Common fixed point of a power graphic contraction pair in partial metric spaces endowed with a graph

Abstract

In this paper, we initiate a study of fixed point results in the setup of partial metric spaces endowed with a graph. The concept of a power graphic contraction pair of two mappings is introduced. Common fixed point results for such maps without appealing to any form of commutativity conditions defined on a partial metric space endowed with a directed graph are obtained. These results unify, generalize and complement various known comparable results from the current literature.

MSC:47H10, 54H25, 54E50.

1 Introduction and preliminaries

Consistent with Jachymski [1], let X be a nonempty set and d be a metric on X. A set $\{(x,x):x\in X\}$ is called a diagonal of $X\times X$ and is denoted by Δ. Let G be a directed graph such that the set $V(G)$ of its vertices coincides with X and $E(G)$ is the set of the edges of the graph with $\mathrm{\Delta }\subseteq E(G)$. Also assume that the graph G has no parallel edges. One can identify a graph G with the pair $(V(G),E(G))$. Throughout this paper, the letters ℝ, ${\mathbb{R}}^{+}$, ω and ℕ will denote the set of real numbers, the set of nonnegative real numbers, the set of nonnegative integers and the set of positive integers, respectively.

Definition 1.1 [1]

A mapping $f:X\to X$ is called a Banach G-contraction or simply G-contraction if

(a₁) for each $x,y\in X$ with $(x,y)\in E(G)$, we have $(f(x),f(y))\in E(G)$,

(a₂) there exists $\alpha \in (0,1)$ such that for all $x,y\in X$ with $(x,y)\in E(G)$ implies that $d(f(x),f(y))\le \alpha d(x,y)$.

Let $X^f:=\{x\in X:(x,f(x))\in E(G)\text{ or }(f(x),x)\in E(G)\}$.

Recall that if $f:X\to X$, then a set $\{x\in X:x=f(x)\}$ of all fixed points of f is denoted by $F(f)$. A self-mapping f on X is said to be

• (1) a Picard operator if $F(f)=\{x^{\ast }\}$ and $f^n(x)\to x^{\ast }$ as $n\to \mathrm{\infty }$ for all $x\in X$;

• (2) a weakly Picard operator if $F(f)\ne \mathrm{\varnothing }$ and for each $x\in X$, we have $f^n(x)\to x^{\ast }\in F(f)$ as $n\to \mathrm{\infty }$;

• (3) orbitally continuous if for all $x,a\in X$, we have

  $$\underset{k\to \mathrm{\infty }}{lim}{f}^{n_k}(x)=a\text{implies}\underset{i\to \mathrm{\infty }}{lim}f(f^{n_k}(x))=f(a).$$

The following definition is due to Chifu and Petrusel [2].

Definition 1.2 An operator $f:X\to X$ is called a Banach G-graphic contraction if

(b₁) for each $x,y\in X$ with $(x,y)\in E(G)$, we have $(f(x),f(y))\in E(G)$,

(b₂) there exists $\alpha \in [0,1)$ such that

$$d(f(x),f^2(x))\le \alpha d(x,f(x))\text{for all }x\in X^f.$$

If x and y are vertices of G, then a path in G from x to y of length $k\in \mathbb{N}$ is a finite sequence $\{x_n\}$, $n\in \{0,1,2,\dots ,k\}$ of vertices such that $x_0=x$, $x_k=y$ and $(x_{i-1},x_i)\in E(G)$ for $i\in \{1,2,\dots ,k\}$.

Notice that a graph G is connected if there is a path between any two vertices and it is weakly connected if $\tilde{G}$ is connected, where $\tilde{G}$ denotes the undirected graph obtained from G by ignoring the direction of edges. Denote by $G^{-1}$ the graph obtained from G by reversing the direction of edges. Thus,

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

Since it is more convenient to treat $\tilde{G}$ as a directed graph for which the set of its edges is symmetric, under this convention, we have that

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

If G is such that $E(G)$ is symmetric, then for $x\in V(G)$, the symbol ${[x]}_G$ denotes the equivalence class of the relation R defined on $V(G)$ by the rule:

$$yRz\text{ if there is a path in }G\text{ from }y\text{ to }z.$$

A graph G is said to satisfy the property (A) (see also [2]) if for any sequence $\{x_n\}$ in $V(G)$ with $x_n\to x$ as $n\to \mathrm{\infty }$ and $(x_n,x_{n+1})\in E(G)$ for $n\in \mathbb{N}$ implies that $(x_n,x)\in E(G)$.

Jachymski [1] obtained the following fixed point result for a mapping satisfying the Banach G-contraction condition in metric spaces endowed with a graph.

