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])
Anzeige
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:
In this case, the pair \((X, d)\) is called a b-metric space (with constant s).
\((bM_{1})\)
\(d(x, y) =0\) 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).
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 \(s=2\).
Anzeige
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
Consequently, we derive that
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.
$$ d(x_{i},x_{j})= \textstyle\begin{cases} 0 & \text{if } i=j, \\ s & \text{if } (i,j)=(1,2) \text{ or } (i,j)=(2,1), \\ 1 & \text{otherwise.} \end{cases} $$
$$ d(x_{i},x_{j}) \leq \frac{s}{2}\bigl[d(x_{i},x_{k}) +d(x_{k},x_{j}) \bigr], $$
For more examples for b-metric, we may refer, e.g., to [1‐7, 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
is a b-metric space. Notice that \(s=2^{1/p}\).
$$ d(x,y):= \biggl( \int _{0}^{1} \bigl\vert x(t)-y(t) \bigr\vert ^{p} \,dt \biggr)^{1/p}, \quad \text{for each } x,y\in L^{p}[0,1], $$
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
The inequality is sharp in the sense that the constant
\(c=1\)
in front of the infimum cannot be replaced by a smaller constant.
$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq \inf_{x\in X} \Biggl[ \sum_{i=1}^{n}p_{i}d ( x _{i},x ) \Biggr]. $$
(1)
Proof
Using the b-triangle inequality, for any \(x\in X\), \(i,j\in \{ 1,2,\dots ,n \} \) we have
If we multiply (2) by \(p_{i}\), \(p_{j}\) and sum over i and j from 1 to n, we get
Note that by symmetry we have
Now, using the condition \(\sum_{i=1}^{n}p_{i}=\frac{1}{s}\), we can easily deduce that
So, from (3) we have
$$ d ( x_{i},x_{j} ) \leq s \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr]. $$
(2)
$$ \sum_{i,j=1}^{n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq s \Biggl[ \sum _{i,j=1}^{n}p_{i}p_{j} \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr] \Biggr]. $$
$$ \sum_{i,j=1}^{n}p_{i}p_{j}d ( x_{i},x_{j} ) =2 \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ). $$
(3)
$$ \sum_{i,j=1}^{n}p_{i}p_{j} \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr] =\frac{2}{s} \sum_{i=1}^{n}p _{i}d ( x_{i},x ). $$
$$ \begin{aligned} \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) &= \frac{1}{2}\sum _{i,j=1}^{n}p_{i}p_{j}d ( x_{i},x_{j} ) \\ &\leqslant \frac{s}{2} \Biggl[ \sum_{i,j=1}^{n}p_{i}p_{j} \bigl[ d ( x_{i},x ) +d ( x,x_{j} ) \bigr] \Biggr] \\ &=\sum_{i=1}^{n}p_{i}d ( x_{i},x ). \end{aligned} $$
Therefore,
for any \(x\in X\). Using the fact that the infimum is the greatest lower bound, we deduce (1).
$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq \sum_{i=1}^{n}p_{i}d ( x_{i},x ) $$
Now, suppose that there exists \(c>0\) such that
and choose \(n=2\), \(p_{1}=p\) and \(p_{2}=1-p\) where \(p\in (0,1)\). Then,
If we let \(x=x_{1}\) in (4), we get
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 \(c\geq 1\). □
$$ \sum_{1\leq i< j\leq n}p_{i}p_{j}d ( x_{i},x_{j} ) \leq c \inf_{x\in X} \Biggl[ \sum_{i=1}^{n}p_{i}d ( x_{i},x ) \Biggr]; $$
$$ p(1-p)d ( x_{1},x_{2} )\leq c \bigl[ pd ( x_{1},x )+(1-p)d ( x,x_{2} ) \bigr]. $$
(4)
$$ p(1-p)d ( x_{1},x_{2} )\leq c(1-p)d ( x_{1},x_{2} ). $$
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,
which prove (Nb3) and (Nb4), respectively. □
$$ \begin{aligned} \Vert x+y \Vert _{b}&=d ( x+y,0 ) \leq s \bigl[ d(x+y,x)+d(x,0) \bigr] \\ & =s \bigl[ d ( y,0 ) +d ( x,0 ) \bigr] \\ & =s \bigl[ \Vert x \Vert _{b}+ \Vert y \Vert _{b} \bigr] , \end{aligned} $$
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:
which completes the proof. □
$$ \begin{aligned} \frac{1}{2}\sum_{i=1}^{n}p_{i} \Vert x_{i}-x_{p} \Vert &= \frac{1}{2}\sum _{i=1}^{n}p_{i} \Biggl\Vert x_{i}-\sum_{j=1}^{n}p_{j}x_{j} \Biggr\Vert \\ &=\frac{1}{2}\sum_{i=1}^{n}p_{i} \Biggl\Vert \sum_{j=1} ^{n} ( x_{i}-p_{j}x_{j} ) \Biggr\Vert \\ &\leq \frac{1}{2}\sum_{i,j=1}^{n}p_{i}s^{j} \Vert x_{i}-p _{j}x_{j} \Vert \\ &\leq \frac{s^{n}}{2}\sum_{i,j=1}^{n}p_{i}p_{j} \Vert x _{i}-x_{j} \Vert \\ &=s^{n}\sum_{1\leq i< j\leq n}p_{i}p_{j} \Vert x_{i}-x_{j} \Vert , \end{aligned} $$
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.
Competing interests
The authors declare that they have no competing interests.
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