Some Results on (s − q)-Graphic Contraction Mappings in b-Metric-Like Spaces

Abstract

In this paper we consider $(s-q)$-graphic contraction mapping in b-metric like spaces. By using our new approach for the proof that a Picard sequence is Cauchy in the context of b-metric-like space, our results generalize, improve and complement several approaches in the existing literature. Moreover, some examples are presented here to illustrate the usability of the obtained theoretical results.

1. Introduction and Preliminaries

First, we present some definitions and basic notions of partial-metric, metric-like, b-metric, partial b-metric and b-metric-like spaces as the generalizations of standard metric spaces. After that, we give a process diagram, where arrows stand for generalization relationships.

Definition 1.

[1] Let X be a nonempty set. A mapping$p_{pm}:X\times X\to [0,+\infty )$is said to be a p-metric if the following conditions hold for all$x,y,z\in X:$

$\left(p_{pm}1\right)$ $x=y$if and only if$p_{pm}\left(x,x\right)=p_{pm}\left(x,y\right)=p_{pm}\left(y,y\right);$

$\left(p_{pm}2\right)$ $p_{pm}\left(x,x\right)\le$$p_{pm}\left(x,y\right);$

$\left(p_{pm}3\right)$ $p_{pm}\left(x,y\right)=$$p_{pm}\left(y,x\right);$

$\left(p_{pm}4\right)$ $p_{pm}\left(x,y\right)\le p_{pm}\left(x,z\right)+p_{pm}\left(z,y\right)-p_{pm}\left(z,z\right).$

Then, the pair$\left(X,p_{pm}\right)$is called a partial metric space.

Definition 2.

[2] Let X be a nonempty set. A mapping$b_{ml}:X\times X\to [0,+\infty )$is said to be metric-like if the following conditions hold for all$x,y,z\in X:$

$\left(b_{l}1\right)$ $b_{ml}\left(x,y\right)=0$implies$x=y;$

$\left(b_{l}2\right)$ $b_{ml}\left(x,y\right)=b_{ml}\left(y,x\right);$

$\left(b_{l}3\right)$ $b_{ml}\left(x,z\right)\le b_{ml}\left(x,y\right)+b_{ml}\left(y,z\right).$

In this case, the pair$\left(X,b_{ml}\right)$is called a metric-like space.

Definition 3.

[3,4] Let X be a nonempty set and$s\ge 1$a given real number. A mapping$b:X\times X\to [0,+\infty )$is called a b-metric on the set X if the following conditions hold for all$x,y,z\in X:$

$\left(b1\right)$ $b\left(x,y\right)=0$if and only if$x=y;$

$\left(b2\right)$ $b\left(x,y\right)=b\left(y,x\right);$

$\left(b3\right)$ $b\left(x,z\right)\le s\left[b\left(x,y\right)+b\left(y,z\right)\right].$

In this case, the pair$\left(X,b\right)$is called a b-metric space (with coefficient$s\ge 1$).

Definition 4.

[5,6] Let X be a nonempty set and$s\ge 1$. A mapping$b_{pb}:X\times X\to [0,+\infty )$is called a partial b-metric on the set X if the following conditions hold for all$x,y,z\in X:$

$\left(b_{pb}1\right)$ $x=y$if and only if$p_{pb}\left(x,x\right)=p_{pb}\left(x,y\right)=p_{pb}\left(y,y\right);$

$\left(b_{pb}2\right)$ $b_{pb}\left(x,x\right)\le b_{pb}\left(x,y\right);$

$\left(b_{pb}3\right)$ $b_{pb}\left(x,y\right)=b_{pb}\left(y,x\right);$

$\left(b_{pb}4\right)$ $b_{pb}\left(x,y\right)\le s\left[b_{pb}\left(x,z\right)+b_{pb}\left(z,y\right)\right]-b_{pb}\left(z,z\right).$

Then, the pair$\left(X,b_{pb}\right)$is called a partial b-metric space.

Definition 5.

[7] Let X be a nonempty set and$s\ge 1$. A mapping$b_{bl}:X\times X\to [0,+\infty )$is called b-metric-like on the set X if the following conditions hold for all$x,y,z\in X$:

$\left(b_{bl}1\right)$ $b_{bl}\left(x,y\right)=0$implies$x=y;$

$\left(b_{bl}2\right)$ $b_{bl}\left(x,y\right)=b_{bl}\left(y,x\right);$

$\left(b_{bl}3\right)$ $b_{bl}\left(x,z\right)\le s\left[b_{bl}\left(x,y\right)+b_{bl}\left(y,z\right)\right].$

In this case, the pair$\left(X,b_{bl}\right)$is called a b-metric-like space with coefficient$s\ge 1$.

Now, we give the process diagram of the classes of generalized metric spaces that were introduced earlier:

$$\begin{array}{ccccc}\mathrm{Metric}\mathrm{space}& \to & \mathrm{Partial}\mathrm{metric}\mathrm{space}& \to & \text{Metric-like}\mathrm{space}\\ \downarrow & & \downarrow & & \downarrow \\ \text{b-Metric}\mathrm{space}& \to & \mathrm{Partial}\text{b-metric}\mathrm{space}& \to & \text{b-Metric-like}\mathrm{space}\end{array}$$

For more details on other generalized metric spaces see [8,9,10,11,12,13,14].

