Dragomir and Gosa type inequalities on b-metric spaces

Abstract

In this paper, we investigate Dragomir and Gosa type inequalities in the setting of b-metric spaces. As an application, we consider some inequalities in b-normed spaces. We prove that the inequalities admit geometrical interpretation.

1 Introduction and preliminaries

It is a natural trend in fixed point theory to refine a standard metric space structure with a weaker one. One of the interesting extensions of the notion of a metric space is the concept of a b-metric space which was introduced by Czerwik [8].

Definition 1.1

([8])

Let X be a nonempty set and $s\geq 1$ a given real number. A mapping $d \colon X \times X\to [0, \infty )$ is said to be a b-metric if for all $x, y, z \in X$ the following conditions are satisfied:

$(bM_{1})$ :

$d(x, y) =0$ if and only if $x = y$;

$(bM_{2})$ :

$d(x, y) = d(y,x)$ (symmetry);

$(bM_{3})$ :

$d(x, z)\leq s[d(x, y) + d(y, z)]$ (b-triangle inequality).

In this case, the pair $(X, d)$ is called a b-metric space (with constant s).

Clearly, any metric space is a b-metric space (with constant $s=1$).

Example 1.2

([10])

Let $X= [ 0,1 ]$ and let $d:X\times X\longrightarrow {}[ 0,\infty )$ be defined by $d ( x,y ) = ( x-y ) ^{2}$. Then, clearly, $( X,d )$ is a b-metric space with $s=2$.

The following is another constructive example of b-metric.

Example 1.3

([1])

Let $X=\{x_{i}: 1\leq i\leq M\}$ for some $M \in \mathbb{N}$ and $s\geq 2$. Define $d: X\times X\to \infty$ as

$$ d(x_{i},x_{j})= \textstyle\begin{cases} 0 & \text{if } i=j, \\ s & \text{if } (i,j)=(1,2) \text{ or } (i,j)=(2,1), \\ 1 & \text{otherwise.} \end{cases} $$

Consequently, we derive that

$$ d(x_{i},x_{j}) \leq \frac{s}{2}\bigl[d(x_{i},x_{k}) +d(x_{k},x_{j}) \bigr], $$

for all $i,j,k \in \{1,M\}$. Thus, $(X,d)$ forms a b-metric for $s >2$ where the ordinary triangle inequality does not hold.

For more examples for b-metric, we may refer, e.g., to [1,2,3,4,5,6,7, 9, 12] and the corresponding references therein.

Example 1.4

(see, e.g., [6])

The space $L^{p}[0,1]$ (where $0< p<1$) of all real functions $x(t)$, $t\in [0,1]$ such that $\int _{0}^{1} |x(t)|^{p} \,dt<\infty$, together with the functional

$$ d(x,y):= \biggl( \int _{0}^{1} \bigl\vert x(t)-y(t) \bigr\vert ^{p} \,dt \biggr)^{1/p}, \quad \text{for each } x,y\in L^{p}[0,1], $$

is a b-metric space. Notice that $s=2^{1/p}$.

2 Main result

We start this section by recalling an interesting inequality that was proposed by Dragomir and Gosa in [11]. In what follows we investigate their inequality in the setting of a more general structure, namely that of b-metric spaces.

Theorem 2.1

Let $( X,d )$ be a b-metric space with constant $s\geq 1$, and $x_{i}\in X$, $p_{i}\geq 0$ ($i\in \{ 1,2, \dots ,n \}$) with $\sum_{i=1}^{n}p_{i}= \frac{1}{s}$. Then we have

$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq \inf_{x\in X} \Biggl[ \sum_{i=1}^{n}p_{i}d ( x _{i},x ) \Biggr]. $$

(1)

The inequality is sharp in the sense that the constant $c=1$ in front of the infimum cannot be replaced by a smaller constant.

