Generalized Dunkl-Williams inequality in 2-inner product spaces

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.

1 Introductions and preliminaries

In 1964, Dunkl and Williams [1] 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 [2] 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

(1.2)

(1.3)

Also, the Maligranda inequality and its reverse in normed linear spaces were proved by Mercer in [3]. In [4] Kato et al. improved the triangle inequality and provided the reverse by showing that

(1.4)

(1.5)

for all non-zero elements $x_1,\dots ,x_n$ of a normed linear space.

Pečarić and Rajić [5] sharpened inequalities (1.4) and (1.5) (when $n>2$) and generalized inequalities (1.2) and (1.3) by showing that

(1.6)

(1.7)

for all non-zero elements $x_1,\dots ,x_n$ of a normed linear space.

Dragomir [6] 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 [3] in inner product spaces.

The concept of 2-normed spaces was introduced by Gähler [7] 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 [10] to readers for more details.

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

• (1) $(x,x|y)⩾0$; $(x,x|y)=0$ if and only if x and y are linearly dependent,

• (2) $(x,x|y)=(y,y|x)$,

• (3) $(x,y|z)=\overline{(y,x|z)}$,

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

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

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

Lemma 1.2 [10]Let X be a 2-inner product space. Then

$$|(x,y|z)|⩽\sqrt{(x,x|z)}\sqrt{(y,y|z)}$$

(1.8)

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

Definition 1.3 [10] Let X be a linear space of dimension greater than 1 on the field $\mathbb{K}$ and let $\parallel \cdot ,\cdot \parallel :X\times X\to [0,+\mathrm{\infty })$ be a function satisfying the following conditions:

• (1) $\parallel x,y\parallel =0$ if and only if x and y are linearly dependent,

• (2) $\parallel x,y\parallel =\parallel y,x\parallel$,

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

• (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 $(X,\parallel \cdot ,\cdot \parallel )$ 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 ℝ. If we define $\parallel x,y\parallel =\sqrt{(x,x|y)}$ for all $x,y\in X$, then $\parallel \cdot ,\cdot \parallel$ is a 2-norm on X and

(see Theorem 3.1.9 of [10]). In this case, (1.8) means

$$|(x,y|z)|⩽\parallel x,z\parallel \parallel y,z\parallel .$$

A linear 2-normed space $(X,\parallel \cdot ,\cdot \parallel )$ 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 $z\notin span\{x,y\}$, 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

(2.1)

(2.2)

Proof For a fixed $1⩽i⩽n$, we have

$$\begin{array}{rl}\parallel \sum _{j=1}^n{a}_j{x}_j,z\parallel & =\parallel \sum _{j=1}^n{a}_i{x}_j+\sum _{j=1}^n(a_j-a_i)x_j,z\parallel \\ ⩽\parallel \sum _{j=1}^n{a}_i{x}_j,z\parallel +\parallel \sum _{j=1}^n(a_j-a_i)x_j,z\parallel \\ ⩽\parallel \sum _{j=1}^n{a}_i{x}_j,z\parallel +\sum _{j=1}^n\parallel (a_j-a_i)x_j,z\parallel \\ =|a_i|\parallel \sum _{j=1}^n{x}_j,z\parallel +\sum _{j=1}^n|a_j-a_i|\parallel x_j,z\parallel .\end{array}$$

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

Now, we have

$$\begin{array}{rl}\parallel \sum _{j=1}^n{a}_j{x}_j,z\parallel & =\parallel \sum _{j=1}^n{a}_i{x}_j-\sum _{j=1}^n(a_i-a_j)x_j,z\parallel \\ ⩾|\parallel \sum _{j=1}^n{a}_i{x}_j,z\parallel -\parallel \sum _{j=1}^n(a_i-a_j)x_j,z\parallel |\\ ⩾\parallel \sum _{j=1}^n{a}_i{x}_j,z\parallel -\sum _{j=1}^n\parallel (a_i-a_j)x_j,z\parallel \\ =|a_i|\parallel \sum _{j=1}^n{x}_j,z\parallel -\sum _{j=1}^n|a_i-a_j|\parallel x_j,z\parallel .\end{array}$$

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

Theorem 2.2 [10]

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

• (1) $(X,\parallel \cdot ,\cdot \parallel )$ is strictly convex.