The next proposition helps us to construct some more examples of b-metric (respectively partial b-metric, b-metric-like) spaces.

Proposition 1.

Let$\left(X,d\right)$(resp.$\left(X,p_{pm}\right),\left(X,b_{ml}\right)$) be a metric (resp. partial metric, metric-like) space and$D\left(x,y\right)={\left(d\left(x,y\right)\right)}^k$(resp.$P_{pm}\left(x,y\right)={\left(p_{pm}\left(x,y\right)\right)}^k,$$B_{ml}\left(x,y\right)={\left(b_{ml}\left(x,y\right)\right)}^k$), where$k>1$is a real number. Then D (resp.$P_{pm},$$B_{pm}$) is b-metric (resp. partial b-metric, b-metric-like) with coefficient$s=2^{k-1}.$

Proof. 

The proof follows from the fact that

$$u^k+v^k\le {\left(u+v\right)}^k\le {\left(a+b\right)}^k\le 2^{k-1}\left(a^k+b^k\right),$$

for all nonnegative real numbers $a,b,u,v$ with $u+v\le a+b$. □

It is clear that each metric-like space, i.e., each partial b-metric space, is a b-metric-like space, while the converse is not true. For more such examples and details see [1,2,5,6,7,15,16,17,18,19,20,21,22,23,24,25,26,27]. Moreover, for various metrics in the context of the complex domain see [28,29].

The definitions of convergent and Cauchy sequences are formally the same in partial metric, metric-like, partial b-metric and b-metric like spaces. Therefore, we give only the definition of convergence and Cauchyness of the sequences in b-metric-like space. Moreover, these two notions are formally the same in metric and b-metric spaces.

Definition 6.

[7] Let$\left\{x_n\right\}$be a sequence in a b-metric-like space$\left(X,b_{bl}\right)$with coefficient s.

(i) The sequence $\left\{x_n\right\}$ is said to be convergent to x if ${lim}_{n\to \infty }{b}_{bl}\left(x_n,x\right)=b_{bl}\left(x,x\right)$;

(ii) The sequence $\left\{x_n\right\}$ is said to be $b_{bl}$-Cauchy in $\left(X,b_{bl}\right)$ if ${lim}_{n,m\to \infty }{b}_{bl}\left(x_n,x_m\right)$ exists and is finite;

(iii) One says that a b-metric-like space $\left(X,b_{bl}\right)$ is $b_{bl}$-complete if for every $b_{bl}$-Cauchy sequence $\left\{x_n\right\}$ in X there exists an $x\in X,$ such that ${lim}_{n,m\to \infty }{b}_{bl}\left(x_n,x_m\right)=b_{bl}\left(x,x\right)={lim}_{n\to \infty }{b}_{bl}\left(x_n,x\right).$

Remark 1.

In a b-metric-like space the limit of a sequence need not be unique and a convergent sequence need not be a$b_{bl}$-Cauchy sequence (see Example 7 in [18]). However, if the sequence$\left\{x_n\right\}$is$b_{bl}$-Cauchy with${lim}_{n,m\to \infty }{b}_{bl}\left(x_n,x_m\right)=0$in the$b_{bl}$-complete b-metric-like space$\left(X,b_{bl}\right)$with coefficient$s\ge 1,$then the limit of such a sequence is unique. Indeed, in such a case if$x_n\to x$$\left(b_{bl}\left(x_n,x\right)\to b_{bl}\left(x,x\right)\right)$as$n\to \infty$we get that$b_{bl}\left(x,x\right)=0.$Now, if$x_n\to x$and$x_n\to y$where$x\ne y,$we obtain that:

$$\frac{1}{s}{b}_{bl}\left(x,y\right)\le b_{bl}\left(x,x_n\right)+b_{bl}\left(x_n,y\right)\to b_{bl}\left(x,x\right)+b_{bl}\left(y,y\right)=0+0=0.$$

(1)

From$\left(b_{bl}1\right)$it follows that$x=y$, which is a contradiction. The same is true as well for partial metric, metric like and partial b-metric spaces.

The next definition and the corresponding proposition are important in the context of fixed point theory.

Definition 7.

[30] The self-mappings$f,g:X\to X$are weakly compatible if$f\left(g\left(x\right)\right)=g\left(f\left(x\right)\right),$whenever$f\left(x\right)=g\left(x\right).$

Proposition 2.

[30] Let T and S be weakly compatible self-maps of a nonempty set$X.$If they have a unique point of coincidence$w=f\left(u\right)=g\left(u\right),$then w is the unique common fixed point of f and$g.$

In this paper we shall use the following result to prove that certain Picard sequences are Cauchy. The proof is completely identical with the corresponding in [31] (see also [25]).

Lemma 1.

Let$\left\{x_n\right\}$be a sequence in a b-metric-like space$\left(X,b_{bl}\right)$with coefficient$s\ge 1$such that

$$b_{bl}\left(x_n,x_{n+1}\right)\le \lambda b_{bl}\left(x_{n-1},x_n\right)$$

(2)

for some$\lambda ,0\le \lambda <\frac{1}{s},$and each$n=1,2,\dots .$Then$\left\{x_n\right\}$is a$b_{bl}$-Cauchy sequence in$\left(X,b_{bl}\right)$such that${lim}_{n,m\to \infty }{b}_{bl}\left(x_n,x_m\right)=0.$

Remark 2.