Proof

Using the b-triangle inequality, for any $x\in X$, $i,j\in \{ 1,2,\dots ,n \}$ we have

$$ d ( x_{i},x_{j} ) \leq s \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr]. $$

(2)

If we multiply (2) by $p_{i}$, $p_{j}$ and sum over i and j from 1 to n, we get

$$ \sum_{i,j=1}^{n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq s \Biggl[ \sum _{i,j=1}^{n}p_{i}p_{j} \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr] \Biggr]. $$

Note that by symmetry we have

$$ \sum_{i,j=1}^{n}p_{i}p_{j}d ( x_{i},x_{j} ) =2 \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ). $$

(3)

Now, using the condition $\sum_{i=1}^{n}p_{i}=\frac{1}{s}$, we can easily deduce that

$$ \sum_{i,j=1}^{n}p_{i}p_{j} \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr] =\frac{2}{s} \sum_{i=1}^{n}p _{i}d ( x_{i},x ). $$

So, from (3) we have

$$ \begin{aligned} \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) &= \frac{1}{2}\sum _{i,j=1}^{n}p_{i}p_{j}d ( x_{i},x_{j} ) \\ &\leqslant \frac{s}{2} \Biggl[ \sum_{i,j=1}^{n}p_{i}p_{j} \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr] \Biggr] \\ &=\sum_{i=1}^{n}p_{i}d ( x_{i},x ). \end{aligned} $$

Therefore,

$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq \sum_{i=1}^{n}p_{i}d ( x_{i},x ) $$

for any $x\in X$. Using the fact that the infimum is the greatest lower bound, we deduce (1).

Now, suppose that there exists $c>0$ such that

$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq c \inf_{x\in X} \Biggl[ \sum_{i=1}^{n}p_{i}d ( x_{i},x ) \Biggr]; $$

and choose $n=2$, $p_{1}=p$ and $p_{2}=1-p$ where $p\in (0,1)$. Then,

$$ p(1-p)d ( x_{1},x_{2} )\leq c \bigl[ pd ( x_{1},x )+(1-p)d ( x,x_{2} ) \bigr]. $$

(4)

If we let $x=x_{1}$ in (4), we get

$$ p(1-p)d ( x_{1},x_{2} )\leq c(1-p)d ( x_{1},x_{2} ). $$

As $d ( x_{1},x_{2} )>0$ and $1-p>0$, so $p\leq c$ for any $p\in (0,1)$. Using the fact that the supremum is the least upper bound, we deduce that $c\geq 1$. □

The following corollary is a generalization of Corollary 1 in [11] to the case of a b-metric space.

Corollary 2.2

Let $( X,d )$ be a b-metric space with constant $s\geq 1$, and $x_{i}\in X$, $i\in \{ 1,2,\dots ,n \}$, then

$$ \sum_{1\leq i< j\leq n}d ( x_{i},x_{j} ) \leq \frac{n}{s} \inf_{x\in X} \Biggl[ \sum _{i=1}^{n}d ( x_{i},x ) \Biggr] . $$

The proof follows directly by taking $p_{i}=\frac{1}{ns}$, $i\in \{ 1,2,\dots ,n \}$ in the previous theorem.

The above corollary can be interpreted geometrically as follows: The sum of all edges and diagonals of a polygon with n vertices in a b-metric space is less than or equal to $\frac{n}{s}$-times the sum of the distances from any arbitrary point in the space to its vertices.

The next corollary is a generalization of Corollary 2 in [11] in the framework of b-metric spaces.

Corollary 2.3

Let $( X,d )$ be a b-metric space with constant s and $x_{i}\in X$, $i\in \{ 1,2,\dots ,n \}$. If there exist $z\in X$ and $r>0$ such that the closed ball $\overline{B} ( z,r ) = \{ y\in X:d ( z,y ) \leq r \}$ contains all the points $x_{i}$, then for any $p_{i}\geq 0$ with $\sum_{i=1}^{n}p_{i}=\frac{1}{s}$ we have

$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq \frac{r}{s}. $$

Proof

Using (1) we have

$$ \begin{aligned} \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) & \leq \inf_{x\in X} \Biggl[ \sum_{i=1}^{n}p_{i}d ( x _{i},x ) \Biggr] \\ & \leq \sum_{i=1}^{n}p_{i}d ( x_{i},z ) \\ &\leq \frac{r}{s}. \end{aligned} $$

 □

3 Applications

In this section we define a new notion of a b-normed space and study some of its properties.