Theorem 1.3 [1]

Let $(X,d)$ be a complete metric space and G be a directed graph and let the triple $(X,d,G)$ have a property (A). Let $f:X\to X$ be a G-contraction. Then the following statements hold:

• 1. $F_f\ne \mathrm{\varnothing }$ if and only if $X_f\ne \mathrm{\varnothing }$;

• 2. if $X_f\ne \mathrm{\varnothing }$ and G is weakly connected, then f is a Picard operator;

• 3. for any $x\in X_f$ we have that $f{|}_{{[x]}_{\tilde{G}}}$ is a Picard operator;

• 4. if $f\subseteq E(G)$, then f is a weakly Picard operator.

Gwodzdz-Lukawska and Jachymski [3] developed the Hutchinson-Barnsley theory for finite families of mappings on a metric space endowed with a directed graph. Bojor [4] obtained a fixed point of a φ-contraction in metric spaces endowed with a graph (see also [5]). For more results in this direction, we refer to [2, 6, 7].

On the other hand, Mathews [8] introduced the concept of a partial metric to obtain appropriate mathematical models in the theory of computation and, in particular, to give a modified version of the Banach contraction principle more suitable in this context. For examples, related definitions and work carried out in this direction, we refer to [9–19] and the references mentioned therein. Abbas et al. [20] proved some common fixed points in partially ordered metric spaces (see also [21]). Gu and He [22] proved some common fixed point results for self-maps with twice power type Φ-contractive condition. Recently, Gu and Zhang [23] obtained some common fixed point theorems for six self-mappings with twice power type contraction condition.

Throughout this paper, we assume that a nonempty set $X=V(G)$ is equipped with a partial metric p, a directed graph G has no parallel edge and G is a weighted graph in the sense that each vertex x is assigned the weight $p(x,x)$ and each edge $(x,y)$ is assigned the weight $p(x,y)$. As p is a partial metric on X, the weight assigned to each vertex x need not be zero and whenever a zero weight is assigned to some edge $(x,y)$, it reduces to a loop $(x,x)$.

Also, the subset $W(G)$ of $V(G)$ is said to be complete if for every $x,y\in W(G)$, we have $(x,y)\in E(G)$.

Definition 1.4 Self-mappings f and g on X are said to form a power graphic contraction pair if

• (a) for every vertex v in G, $(v,fv)$ and $(v,gv)\in E(G)$,

• (b) there exists $\varphi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ an upper semi-continuous and nondecreasing function with $\varphi (t)<t$ for each $t>0$ such that

  $$p^{\delta }(fx,gy)\le \varphi (p^{\alpha }(x,y)p^{\beta }(x,fx)p^{\gamma }(y,gy))$$

  (1.1)

for all $(x,y)\in E(G)$ holds, where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in (0,\mathrm{\infty })$.

If we take $f=g$, then the mapping f is called a power graphic contraction.

The aim of this paper is to investigate the existence of common fixed points of a power graphic contraction pair in the framework of complete partial metric spaces endowed with a graph. Our results extend and strengthen various known results [8, 12, 13, 24].

2 Common fixed point results

We start with the following result.

Theorem 2.1 Let $(X,p)$ be a complete partial metric space endowed with a directed graph G. If $f,g:X\to X$ form a power graphic contraction pair, then the following hold:

• (i) $F(f)\ne \mathrm{\varnothing }$ or $F(g)\ne \mathrm{\varnothing }$ if and only if $F(f)\cap F(g)\ne \mathrm{\varnothing }$.

• (ii) If $u\in F(f)\cap F(g)$, then the weight assigned to the vertex u is 0.

• (iii) $F(f)\cap F(g)\ne \mathrm{\varnothing }$ provided that G satisfies the property (A).

• (iv) $F(f)\cap F(g)$ is complete if and only if $F(f)\cap F(g)$ is a singleton.

Proof To prove (i), let $u\in F(f)$. By the given assumption, $(u,gu)\in E(G)$. Assume that we assign a non-zero weight to the edge $(u,gu)$. As $(u,u)\in E(G)$ and f and g form a power graphic contraction, we have

$$\begin{array}{rcl}{p}^{\delta }(u,gu)& =& p^{\delta }(fu,gu)\\ \le & \varphi (p^{\alpha }(u,u)p^{\beta }(u,fu)p^{\gamma }(u,gu))\\ =& \varphi (p^{\alpha +\beta }(u,u)p^{\gamma }(u,gu))\\ \le & \varphi (p^{\alpha +\beta }(u,gu)p^{\gamma }(u,gu))\\ =& \varphi (p^{\delta }(u,gu))\\ <& p^{\delta }(u,gu),\end{array}$$

a contradiction. Hence, the weight assigned to the edge $(u,gu)$ is zero and so $u=gu$. Therefore, $u\in F(f)\cap F(g)\ne \mathrm{\varnothing }$. Similarly, if $u\in F(g)$, then we have $u\in F(f)$. The converse is straightforward.