It is worth noting that the previous lemma holds in the context of b-metric-like spaces for each$\lambda \in [0,1).$For more details see [6,32].

2. Main Results

In line with Jachymski [33], let $\left(X,b_{bl}\right)$ be a b-metric-like space and $\mathcal{D}$ denote the diagonal of the Cartesian product $X\times X.$ Consider a directed graph G such that the set $V\left(G\right)$ of its vertices coincides with $X,$ and the set $E\left(G\right)$ of its edges contains all loops, i.e., $E\left(G\right)\supseteq \mathcal{D}.$ We also assume that G has no parallel edges, so we can identify G with the pair $\left(V\left(G\right),E\left(G\right)\right).$ Moreover, we may treat G as a weighted graph by assigning the distance between its vertices to each edge (see [33]).

By $G^{-1}$ we denote the conversion of a graph $G,$ i.e., the graph obtained from G by reversing the direction of edges. Thus, we have

$$E\left(G^{-1}\right)=\left\{\left(x,y\right)\in X\times X:\left(y,x\right)\in E\left(G\right)\right\}.$$

(3)

The letter $\tilde{G}$ denotes the undirected graph obtained from G by ignoring the direction of edges. Actually, it will be more convenient for us to treat $\tilde{G}$ as a directed graph for which the set of its edges is symmetric under the convention

$$E\left(\tilde{G}\right)=E\left(G\right)\cup E\left(G^{-1}\right).$$

(4)

If x and y are vertices in a graph $G,$ then a path in G from x to y of length N $\left(N\in N\right)$ is a sequence ${\left\{x_i\right\}}_{i=0}^N$ of $N+1$ vertices such that $x_0=x,x_N=y$ and $\left(x_{i-1},x_i\right)\in E\left(G\right)$ for $i=1,\dots ,N.$ A graph G is connected if there is a path between any two vertices. G is weakly connected if $\tilde{G}$ is connected.

Recently, some results have appeared providing sufficient conditions for a self mapping of X to be a Picard operator when $\left(X,d\right)$ is endowed with a graph. The first result in this direction was given by Jachymski [33]. Moreover, see [34,35,36].

Definition 8.

[33] We say that a mapping$f:X\to X$is a Banach G-contraction or simply a G-contraction if f preserves edges of$G,$i.e.,

$$\mathit{for}\mathit{all}x,y\in X:\left(x,y\right)\in E\left(G\right)\mathit{implies}\left(f\left(x\right),f\left(y\right)\right)\in E\left(G\right)$$

(5)

and f decreases the weights of edges of G as for all$x,y\in X,$there exists$\lambda \in \left(0,1\right),$such that

$$\left(x,y\right)\in E\left(G\right)\mathit{implies}d\left(f\left(x\right),f\left(y\right)\right)\le \lambda d\left(x,y\right).$$

(6)

Definition 9.

[37] A mapping$g:X\to X$is called orbitally continuous, if given$x\in X$and any sequence$\left\{k_n\right\}$of positive integers,

$$g^{k_n}\left(x\right)\to y\mathit{as}n\to \infty \mathit{implies}g\left(g^{k_n}\left(x\right)\right)\to g\left(y\right)\mathit{as}n\to \infty .$$

(7)

Definition 10.

[33] A mapping$g:X\to X$is called G-continuous, if for any given$x\in X$and any sequence${\left\{x_n\right\}}_{n\in \mathbb{N}}\subset X$with the properties that for all$n\in \mathbb{N}$the pair$\left(x_n,x_{n+1}\right)\in E\left(G\right)$and that$x_n\to x$as$n\to \infty$it follows that$g\left(x_n\right)\to g\left(x\right)$.

Definition 11.

[33] A mapping$g:X\to X$is called orbitally G-continuous, if given$x,y\in X$and any sequence$\left\{k_n\right\}$of positive integers for all$n\in N,$

$$g^{k_n}x\to y\mathit{and}\left(g^{k_n}\left(x\right),g^{k_n+1}\left(x\right)\right)\in E\left(G\right)\mathit{implies}g\left(g^{k_n}\left(x\right)\right)\to g\left(y\right)\mathit{as}n\to \infty .$$

(8)

In this section, we consider self-mappings $f,g:X\to X$ with $f\left(X\right)\subset g\left(X\right)$. Let $x_0\in X$ be an arbitrary point, then there exists $x_1\in X$ such that $z_0=f\left(x_0\right)=g\left(x_1\right)$. By repeating this step we can build a sequence $\left\{z_n\right\}$ such that $z_n=f\left(x_n\right)=g\left(x_{n+1}\right)$ and the following property:

The property $G_{f,g\left(x_n\right)}.$ If ${\left\{g\left(x_n\right)\right\}}_{n\in \mathbb{N}}$ is a sequence in X such that $\left(g\left(x_n\right),g\left(x_{n+1}\right)\right)\in E\left(G\right)$ for all $n\ge 1$ and $g\left(x_n\right)\to x$, then there is a subsequence ${\left\{g\left(x_{n_i}\right)\right\}}_{i\in \mathbb{N}}$ of ${\left\{g\left(x_n\right)\right\}}_{n\in \mathbb{N}}$ such that $\left(g\left(x_{n_i}\right),x\right)\in E\left(G\right)$ for all $i\ge 1$. Note that the property $G_{f,g\left(x_n\right)}$ depends only on the pair of mappings f and g, and does not depend on the sequence $\left\{x_n\right\}$. Here, we use notation $G_{gf}$ in the following sense: $x\in X$ belongs to $G_{gf}$ if and only if there exists a sequence ${\left\{x_n\right\}}_{n\in \mathbb{N}}$ in X such that $x_0=x$, $f\left(x_{n-1}\right)=g\left(x_n\right)$ for $n\in \mathbb{N}$, and $\left(g\left(x_n\right),g\left(x_m\right)\right)\in E\left(G\right)$ for all $m,n\in \mathbb{N}$.

