On Fractional Power Inequalities In b-Metric-Type Spaces

1. Introduction and Preliminaries

We first recall some basic definitions in literature.

Definition 1.1. 

A metric space is defined as a pair $(X,\psi )$, where X is a non empty set and $\psi :X\times X\to \mathbb{R}$ is a function. This function ψ satisfies the following properties for any elements $\xi ,\phi ,\varrho$ in X:

• $\psi (\xi ,\phi )\ge 0$ and $\rho (\xi ,\phi )=0$ if and only if $\xi =\phi$ (Non-negative);
• $\psi (\xi ,\phi )=\psi (\phi ,\xi )$ (symmetry);
• $\psi (\xi ,\phi )\le \psi (\xi ,\varrho )+\psi (\varrho ,\phi )$ (Triangle inequality).

The study of metric space involving distance function provides a powerful tool in mathematics and other science such as fixed point theory, topology and operator theory, see [1,3,11,13]. In 1998, Czerwik [6] (see also [4]) constructed a lemma to obtain some generalizations of the well known Banach’s inequality contraction in b-meric spaces using $\alpha$-relaxed triangle inequality as follows:

$$\psi (\xi ,\phi )\le \alpha \left[\psi \right(\xi ,\varrho )+\psi (\varrho ,\phi \left)\right],$$

(1.1)

where $\alpha \ge 1$. We note that in the case where $\alpha =1$, every b-metric space is a metric space. Some examples of b-metric are given below:

Example 1.2.  

Let $X=[0,1]$ and $\psi :X\times X\to [0,\infty ]$ is defined by $\psi (\varsigma ,\phi )={(\varsigma -\phi )}^2$, for all $\varsigma ,\phi \in X$. Clearly, $(X,\psi )$ is a b-metric space with $k=2$.

Example 1.3.  

The set $l_p\left(R\right)$ with $0<p<1$, where $l_p\left(R\right):=\{\left({\varsigma }_n\right)\subset R|\sum _{n=1}^{\infty }|{\varsigma }_n{|}^p,\infty \}$, is defined by the function $\psi :l_p\left(R\right)X{l}_p\left(R\right)\to R$,

$$\psi (\varsigma ,\phi )={\left(\sum _{n=1}^{\infty }{|{\varsigma }_n-{\phi }_n|}^p\right)}^{{\displaystyle \frac{1}{p}}},$$

where $\varsigma ={\varsigma }_n$, $\phi ={\phi }_n\in l_p\left(R\right)$. Then $(l_p\left(R\right),\psi )$is a b-metric space such that $\psi (\varsigma ,\varrho )\le 2^{{\displaystyle \frac{1}{p}}}[\psi (\varsigma ,\phi )+\psi (\phi ,\varrho )]$.

Definition 1.4.  

Let X be a vector space over a field $\mathbb{K}$ and let $s\ge 1$ be a constant. A function ${\parallel ·\parallel }_b$ defined by ${\parallel ·\parallel }_b:X\to [0,\infty ]$ is said to be a b-norm space if the following conditions are satisfied for all $\varsigma ,\phi \in X$:

• (bN1) ${\parallel \varsigma \parallel }_b\ge 0$;
• (bN2)${\parallel \varsigma \parallel }_b=0$ if and only if $\varsigma =0$;
• (bN3) ${\parallel K\varsigma \parallel }_b={\left|K\right|}^{{log}_{2}s+1}{\parallel \varsigma \parallel }_b$;
• (bN4)${\parallel \varsigma +\phi \parallel }_b\le s\left[{\parallel \varsigma \parallel }_b+{\parallel \phi \parallel }_b\right]$.

In this case $(X,\parallel ·\parallel )$ is called a b-normed space with constant s.

Remark 1.5.  

Clearly, when $s=1$, we recover the definition of a norm linear space, see [6] .

Example 1.6.  

Let $X=\mathbb{R}$ and define ${\parallel ·\parallel }_b:X\to [0,\infty ]$ by ${\parallel \xi \parallel }_b={\left|\xi \right|}^p$ where $p\in (1,\infty )$ then, using the relation ${(\xi +\phi )}^p\le 2^{p-1}({\xi }^p+{\phi }^p)$, we can easily deduce that $(X,\parallel ·\parallel )$ is a b-normed space with constant $s=2^{p-1}$ for all $\xi ,\phi \in X$.

One of the important properties of a (classical) distance function on any abstract set X is the triangle inequality, i.e., $\psi (\xi ,\phi )\le \psi (\xi ,\varrho )+\psi (\varrho ,\phi )$. Several generalizations and refinements of the concept of a distance have been achieved by relaxing the triangle inequality, see [10].