• (2) If $\parallel x+y,z\parallel =\parallel x,z\parallel +\parallel y,z\parallel$ and $z\notin span\{x,y\}$, then $x=\lambda y$ for some $\lambda >0$.

Let X be a real linear 2-normed space. If $z\notin span\{x,y\}$ and $|(x,y|z)|=\parallel x,z\parallel \parallel y,z\parallel$, 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 ℝ. For non-zero elements $x_1,\dots ,x_n\in X$ satisfying ${\sum }_{j=1}^n{x}_j\ne 0$ , $z\notin span\{x_1,\dots ,x_n\}$ and non-zero elements $a_1,\dots ,a_n$ of ℝ such that $a_i\ne a_j$ for some i, j, the following statements are equivalent:

• (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) ${\sum }_{j=1}^n(x_j,x_k|z)a_i(a_k-a_i)=\parallel {\sum }_{j=1}^n{x}_j,z\parallel |a_i||a_k-a_i|\parallel x_k,z\parallel$.

Proof Let (1) hold. For a fixed $1⩽i⩽n$, we have

$$\begin{array}{rcl}\parallel \sum _{j=1}^n{a}_j{x}_j,z\parallel & =& \parallel \sum _{j=1}^n{a}_i{x}_j+\sum _{j=1}^n(a_j-a_i)x_j,z\parallel \\ =& |a_i|\parallel \sum _{j=1}^n{x}_j,z\parallel +\sum _{j=1}^n|a_j-a_i|\parallel x_j,z\parallel .\end{array}$$

(2.3)

From the assumption, there exists a non-empty maximal subset $\{j_1,\dots ,j_m\}$ of $\{1,\dots ,n\}$ 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

$$\parallel \sum _{j=1}^n{a}_i{x}_j+\sum _{k=1}^m(a_{j_k}-a_i)x_{j_k},z\parallel =|a_i|\parallel \sum _{j=1}^n{x}_j,z\parallel +\sum _{k=1}^m|a_{j_k}-a_i|\parallel x_{j_k},z\parallel .$$

Using Theorem 2.2, we deduce that there exists ${\beta }_{j_k}>0$ such that ${\sum }_{j=1}^n{a}_i{x}_j={\beta }_{j_k}(a_{j_k}-a_i)x_{j_k}$ which is equivalent to

$$(\sum _{j=1}^n{a}_i{x}_j,(a_{j_k}-a_i)x_{j_k}|z)=\parallel \sum _{j=1}^n{a}_i{x}_j,z\parallel \parallel (a_{j_k}-a_i)x_{j_k},z\parallel .$$

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 ℝ. For non-zero elements $x_1,\dots ,x_n\in X$ satisfying ${\sum }_{j=1}^n{x}_j=0$, $z\notin span\{x_1,\dots ,x_n\}$ and non-zero elements $a_1,\dots ,a_n$ of ℝ such that $a_i\ne a_j$ for some i, j, the following statements are equivalent:

• (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) There exist $1⩽i,l⩽n$ such that $a_l\ne a_i$ and $(x_k,x_l|z)(a_k-a_i)(a_l-a_i)=\parallel x_k,z\parallel \parallel x_l,z\parallel |a_k-a_i||a_l-a_i|$ for all k.

Proof Let i ($1⩽i⩽n$) be fixed. By ${\sum }_{j=1}^n{x}_j=0$, we get

$$\sum _{j=1}^n{a}_j{x}_j=\sum _{j=1}^n(a_j-a_i)x_j.$$

Now, by the above equality and (1), we get

$$\parallel \sum _{j=1}^n(a_j-a_i)x_j,z\parallel =\parallel \sum _{j=1}^n{a}_j{x}_j,z\parallel =\sum _{j=1}^n|a_j-a_i|\parallel x_j,z\parallel .$$

Let $\{j_1,\dots ,j_m\}$ be as in the proof of Lemma 2.3. So,

$$\parallel \sum _{k=1}^m(a_{j_k}-a_i)x_{j_k},z\parallel =\sum _{k=1}^m|a_{j_k}-a_i|\parallel x_{j_k},z\parallel .$$

Now, let $1⩽l⩽m$. Then we get

$$\parallel \underset{k\ne l}{\overset{m}{\sum _{k=1}}}(a_{j_k}-a_i)x_{j_k}+(a_{j_l}-a_i)x_{j_l},z\parallel =\underset{k\ne l}{\overset{m}{\sum _{k=1}}}|a_{j_k}-a_i|\parallel x_{j_k},z\parallel +|a_{j_l}-a_i|\parallel x_{j_l},z\parallel .$$