Now, we present the first result of this section.

Theorem 1.

(Hardy-Rogers) Let$f,g:X\to X$be self-mappings defined on a b-metric-like space$\left(X,b_{bl}\right)$(with coefficient$s\ge 1)$endowed with a graph$G,$and which satisfy

$$\begin{array}{ccc}\hfill s^q{b}_{bl}\left(f\left(x\right),f\left(y\right)\right)& \le & c_1{b}_{bl}\left(g\left(x\right),g\left(y\right)\right)+c_2{b}_{bl}\left(g\left(x\right),f\left(x\right)\right)+c_3{b}_{bl}\left(g\left(y\right),f\left(y\right)\right)\hfill \\ & & +c_4{b}_{bl}\left(g\left(x\right),f\left(y\right)\right)+c_5{b}_{bl}\left(g\left(y\right),f\left(x\right)\right),\hfill \end{array}$$

(9)

for all$x,y\in X$with$\left(g\left(x\right),g\left(y\right)\right)\in E\left(G\right)$where$q\ge 2,c_i\ge 0,i=1,\dots ,5$and either

$$c_1+c_2+c_3+2c_4+2c_5<\frac{1}{s}$$

(10)

or

$$c_1+2c_2+2c_3+c_4+c_5<\frac{1}{s}.$$

(11)

Suppose that$f\left(X\right)\subset g\left(X\right)$and at least one of$f\left(X\right),g\left(X\right)$is$b_{bl}$-complete subspace of$\left(X,b_{bl}\right).$Then:

(i) If the pair$(f,g)$has property$G_{f,g\left(x_n\right)}$and$G_{gf}\ne \varnothing ,$then f and g have a point of coincidence in$X.$

(ii) If x and y in X are points of coincidence of f and g such that$\left(x,y\right)\in E\left(G\right)$, then$x=y$. Hence, points of coincidence of f and g are unique in X. Moreover, if the pair$\left(f,g\right)$is weakly compatible, then f and g have a unique common fixed point in X.

Proof. 

(i) Assume that $G_{gf}\ne \varnothing ,$ there exists $x_0\in G_{gf}.$ Since $f\left(X\right)\subset g\left(X\right),$ there exists $x_1\in X$ such that $f\left(x_0\right)=g\left(x_1\right),$ again we can find $x_2\in X$ such that $f\left(x_1\right)=g\left(x_2\right).$ Repeating this step, we can build a sequence $z_n=f\left(x_n\right)=g\left(x_{n+1}\right)$ such that $\left(z_n,z_m\right)\in E\left(G\right).$ If $z_k=z_{k+1}$ for some $k\in \mathbb{N},$ then $f\left(x_{k+1}\right)=g\left(x_{k+1}\right)$ is a point of coincidence of f and $g.$ Therefore, let $z_n\ne z_{n+1}$ for all $n\in \mathbb{N}.$ By Condition (9), we can get that

$$\begin{array}{ccc}\hfill b_{bl}\left(z_n,z_{n+1}\right)& \le & s^q{b}_{bl}\left(z_n,z_{n+1}\right)=s^q{b}_{bl}\left(f\left(x_n\right),f\left(x_{n+1}\right)\right)\hfill \\ & \le & c_1{b}_{bl}\left(g\left(x_n\right),g\left(x_{n+1}\right)\right)+c_2{b}_{bl}\left(g\left(x_n\right),f\left(x_n\right)\right)+c_3{b}_{bl}\left(g\left(x_{n+1}\right),f\left(x_{n+1}\right)\right)\hfill \\ & & +c_4{b}_{bl}\left(g\left(x_n\right),f\left(x_{n+1}\right)\right)+c_5{b}_{bl}\left(g\left(x_{n+1}\right),f\left(x_n\right)\right).\hfill \end{array}$$

(12)

Since $z_n=f\left(x_n\right)=g\left(x_{n+1}\right)$ then Condition (12) becomes

$$\begin{array}{ccc}\hfill b_{bl}\left(z_n,z_{n+1}\right)& \le & c_1{b}_{bl}\left(z_{n-1},z_n\right)+c_2{b}_{bl}\left(z_{n-1},z_n\right)+c_3{b}_{bl}\left(z_n,z_{n+1}\right)\hfill \\ & & +c_4{b}_{bl}\left(z_{n-1},z_{n+1}\right)+c_5{b}_{bl}\left(z_n,z_n\right)\hfill \\ & \le & c_1{b}_{bl}\left(z_{n-1},z_n\right)+c_2{b}_{bl}\left(z_{n-1},z_n\right)+c_3{b}_{bl}\left(z_n,z_{n+1}\right)+s{c}_4{b}_{bl}\left(z_{n-1},z_n\right)\hfill \\ & & +s{c}_4{b}_{bl}\left(z_n,z_{n+1}\right)+2s{c}_5{b}_{bl}\left(z_{n-1},z_n\right),\hfill \end{array}$$