Definition 3.1

Let X be a vector space over a field K and let $s\geq 1$ be a constant. A function $\Vert \cdot \Vert _{b}:X\longrightarrow {}[ 0,\infty )$ is said to be a b-norm if the following conditions hold for every $x,y\in X$, $c\in K$:

(Nb1):

$\Vert x \Vert _{b}\geq 0$;

(Nb2):

$\Vert x \Vert _{b}=0\Longleftrightarrow x=0$;

(Nb3):

$\Vert cx \Vert _{b}=|c|^{\log _{2}s+1} \Vert x \Vert _{b}$ (b-homogeneity);

(Nb4):

$\Vert x+y \Vert _{b}\leq s [ \Vert x \Vert _{b}+ \Vert y \Vert _{b} ]$ (b-norm triangle inequality).

In this case $( X, \Vert \cdot \Vert _{b} )$ is called a b-normed space with constant s.

Here we give an example of a b-normed space.

Example 3.2

Let $X=\mathbb{R}$ and define $\Vert \cdot \Vert _{b}:X \longrightarrow {}[ 0,\infty )$ by $\Vert x \Vert _{b}=|x|^{p}$ where $p\in (1,\infty )$, then, using the relation $( x+y ) ^{p}\leq 2^{p-1} ( x+y )$, we can easily deduce that $( X, \Vert \cdot \Vert _{b} )$ is a b-normed space with constant $s=2^{p-1}$.

Remark 3.3

Let $( X, \Vert \cdot \Vert _{b} )$ be a b-normed space with constant $s\geq 1$, $x_{i}\in X$, $i\in \{ 1,\dots ,n \}$. Then it is easy to prove the following generalized b-triangle inequality:

$$ \Biggl\Vert \sum_{i=1}^{n}x_{i} \Biggr\Vert \leq \sum_{i=1}^{n}s^{i} \Vert x_{i} \Vert . $$

Remark 3.4

Any b-norm with $s\geq 1$ defines a b-metric as follows:

$$ d ( x,y ) = \Vert x-y \Vert _{b}. $$

The question now is the following: Is any b-metric induced from a b-norm? The following remark can answer this question.

Remark 3.5

Let X be a vector space over a field K. Any b-metric $d:X\times X\longrightarrow {}[ 0,\infty )$ with constant $s\geq 1$ induced from a b-norm must satisfy the following properties for each $x,y,z\in X$, $c\in K$:

• (i) $d ( x+z,y+z ) =d ( x,y )$ (translation invariance);

• (ii) $d ( cx,cy ) =|c|^{\log _{2}s+1}d ( x,y )$ (b-homogeneity).

Proposition 3.6

A b-homogeneous translation invariant b-metric $d:X\times X\longrightarrow {}[ 0,\infty )$ with constant $s\geq 1$ can define a b-norm $\Vert \cdot \Vert _{b}:X\longrightarrow {}[ 0, \infty )$ as follows:

$$ \Vert x \Vert _{b}=d ( x,0 ) \quad \forall x\in X. $$

Proof

Clearly, (Nb1) and (Nb2) are satisfied.

As d is homogeneous, $\Vert cx \Vert =d ( cx,0 ) =|c|^{\log _{2}s+1}d ( x,0 ) =|c|^{\log _{2}s+1} \Vert x \Vert _{b}$.

As d is translation invariant,

$$ \begin{aligned} \Vert x+y \Vert _{b}&=d ( x+y,0 ) \leq s \bigl[ d(x+y,x)+d(x,0) \bigr] \\ & =s \bigl[ d ( y,0 ) +d ( x,0 ) \bigr] \\ & =s \bigl[ \Vert x \Vert _{b}+ \Vert y \Vert _{b} \bigr] , \end{aligned} $$

which prove (Nb3) and (Nb4), respectively. □

Now, we rewrite inequality (1) in the sense of b-normed spaces and obtain some corollaries.