Dragomir and Gosa [7] established the polygonal inequality in the metric space setting by obtaining the following result:

Theorem 1.7.  

Let $(X,\psi )$ be a metric space and ${\xi }_i\in X,{\rho }_i\ge 0,i\in \{1,...,N\}$ with ${\sum }_{i=1}^N{\rho }_i=1$. Then we have the inequality

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j\psi ({\xi }_i,{\xi }_j)\le \underset{\xi \in X}{inf}\left[\sum _{i=1}^N{\rho }_i\psi ({\xi }_i,\xi )\right].$$

(1.2)

In a recent paper, Karapinar and Noorwali [10] gave an improved version of Dragomir and Gosa’s result as follows:

Theorem 1.8.  

Let $(X,\psi )$ be a b-metric space with constant $m\ge 1$ and ${\xi }_i\in X,{\rho }_i\ge 0$, $i\in \{1,2,...,N\}$ with ${\sum }_{i=1}^N{\rho }_i=1$. Then we have the inequality

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j\psi ({\xi }_i,{\xi }_j)\le {\displaystyle \frac{N}{m}}\underset{\xi \in X}{inf}\left[\sum _{i=1}^N{\rho }_i\psi ({\xi }_i,\xi )\right].$$

(1.3)

The aim of this paper is to obtain some upper bounds for the distance on b-metric spaces. Thus, our results are generalizations of [2,7,8,10]. Before we give our main result, we can state the following result that is regarded as a generalization of the binomial theorem.

Theorem 1.9.  

(Neo-classical inequality; Theorem 1.2 in [9]) For $k\in \mathbb{N}$ and $0<s<1$, we have

$$s\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)x^{sk}{y}^{s(m-k)}\le {(x+y)}^{sm},x,y\ge 0.$$

(1.4)

Remark 1.10.  

When $s=1$, the equality holds in (1.4), which is just the conventional binomial theorem.

2. Main Results

Now, we first discuss the following new concept:

Definition 2.1.  

Let X be a non empty set and $\alpha ,\beta \in \mathbb{N}$ be a given real number. A mapping ${\psi }_b:X\times X\to {\mathbb{R}}^{+}$ is said to be a variant of b-metric if for all $\varsigma ,\varrho ,\phi$ in X, the following conditions are satisfied:

• (b1): ${\psi }_b(\varsigma ,\phi )=0$ if and only if $\varsigma =\phi$;
• (b2): ${\psi }_b(\varsigma ,\phi )={\psi }_b(\phi ,\varsigma )$ (symmetry);
• (b3): ${\psi }_b(\varsigma ,\phi )\le \alpha [{\psi }_b(\varsigma ,\varrho )+\beta {\psi }_b(\varrho ,\phi )]$ (Triangle inequality).

It is easy to see that when $\beta =1$ then it is b-metric space [6] which in turn is a generalization of the standard metric space [13]. It may be of interest to study this new variant of b-metric space for the case $\beta \ge 1$

The following example may be stated to support Definition 2.1.

Example 2.2.  

Consider the set X of all continuous functions $\varsigma :[0,1]\to \mathbb{R}$ defined by the distance function ${\psi }_b:X\times X\to {\mathbb{R}}^{+}$ as:

$${\psi }_b(\varsigma ,\phi )=\underset{0}{\overset{1}{\int }}{|\varsigma \left(t\right)-\phi \left(t\right)|}^{2}dt.$$

This is a metric space that satisfies the condition ${\psi }_b(\varsigma ,\phi )\le \alpha [{\psi }_b(\varsigma ,\varrho )+\beta {\psi }_b(\varrho ,\phi )]$ with $\alpha =2$ and $\beta =2$, since

• (b1): ${\psi }_b(\varsigma ,\phi )=0$ if and only if $\varsigma =\phi$;
• (b2): ${\psi }_b(\varsigma ,\phi )={\psi }_b(\phi ,\varsigma )$ (symmetry);
• (b3): ${\psi }_b(\varsigma ,\phi )\le 2\underset{0}{\overset{1}{\int }}{|\varsigma \left(t\right)-\varrho \left(t\right)|}^{2}dt+4\underset{0}{\overset{1}{\int }}{|\varrho \left(t\right)-\phi \left(t\right)|}^{2}dt=2[{\psi }_b(\varsigma ,\varrho )+2{\psi }_b(\varrho ,\phi )]$.

Example 2.3.  