(13)

or equivalently:

$$b_{bl}\left(z_n,z_{n+1}\right)\le \lambda b_{bl}\left(z_{n-1},z_n\right),$$

(14)

where $\lambda =\frac{c_1+c_2+s{c}_4+2s{c}_5}{1-c_3-s{c}_4}.$ Since, $c_1+c_2+c_3+s{c}_4+2s{c}_5\le s{c}_1+s{c}_2+s{c}_3+2s{c}_4+2s{c}_5<1,$ it follows that $\lambda <1.$ Therefore, by Remark 2 of Lemma 1, the sequence $z_n=f\left(x_n\right)=g\left(x_{n+1}\right)$ is a $b_{bl}$-Cauchy sequence. The $b_{bl}$-completeness of $f\left(X\right)$ leads to $u\in f\left(X\right)\subset g\left(X\right)$ such that $z_n\to u=g\left(v\right)$ for some $v\in X.$ As $z_0\in G_{gf}$, this implies that $\left(z_n,z_m\right)\in E\left(G\right)$ for $n,m=1,2,\dots$ and so $\left(z_n,z_{n+1}\right)\in E\left(G\right).$ By property $G_{f,g\left(x_n\right)}$, there is a subsequence ${\left\{z_{n_i}\right\}}_{i\in N}$ of ${\left\{z_n\right\}}_{n\in N}$ such that $\left(z_{n_i},u\right)\in E\left(G\right).$ Applying $\left(b_{bl}3\right)$, we get

$$\begin{array}{ccc}\hfill b_{bl}\left(f\left(v\right),g\left(v\right)\right)& \le & s{b}_{bl}\left(f\left(v\right),f\left(x_{n_i}\right)\right)+s{b}_{bl}\left(f\left(x_{n_i}\right),g\left(v\right)\right)\hfill \\ & \le & s^q{b}_{bl}\left(f\left(v\right),f\left(x_{n_i}\right)\right)+s{b}_{bl}\left(f\left(x_{n_i}\right),g\left(v\right)\right)\hfill \\ & \le & c_1{b}_{bl}\left(g\left(v\right),g\left(x_{n_i}\right)\right)+c_2{b}_{bl}\left(g\left(v\right),f\left(v\right)\right)+c_3{b}_{bl}\left(g\left(x_{n_i}\right),f\left(x_{n_i}\right)\right)\hfill \\ & \le & +c_4{b}_{bl}\left(g\left(v\right),f\left(x_{n_i}\right)\right)+c_5{b}_{bl}\left(g\left(x_{n_i}\right),f\left(v\right)\right)+s{b}_{bl}\left(f\left(x_{n_i},g\left(v\right)\right)\right)\hfill \\ & =& c_1{b}_{bl}\left(g\left(v\right),z_{n_i-1}\right)+c_2{b}_{bl}\left(g\left(v\right),f\left(v\right)\right)+c_3{b}_{bl}\left(z_{n_i-1},z_{n_i}\right)\hfill \\ & & +c_4{b}_{bl}\left(g\left(v\right),z_{n_i}\right)+c_5{b}_{bl}\left(z_{n_i-1},f\left(v\right)\right)+s{b}_{bl}\left(z_{n_i},g\left(v\right)\right).\hfill \end{array}$$

(15)

Since $b_{bl}\left(z_{n_i-1},f\left(v\right)\right)\le s{b}_{bl}\left(z_{n_i-1},g\left(v\right)\right)+s{b}_{bl}\left(g\left(v\right),f\left(v\right)\right)$, Condition (15) becomes

$$\begin{array}{cc}\hfill & \left(1-c_2-c_{5}s\right)b_{bl}\left(f\left(v\right),g\left(v\right)\right)\hfill \\ \le & c_1{b}_{bl}\left(g\left(v\right),z_{n_i-1}\right)+c_3{b}_{bl}\left(z_{n_i-1},z_{n_i}\right)+c_4{b}_{bl}\left(g\left(v\right),z_{n_i}\right)\hfill \\ & +c_{5}s{b}_{bl}\left(z_{n_i-1},g\left(v\right)\right)+s{b}_{bl}\left(z_{n_i},g\left(v\right)\right).\hfill \end{array}$$

(16)

Taking the limit in Condition (16) as $i\to \infty$ we obtain that $b_{bl}\left(f\left(v\right),g\left(v\right)\right)=0,$ because $c_2+c_{5}s\le c_{1}s+c_{2}s+c_{3}s+2c_{4}s+2c_{5}s<1.$ That is, $f\left(v\right)=g\left(v\right)=u$ is a point of coincidence for the mappings f and $g,$ i.e., (i) is proved in the case if $f\left(X\right)$ is $b_{bl}$-complete. The proof for the case if $g\left(X\right)$ is $b_{bl}$-complete is similar.

