallel x-y,z\parallel }^{2}}{{\left(\parallel y,z\parallel +\parallel x,z\parallel \right)}^{2}}⩽0.$
So, α is between two roots of the quadratic equation
${\lambda }^{2}-2\lambda {\left(\frac{\parallel y,z\parallel -\parallel x,z\parallel }{\parallel y,z\parallel +\parallel x,z\parallel }\right)}^{2}+4\frac{{\left(\parallel y,z\parallel -\parallel x,z\parallel \right)}^{2}-{\parallel x-y,z\parallel }^{2}}{{\left(\parallel y,z\parallel +\parallel x,z\parallel \right)}^{2}}=0.$

Hence, we get (3.3). □

## Declarations

### Acknowledgements

The authors wish to thank the anonymous referees for their valuable comments. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (grant number 2012003499).

## Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Mathematical Sciences, University of Mohaghegh Ardabili, Ardabil, 56199-11367, Iran
(2)
Graduate School of Education, Kyung Hee University, Yongin, 446-701, Republic of Korea

## References 