Let $X=\{1,2,3\}$ be a discrete set and let ${\psi }_b:X\times X\to {\mathbb{R}}^{+}$ be a function defined by

$${\psi }_b(1,1)={\psi }_b(2,2)={\psi }_b(3,3)=0$$

$${\psi }_b(1,2)={\psi }_b(2,1)={\displaystyle \frac{1}{3}}$$

$${\psi }_b(1,3)={\psi }_b(3,1)=3$$

$${\psi }_b(2,3)={\psi }_b(3,2)=4$$

By Definition 2.1, (b1) and (b2) clearly holds. For all $\xi ,\varrho ,\phi \in X$ it follows that

$${\psi }_b(\xi ,\phi )\le 2[{\psi }_b(\xi ,\varrho )+3{\psi }_b(\varrho ,\phi )]$$

The following result may be stated:

Theorem 2.4.  

Let $(X,{\psi }_b)$ be a metric space and $\alpha ,m\in \mathbb{N}$, $\beta \ge 1$, $0<s<1$, ${\upsilon }_i\in X$, ${\rho }_i\ge 0$, $i\in \{1,...,N\}$ with ${\sum }_{i=1}^N{\rho }_i=1$, then

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)$$

$$\ge {\displaystyle \frac{s{\alpha }^{sm}}{2}}\underset{\upsilon \in X}{sup}\left[2s\sum _{i=1}^N{\rho }_i{\psi }_b^{sm}({\upsilon }_i,\upsilon )+\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{sk}(\upsilon ,{\upsilon }_i)\right)\left(\sum _{i=1}^N{\rho }_i{\beta }^{sk}{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\right)\right].$$

(2.1)

Proof. 

Using the b-triangle inequality in metric space, we have that for any $\upsilon \in X$ and $i,j\in \{1,...,N\}$ that

$${\psi }_b({\upsilon }_i,{\upsilon }_j)\le \alpha \left[{\psi }_b({\upsilon }_i,\upsilon )+\beta {\psi }_b(\upsilon ,{\upsilon }_j)\right].$$

(2.2)

Taking the power $sm$, where $0<s<1$, $m=1,2,...$ to have

$${\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)\ge {\alpha }^{sm}{\left[{\psi }_b({\upsilon }_i,\upsilon )+\beta {\psi }_b(\upsilon ,{\upsilon }_j)\right]}^{sm}.$$

(2.3)

By expanding the RHS of (2.3) using Proposition 1.13 we have

$${\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)\ge s{\alpha }^{sm}\left[\sum _{k=0}^{sm}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right){\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon ){\beta }^k{\psi }_b^k(\upsilon ,{\upsilon }_j)\right]$$

(2.4)

where

$$\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)={\displaystyle \frac{m!}{s(m-k)!k!}}.$$

Multiplying (2.4) by ${\rho }_i{\rho }_j$ and summing over i and j from 1 to N, we get

$$\sum _{1\le i,j\le N}{\rho }_i{\rho }_j{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)\ge s{\alpha }^{sm}\sum _{1\le i,j\le N}{\rho }_i{\rho }_j\left(\sum _{k=0}^{sm}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right){\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon ){\beta }^{(}sk){\psi }_b^{sk}(\upsilon ,{\upsilon }_j)\right).$$

(2.5)

i.e.,

$$\begin{array}{cc}\hfill \sum _{1\le i,j\le N}{\rho }_i{\rho }_j{\psi }_b^{sk}({\upsilon }_i,{\upsilon }_j)& \ge s{\alpha }^{sm}\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\sum _{i=1}^N{\rho }_i{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\sum _{j=1}^N{\rho }_j{\beta }^{sk}{\psi }_b^{sk}(\upsilon ,{\upsilon }_j)\hfill \\ \hfill & =s{\alpha }^{sm}\left[\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\beta }^{sk}{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{s(m-k)}(\upsilon ,{\upsilon }_i)\right)\right]\hfill \\ \hfill & =2s{\alpha }^{sm}\sum _{i=1}^N{\rho }_i{\psi }_b^{sm}({\upsilon }_i,\upsilon )+{\alpha }^{sm}\left[\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\right)\left(\sum _{i=1}^N{\rho }_i{\beta }^{sk}{\psi }_b^{sk}({\upsilon }_i,\upsilon )\right)\right].\hfill \end{array}$$

(2.6)

It is easy to see that

$$\sum _{1\le i,j\le N}{\rho }_i{\rho }_j{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)=2\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j).$$

(2.7)