(ii) Assume that x and y are two different points of coincidence of f and g with $\left(x,y\right)\in E\left(G\right).$ This means that there are different points $x_1$ and $y_1$ from X such that: $f\left(x_1\right)=g\left(x_1\right)=x$ and $f\left(y_1\right)=g\left(y_1\right)=y.$ Now, according to Condition (9) we get

$$\begin{array}{ccc}\hfill s{b}_{bl}\left(x,y\right)& \le & s^q{b}_{bl}\left(x,y\right)=s^q{b}_{bl}\left(f\left(x_1\right),f\left(y_1\right)\right)\hfill \\ & \le & c_1{b}_{bl}\left(g\left(x_1\right),g\left(y_1\right)\right)+c_2{b}_{bl}\left(g\left(x_1\right),f\left(y_1\right)\right)+c_3{b}_{bl}\left(g\left(y_1\right),f\left(y_1\right)\right)\hfill \\ & & +c_4{b}_{bl}\left(g\left(x_1\right),f\left(y_1\right)\right)+c_5{b}_{bl}\left(g\left(y_1\right),f\left(x_1\right)\right)\hfill \\ & =& c_1{b}_{bl}\left(x,y\right)+c_2{b}_{bl}\left(x,y\right)+c_3{b}_{bl}\left(y,y\right)\hfill \\ & & +c_4{b}_{bl}\left(x,y\right)+c_5{b}_{bl}\left(y,x\right)\hfill \\ & \le & \left(c_1+c_2+2c_{3}s+c_4+c_5\right)b_{bl}\left(y,x\right)\hfill \\ & \le & \left(c_{1}s+2c_{2}s+2c_{3}s+c_{4}s+c_{5}s\right)b_{bl}\left(y,x\right)<b_{bl}\left(y,x\right).\hfill \end{array}$$

(17)

Hence, if $x\ne y$ we get a contradiction.

If f and g are weakly compatible, then by Proposition 2 f and g have a unique common fixed point. □

Example 1.

Let$X=[0,+\infty )$and$f,g:X\to X$be the mappings such that

$$f\left(x\right)=e^x-1\mathit{and}g\left(x\right)=e^{4x}-1.$$

Consider b-metric-like space$\left(X,b_{bl}\right)$under the distance$b_{bl}(x,y)={(x+y)}^2$with coefficient$s=2$, and the graph$G=(V,E)$with$V=X$and$E=\left\{(x,x):x\in X\right\}\cup \left\{(0,x):x\in X\right\}$. Assume that$c_1=\frac{1}{4}$and$c_2=c_3=c_4=c_5=\frac{1}{25}$for which Inequalities (10) and (11) hold. Note that$\left(g\left(x\right),g\left(y\right)\right)\in E$if and only if$x=y,x\ge 0$or$x=0,y>0$or$y=0,x>0$. For$q=2$let us check whether Condition (9) holds in these cases.

Case 1:$x=y,x\ge 0$;

$$\begin{array}{cc}& c_1{b}_{bl}(g\left(x\right),g\left(x\right))+c_2{b}_{bl}(g\left(x\right),f\left(x\right))+c_3{b}_{bl}(g\left(x\right),f\left(x\right))+c_4{b}_{bl}(g\left(x\right),f\left(x\right))+c_5{b}_{bl}(g\left(x\right),f\left(x\right))\hfill \\ =& c_1{\left(e^{4x}-1+e^{4x}-1\right)}^2+(c_2+c_3+c_4+c_5){\left(e^{4x}-1+e^x-1\right)}^2\hfill \\ =& 4c_1{\left(e^x-1\right)}^2{\left(e^{3x}+e^{2x}+e^x+1\right)}^2+(c_2+c_3+c_4+c_5){\left(e^x-1\right)}^2{\left(e^{3x}+e^{2x}+e^x+2\right)}^2\hfill \\ \ge & 4c_1{\left(e^x-1\right)}^2{4}^2+(c_2+c_3+c_4+c_5){\left(e^x-1\right)}^2{5}^2=\left(\frac{1}{4}·64+\frac{4}{25}·25\right){\left(e^x-1\right)}^2\hfill \\ >& 4{\left(e^x-1+e^x-1\right)}^2=s^q{b}_{bl}(f\left(x\right),f\left(x\right)).\hfill \end{array}$$

Case 2:$x=0,y>0$(similarly for$y=0,x>0$);

$$\begin{array}{cc}& c_1{b}_{bl}(g\left(0\right),g\left(y\right))+c_2{b}_{bl}(g\left(0\right),f\left(0\right))+c_3{b}_{bl}(g\left(y\right),f\left(y\right))+c_4{b}_{bl}(g\left(0\right),f\left(y\right))+c_5{b}_{bl}(g\left(y\right),f\left(0\right))\hfill \\ =& c_1{\left(e^{4y}-1\right)}^2+c_2{(0+0)}^2+c_3{\left(e^{4y}-1+e^y-1\right)}^2+c_4{\left(e^y-1\right)}^2+c_5{\left(e^{4y}-1\right)}^2\hfill \\ =& (c_1+c_5){\left(e^y-1\right)}^2{\left(e^{3y}+e^{2y}+e^y+1\right)}^2+c_3{\left(e^y-1\right)}^2{\left(e^{3y}+e^{2y}+e^y+2\right)}^2+c_4{\left(e^y-1\right)}^2\hfill \\ >& (c_1+c_5){\left(e^y-1\right)}^2{4}^2+c_3{\left(e^y-1\right)}^2{5}^2+c_4{\left(e^y-1\right)}^2=\left(\frac{29}{100}·16+\frac{1}{25}·25+\frac{1}{25}\right){\left(e^y-1\right)}^2\hfill \\ >& 4{\left(e^y-1\right)}^2=s^q{b}_{bl}(f\left(0\right),f\left(y\right)).\hfill \end{array}$$

Hence, f and g satisfy Condition (9) for all$x,y\in X$such that$\left(g\left(x\right),g\left(y\right)\right)\in E$.

