# Dragomir and Gosa type inequalities on b-metric spaces

• Erdal Karapınar
• Maha Noorwali
Open Access
Research

## 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.

## Keywords

Dragomir and Gosa type inequalities b-metric space Inequality

## 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 ab-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 ab-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 ab-metric space with constantsand$$x_{i}\in X$$, $$i\in \{ 1,2,\dots ,n \}$$. 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 $$x,y,z\in X$$, $$c\in K$$:
1. (i)

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

2. (ii)

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

### Proposition 3.6

Ab-homogeneous translation invariantb-metric$$d:X\times X\longrightarrow {}[ 0,\infty )$$with constant$$s\geq 1$$can define ab-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 ab-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 ab-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.

## Notes

### Authors’ contributions

All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.

### Funding

We declare that funding is not applicable for our paper.

### Competing interests

The authors declare that they have no competing interests.

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