So, (2.6) becomes

$$2\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j{\psi }_b({\upsilon }_i,{\upsilon }_j)$$

$$\ge 2s{\alpha }^{sm}\sum _{i=1}^N{\rho }_i{\psi }_b^{sm}({\upsilon }_i,\upsilon )+{\alpha }^{sm}\left[\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{s(m-k)}({\upsilon }_i,{\upsilon }_j)\right)\left(\sum _{i=1}^N{\rho }_i{\beta }^{sk}{\psi }_b^{sk}(\upsilon ,{\upsilon }_i)\right)\right].$$

That is,

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)$$

$$\ge {\displaystyle \frac{{\alpha }^{sm}}{2}}\left[2s\sum _{i=1}^N{\rho }_i{\psi }_b^{sm}(\upsilon ,{\upsilon }_i)+\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{sk}(\upsilon ,{\upsilon }_i)\right)\left(\sum _{i=1}^N{\rho }_i{\beta }^{sk}{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\right)\right].$$

Hence the result in (2.1) is proved for all $\upsilon \in X$. □

Theorem 2.4 has proven to be an extension of the result in ([2,10]) in a more general setting.

Corollary 2.5.  

Let $(X,{\psi }_b)$ be a b-metric space and $\alpha ,m\in \mathbb{N},N\ge 2$, $0<s<1$, ${\upsilon }_i\in X$ for all $i\in \{1,...,N\}$ with ${\sum }_{i=1}^N{\rho }_i=1$. Then

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)$$

(2.8)

$$\ge {\alpha }^{sm}\underset{\upsilon \in X}{sup}\left[s\sum _{i=1}^N{\rho }_i{\rho }_j{\psi }_b^{sm}(\upsilon ,{\upsilon }_i)+{\displaystyle \frac{1}{2}}\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{sk}(\upsilon ,{\upsilon }_i)\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\right)\right].$$

.

Corollary 2.6.  

Let $(X,{\psi }_b)$ be a b-metric space and $m\in \mathbb{N}$, $N\ge 2$, $0<s<1$, ${\upsilon }_i\in X$ and $i\in \{1,...,N\}$ with ${\sum }_{i=1}^N{\rho }_i=1$. Then

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\psi }_b^{sm}({\upsilon }_i,{\upsilon }_j)$$

$$\ge {\displaystyle \frac{1}{2}}\left[2s\sum _{i=1}^N{\rho }_i{\psi }_b^{sm}(\upsilon ,{\upsilon }_i)+{\displaystyle \frac{1}{2}}\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{sk}(\upsilon ,{\upsilon }_i)\right)\left(\sum _{i=1}^N{\rho }_i{\psi }_b^{s(m-k)}({\upsilon }_i,\upsilon )\right)\right].$$

We can state the following result that give an application in b-normed linear spaces using ${\psi }_b({\upsilon }_i,{\upsilon }_j)={\parallel {\upsilon }_i-{\upsilon }_j\parallel }_b$ to give an application in b-normed linear spaces.

Proposition 2.7.  

Given that $(X,\parallel .{\parallel }_b)$ is a b-normed linear space and $\upsilon \in X$, $\alpha ,m\in \mathbb{N},N\ge 2$, $\beta \ge 1$$0<s<1$. If ${\rho }_i\ge 0,i,j=\{1,...,N\}$ with

$$\sum _{i=1}^N{\rho }_i=1.$$

Indeed, we have by Theorem 2.3 that

$$\sum _{i=1}^{N-1}\sum _{j=i+1}^N{\rho }_i{\rho }_j{\parallel {\upsilon }_i-{\upsilon }_j\parallel }_b^{sm}$$

$$\ge {\displaystyle \frac{{\alpha }^{sm}}{2}}\left[2s\sum _{i=1}^N{\rho }_i\parallel {\upsilon }_i{-\upsilon \parallel }_b^{sm}+\sum _{k=1}^{m-1}\left({\displaystyle \genfrac{}{}{0pt}{}{sm}{sk}}\right)\sum _{i=1}^N{\rho }_i\parallel {\upsilon }_i{-\upsilon \parallel }_b^{s(m-k)}\sum _{i=1}^N{\rho }_i{\beta }^{sk}{\parallel {\upsilon }_i-\upsilon \parallel }_b^{sk}\right],$$

for all $\upsilon \in X$.

Proof. 

It follows from the proof of Theorem 2.2 if we set ${\psi }_b({\upsilon }_i,{\upsilon }_j)={\parallel {\upsilon }_i-{\upsilon }_j\parallel }_b$. □

3. Conclusion

In conclusion, we have provided a fractional power inequality in b-metric-type spaces based on Definition 2.1, condition (b3) for $\beta \ge 1$. Possible consideration of this paper for the case $\beta <1$ maybe of interest.