Moreover, there is$x_1=\frac{x_0}{4}$such that$g\left(x_1\right)=f\left(x_0\right)$,$x_2=\frac{x_0}{4^2}$such that$g\left(x_2\right)=f\left(x_1\right)$, and so on. In this way, we can built the sequence$x_n=\frac{x_0}{4^n},n\in \mathbb{N}$such that$g\left(x_n\right)=f\left(x_{n-1}\right)$. For$x_0\ne 0$it is clear that$(g\left(x_n\right),g\left(x_m\right))\notin E$. For$x_0=0$,$x_n=0,n\in \mathbb{N}$is obtained. Thus, the constant sequence$x_n=0$is only convergent sequence such that$(g\left(x_n\right),g\left(x_m\right))=(0,0)\in E$, and for each subsequence${\left(g\left(x_{n_i}\right)\right)}_{i\in \mathbb{N}}$of${\left(g\left(x_n\right)\right)}_{n\in \mathbb{N}}$holds$(g\left(x_{n_i}\right),0)=(0,0)\in E$. This means that$x_0\in G_{gf}\ne \varnothing$and the pair$(f,g)$possesses the property$G_{f,g\left(x_n\right)}$.

It is obvious that$f\left(X\right)\subset g\left(X\right)$and$g\left(X\right)=X$is$b_{bl}$-complete. Since the mappings f and g are weakly compatible at$x=0$($f\left(0\right)=g\left(0\right)$implies$g\left(f\right(0\left)\right)=f\left(g\right(0\left)\right)$), all conditions of Theorem 1 are satisfied. So, 0 is the unique common fixed point of mappings f and g in X.

Example 2.

Now consider the same b-metric-like space$\left(X,b_{bl}\right)$endowed with the graph G as in Example 1, and the mappings$f,g:X\to X$such that

$$f\left(x\right)=\left\{\begin{array}{ccc}\hfill e^x-1& \hfill ,& \hfill x\ne 0\\ \hfill 1& \hfill ,& \hfill x=0\end{array}\right.\mathit{and}g\left(x\right)=\left\{\begin{array}{ccc}\hfill e^{4x}-1& \hfill ,& \hfill x\ne 0\\ \hfill 2& \hfill ,& \hfill x=0\end{array}\right..$$

In this case we have$G_{gf}=\varnothing$. Namely, for$x_0=0$,$x_n=\frac{1}{4^n}ln2,n\in \mathbb{N}$is now obtained, and$(g\left(x_n\right),g\left(x_m\right))\notin E$. Hence, the conditions of Theorem 1 are not satisfied. Moreover, we can easily see that the mappings f and g have no coincidence point nor common fixed points.

As corollaries of our Theorem 1, we obtain the next results in the context of b-metric-like spaces endowed with a graph:

Corollary 1.

(Jungck) Let$f,g:X\to X$be self-mappings defined on a b-metric-like space$\left(X,b_{bl}\right)$(with coefficient$s\ge 1)$endowed with a graph$G,$and satisfy

$$s^q{b}_{bl}\left(f\left(x\right),f\left(y\right)\right)\le c_1{b}_{bl}\left(g\left(x\right),g\left(y\right)\right)$$

(18)

for all$x,y\in X$with$\left(g\left(x\right),g\left(y\right)\right)\in E\left(G\right)$when$c_1<\frac{1}{s}$. Suppose that$f\left(X\right)\subset g\left(X\right)$and at least one of$f\left(X\right),g\left(X\right)$is a$b_{bl}$-complete subspace of$\left(X,b_{bl}\right).$Then

(i) If the property$G_{f,g\left(x_n\right)}$is satisfied and$G_{gf}\ne \varnothing ,$then f and g have a point of coincidence in$X.$

(ii) If x and y in X are points of coincidence of f and g such that$\left(x,y\right)\in E\left(G\right)$, then$x=y$. Hence, points of coincidence of f and g are unique in X. Moreover, if the pair$\left(f,g\right)$is weakly compatible, then f and g have a unique common fixed point in$X.$

Corollary 2.

(Kannan) Let$f,g:X\to X$be self-mappings defined on a b-metric-like space$\left(X,b_{bl}\right)$(with coefficient$s\ge 1)$endowed with a graph$G,$and satisfy

$$s^q{b}_{bl}\left(f\left(x\right),f\left(y\right)\right)\le c_2{b}_{bl}\left(g\left(x\right),f\left(x\right)\right)+c_3{b}_{bl}\left(g\left(y\right),f\left(y\right)\right)$$

(19)

for all$x,y\in X$with$\left(g\left(x\right),g\left(y\right)\right)\in E\left(G\right)$when

$$c_2+c_3<\frac{1}{2s}.$$

(20)

Suppose that$f\left(X\right)\subset g\left(X\right)$and at least one of$f\left(X\right),g\left(X\right)$is a$b_{bl}$-complete subspace of$\left(X,b_{bl}\right).$Then

(i) If the property$G_{f,g\left(x_n\right)}$is satisfied and$G_{gf}\ne \varnothing ,$then f and g have a point of coincidence in$X.$

(ii) If x and y in X are points of coincidence of f and g such that$\left(x,y\right)\in E\left(G\right)$, then$x=y$. Hence, points of coincidence of f and g are unique in X. Moreover, if the pair$\left(f,g\right)$is weakly compatible, then f and g have a unique common fixed point in$X.$

Corollary 3.