If $( X, \Vert \cdot \Vert _{b} )$ is a b-normed space with constant $s\geq 1$, $x_{i}\in X$, and $p_{i}\geq 0$, $i\in \{ 1,\dots ,n \}$ with $\sum_{i=1}^{n}p_{i}=\frac{1}{s}$, then by (1) we have

$$ \sum_{1\leq i< j\leq n}p_{i}p_{j} \Vert x_{i}-x_{j} \Vert \leq \inf_{x\in X} \Biggl[ \sum_{i=1}^{n}p_{i} \Vert x_{i}-x _{j} \Vert \Biggr] . $$

(5)

The following proposition is a generalization of Proposition 2 in [11] to the case of a b-normed space.

Proposition 3.7

Let $( X, \Vert \cdot \Vert _{b} )$ be a b-normed space with constant $s\geq 1$, $x_{i}\in X$ and $p_{i} \geq 0$, $i\in \{ 1,\dots ,n \}$ with $\sum_{i=1} ^{n}p_{i}=\frac{1}{s}$. Let $x_{p}=\sum_{i=1}^{n}p_{i}x_{i}$, then

$$ \frac{1}{2}\sum_{i=1}^{n}p_{i} \Vert x_{i}-x_{p} \Vert \leq s^{n}\sum _{1\leq i< j\leq n}p_{i}p_{j} \Vert x_{i}-x _{j} \Vert \leq s^{n}\sum_{i=1}^{n}p_{i} \Vert x _{i}-x_{p} \Vert . $$

(6)

Proof

As the infimum is a lower bound, the second part of inequality (6) is trivial. For the first part, we use a generalized b-norm inequality as follows:

$$ \begin{aligned} \frac{1}{2}\sum_{i=1}^{n}p_{i} \Vert x_{i}-x_{p} \Vert &= \frac{1}{2}\sum _{i=1}^{n}p_{i} \Biggl\Vert x_{i}-\sum_{j=1}^{n}p_{j}x_{j} \Biggr\Vert \\ &=\frac{1}{2}\sum_{i=1}^{n}p_{i} \Biggl\Vert \sum_{j=1} ^{n} ( x_{i}-p_{j}x_{j} ) \Biggr\Vert \\ &\leq \frac{1}{2}\sum_{i,j=1}^{n}p_{i}s^{j} \Vert x_{i}-p _{j}x_{j} \Vert \\ &\leq \frac{s^{n}}{2}\sum_{i,j=1}^{n}p_{i}p_{j} \Vert x _{i}-x_{j} \Vert \\ &=s^{n}\sum_{1\leq i< j\leq n}p_{i}p_{j} \Vert x_{i}-x_{j} \Vert , \end{aligned} $$

which completes the proof. □

We have the following corollary, which has a nice geometric interpretation.

Corollary 3.8

Let $( X, \Vert \cdot \Vert _{b} )$ be a b-normed space with constant $s\geq 1$ and $x_{i}\in X$, $i\in \{ 1,\dots ,n \}$. If $\overline{x}=\frac{x_{1}+\cdots +x_{n}}{n}$ is the gravity center of the vectors $\{ x_{1}, \dots ,x_{n} \}$, then we have

$$ \frac{n}{2}\sum_{i=1}^{n} \Vert x_{i}-\overline{x} \Vert \leq s^{n}\sum _{1\leq i< j\leq n} \Vert x_{i}-x_{j} \Vert \leq ns^{n}\sum_{i=1}^{n} \Vert x_{i}-\overline{x} \Vert . $$

Geometrically, the last corollary means that the sum of the edges and diagonals of a polygon with n vertices in a b-normed space is less than or equal to n-times the sum of the distances from the gravity center to its vertices and greater than or equal to $\frac{n}{2s^{n}}$-times this quantity.

4 Conclusion

Similarly, we can generalize more inequalities on metric and normed spaces.