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Dragomir and Gosa type inequalities on b-metric spaces

  • Erdal KarapınarEmail author
  • Maha Noorwali
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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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Authors and Affiliations

  1. 1.Department of Medical Research, China Medical University HospitalChina Medical UniversityTaichungTaiwan
  2. 2.Nonlinear Analysis Research Group (NAAM)King Abdulaziz UniversityJeddahSaudi Arabia

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