(Chatterjea) Let$f,g:X\to X$be self-mappings defined on a b-metric-like space$\left(X,b_{bl}\right)$(with coefficient$s\ge 1)$endowed with a graph$G,$and satisfy

$$s^q{b}_{bl}\left(f\left(x\right),f\left(y\right)\right)\le c_4{b}_{bl}\left(g\left(x\right),f\left(y\right)\right)+c_5{b}_{bl}\left(g\left(y\right),f\left(x\right)\right),$$

(21)

for all$x,y\in X$with$\left(g\left(x\right),g\left(y\right)\right)\in E\left(G\right)$when

$$c_4+c_5<\frac{1}{2s}.$$

(22)

Suppose that$f\left(X\right)\subset g\left(X\right)$and at least one of$f\left(X\right),g\left(X\right)$is a$b_{bl}$-complete subspace of$\left(X,b_{bl}\right).$Then

(i) If the property$G_{f,g\left(x_n\right)}$is satisfied and$G_{gf}\ne \varnothing ,$then f and g have a point of coincidence in$X.$

(ii) If x and y in X are points of coincidence of f and g such that$\left(x,y\right)\in E\left(G\right)$, then$x=y$. Hence, points of coincidence of f and g are unique in X. Moreover, if the pair$\left(f,g\right)$is weakly compatible, then f and g have a unique common fixed point in$X.$

Corollary 4.

(Reich) Let$f,g:X\to X$be self-mappings defined on a b-metric-like space$\left(X,b_{bl}\right)$(with coefficient$s\ge 1)$endowed with a graph$G,$and satisfy

$$s^q{b}_{bl}\left(f\left(x\right),f\left(y\right)\right)\le c_1{b}_{bl}\left(g\left(x\right),g\left(y\right)\right)+c_2{b}_{bl}\left(g\left(x\right),f\left(x\right)\right)+c_3{b}_{bl}\left(g\left(y\right),f\left(y\right)\right)$$

(23)

for all$x,y\in X$with$\left(g\left(x\right),g\left(y\right)\right)\in E\left(G\right)$when

$$c_1+2c_2+2c_3<\frac{1}{s}$$

(24)

Suppose that$f\left(X\right)\subset g\left(X\right)$and at least one of$f\left(X\right),g\left(X\right)$is a$b_{bl}$-complete subspace of$\left(X,b_{bl}\right).$Then

(i) If the property$G_{f,g\left(x_n\right)}$is satisfied and$G_{gf}\ne \varnothing ,$then f and g have a point of coincidence in$X.$

(ii) If x and y in X are points of coincidence of f and g such that$\left(x,y\right)\in E\left(G\right)$, then$x=y$. Hence, points of coincidence of f and g are unique in X. Moreover, if the pair$\left(f,g\right)$is weakly compatible, then f and g have a unique common fixed point in$X.$

Now, we announce our last result in this section in the context of b-metric-like spaces endowed with the graph. The proof is similar enough with the corresponding proof of Theorem 1 and therefore we omit it.

Theorem 2.

(Das-Naik-Ćirić) Let$f,g:X\to X$be self-mappings defined on a b-metric-like space$\left(X,b_{bl}\right)$(with coefficient$s\ge 1)$endowed with a graph$G,$and satisfy

$$\begin{array}{ccc}\hfill s^q{b}_{bl}\left(f\left(x\right),f\left(y\right)\right)& \le & \lambda max\left\{b_{bl}\left(g\left(x\right),g\left(y\right)\right),b_{bl}\left(g\left(x\right),f\left(x\right)\right),b_{bl}\left(g\left(y\right),f\left(y\right)\right),\right.\hfill \\ & & \left.b_{bl}\left(g\left(x\right),f\left(y\right)\right),b_{bl}\left(g\left(y\right),f\left(x\right)\right)\right\}\hfill \end{array}$$

(25)

for all$x,y\in X$with$\left(g\left(x\right),g\left(y\right)\right)\in E\left(G\right)$when$\lambda \in [0,\frac{1}{s})$. Suppose that$f\left(X\right)\subset g\left(X\right)$and at least one of$f\left(X\right),g\left(X\right)$is a$b_{bl}$-complete subspace of$\left(X,b_{bl}\right).$Then

(i) If the property$G_{f,g\left(x_n\right)}$is satisfied and$G_{gf}\ne \varnothing ,$then f and g have a point of coincidence in$X.$

(ii) If x and y in X are points of coincidence of f and g such that$\left(x,y\right)\in E\left(G\right)$, then$x=y$. Hence, points of coincidence of f and g are unique in X. Moreover, if the pair$\left(f,g\right)$is weakly compatible, then f and g have a unique common fixed point in$X.$