Hence, for some ${\beta }_{jk}>0$, $(a_{j_k}-a_i)x_{j_k}={\beta }_{jk}(a_{j_l}-a_i)x_{j_l}$. This is equivalent to

$$(x_{j_k},x_{j_l}|z)(a_{j_k}-a_i)(a_{j_l}-a_i)=\parallel x_{j_k},z\parallel \parallel x_{j_l},z\parallel |a_{j_k}-a_i||a_{j_l}-a_i|,$$

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 ℝ. Also, let $x_1,\dots ,x_n\in X$ be non-zero elements, $z\notin span\{x_1,\dots ,x_n\}$ 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) If ${\sum }_{j=1}^n{x}_j\ne 0$, then

  $$\parallel \sum _{j=1}^n{a}_j{x}_j,z\parallel =\underset{1⩽k⩽n}{min}\{|a_i|\parallel \sum _{j=1}^n{x}_j,z\parallel +\sum _{j=1}^n|a_j-a_i|\parallel x_j,z\parallel \}$$

if and only if there exists $1⩽i⩽n$ such that

$$\sum _{j=1}^n(x_j,x_k|z)a_i(a_k-a_i)=\parallel \sum _{j=1}^n{x}_j,z\parallel |a_i||a_k-a_i|\parallel x_k,z\parallel .$$

• (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

$$(x_k,x_l|z)(a_k-a_i)(a_l-a_i)=\parallel x_k,z\parallel \parallel x_l,z\parallel |a_k-a_i||a_l-a_i|$$

for all k.

3 Improvement of Dunkl-Williams inequality with two elements

In [1], 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 [11]For non-zero vectors x, y, z in a 2-inner product space X with $z\notin span\{x,y\}$,

$$\parallel \frac{x}{\parallel x,z\parallel }-\frac{y}{\parallel y,z\parallel },z\parallel ⩽\frac{2\parallel x-y,z\parallel }{\parallel x,z\parallel +\parallel y,z\parallel }.$$

(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 [3]. 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 $z\notin span\{x,y\}$. Then we have

(3.3)

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

$$\alpha ⩾\frac{\parallel x-y,z\parallel -|\parallel x,z\parallel -\parallel y,z\parallel |}{min(\parallel x,z\parallel ,\parallel y,z\parallel )}.$$

Clearly, we have

So,

$$\parallel y,z\parallel +\parallel x,z\parallel -\parallel x-y,z\parallel ⩾(2-\alpha )min(\parallel x,z\parallel ,\parallel y,z\parallel ).$$

A simple computation shows that

$$\frac{Re(x,y|z)}{\parallel x,z\parallel \parallel y,z\parallel }=1-\frac{1}{2}{\alpha }^2,{\alpha }^2=\frac{{\parallel x-y,z\parallel }^2-{(\parallel y,z\parallel -\parallel x,z\parallel )}^2}{\parallel x,z\parallel \parallel y,z\parallel }.$$

Therefore,

Hence,

$${\parallel x-y,z\parallel }^2-{\left(\frac{\parallel y,z\parallel +\parallel x,z\parallel }{2}\right)}^2{\alpha }^2⩾\frac{{(\parallel y,z\parallel -\parallel x,z\parallel )}^2}{2}(2-\alpha ).$$

Therefore,

$${\alpha }^2-2\alpha {\left(\frac{\parallel y,z\parallel -\parallel x,z\parallel }{\parallel y,z\parallel +\parallel x,z\parallel }\right)}^2+4\frac{{(\parallel y,z\parallel -\parallel x,z\parallel )}^2-{\parallel x-y,z\parallel }^2}{{(\parallel y,z\parallel +\parallel x,z\parallel )}^2}⩽0.$$

So, α is between two roots of the quadratic equation

$${\lambda }^2-2\lambda {\left(\frac{\parallel y,z\parallel -\parallel x,z\parallel }{\parallel y,z\parallel +\parallel x,z\parallel }\right)}^2+4\frac{{(\parallel y,z\parallel -\parallel x,z\parallel )}^2-{\parallel x-y,z\parallel }^2}{{(\parallel y,z\parallel +\parallel x,z\parallel )}^2}=0.$$

Hence, we get (3.3). □