Now, let $u\in F(f)\cap F(g)$. Assume that the weight assigned to the vertex u is not zero, then from (1.1), we have

$$\begin{array}{rcl}{p}^{\delta }(u,u)& =& p^{\delta }(fu,gu)\\ \le & \varphi (p^{\alpha }(u,u)p^{\beta }(u,fu)p^{\gamma }(u,gu))\\ =& \varphi (p^{\alpha +\beta +\gamma }(u,u))\\ =& \varphi (p^{\delta }(u,u))\\ <& p^{\delta }(u,u),\end{array}$$

a contradiction. Hence, (ii) is proved.

To prove (iii), we will first show that there exists a sequence $\{x_n\}$ in X with $f{x}_{2n}=x_{2n+1}$ and $g{x}_{2n+1}=x_{2n+2}$ for all $n\in \mathbb{N}$ with $(x_n,x_{n+1})\in E(G)$, and ${lim}_{n\to \mathrm{\infty }}p(x_n,x_{n+1})=0$.

Let $x_0$ be an arbitrary point of X. If $f{x}_0=x_0$, then the proof is finished, so we assume that $f{x}_0\ne x_0$. As $(x_0,f{x}_0)\in E(G)$, so $(x_0,x_1)\in E(G)$. Also, $(x_1,g{x}_1)\in E(G)$ gives $(x_1,x_2)\in E(G)$. Continuing this way, we define a sequence $\{x_n\}$ in X such that $(x_n,x_{n+1})\in E(G)$ with $f{x}_{2n}=x_{2n+1}$ and $g{x}_{2n+1}=x_{2n+2}$ for $n\in \mathbb{N}$.

We may assume that the weight assigned to each edge $(x_{2n},x_{2n+1})$ is non-zero for all $n\in \mathbb{N}$. If not, then $x_{2k}=x_{2k+1}$ for some k, so $f{x}_{2k}=x_{2k+1}=x_{2k}$, and thus $x_{2k}\in F(f)$. Hence, $x_{2k}\in F(f)\cap F(g)$ by (i). Now, since $(x_{2n},x_{2n+1})\in E(G)$, so from (1.1), we have

$$\begin{array}{rcl}{p}^{\delta }(x_{2n+1},x_{2n+2})& =& p^{\delta }(f{x}_{2n},g{x}_{2n+1})\\ \le & \varphi (p^{\alpha }(x_{2n},x_{2n+1})p^{\beta }(x_{2n},f{x}_{2n})p^{\gamma }(x_{2n+1},g{x}_{2n+1}))\\ =& \varphi (p^{\alpha }(x_{2n},x_{2n+1})p^{\beta }(x_{2n},x_{2n+1})p^{\gamma }(x_{2n+1},x_{2n+2}))\\ =& \varphi (p^{\alpha +\beta }(x_{2n},x_{2n+1})p^{\gamma }(x_{2n+1},x_{2n+2}))\\ <& p^{\alpha +\beta }(x_{2n},x_{2n+1})p^{\gamma }(x_{2n+1},x_{2n+2}),\end{array}$$

which implies that

$$p^{\alpha +\beta }(x_{2n+1},x_{2n+2})<p^{\alpha +\beta }(x_{2n},x_{2n+1}),$$

a contradiction if $\alpha +\beta =0$. So, take $\alpha +\beta >0$, and we have

$$p(x_{2n+1},x_{2n+2})<p(x_{2n},x_{2n+1})$$

for all $n\in \mathbb{N}$. Again from (1.1), we have

$$\begin{array}{rcl}{p}^{\delta }(x_{2n+2},x_{2n+3})& =& p^{\delta }(g{x}_{2n+1},f{x}_{2n+2})\\ =& p^{\delta }(f{x}_{2n+2},g{x}_{2n+1})\\ \le & \varphi (p^{\alpha }(x_{2n+2},x_{2n+1})p^{\beta }(x_{2n+2},f{x}_{2n+2})p^{\gamma }(x_{2n+1},g{x}_{2n+1}))\\ =& \varphi (p^{\alpha }(x_{2n+1},x_{2n+2})p^{\beta }(x_{2n+2},x_{2n+3})p^{\gamma }(x_{2n+1},x_{2n+2}))\\ =& \varphi (p^{\alpha +\gamma }(x_{2n+1},x_{2n+2})p^{\beta }(x_{2n+2},x_{2n+3}))\\ <& p^{\alpha +\gamma }(x_{2n+1},x_{2n+2})p^{\beta }(x_{2n+2},x_{2n+3}),\end{array}$$

which implies that

$$p^{\alpha +\gamma }(x_{2n+2},x_{2n+3})<p^{\alpha +\gamma }(x_{2n+1},x_{2n+2}).$$

We arrive at a contradiction in case $\alpha +\gamma =0$. Therefore, we must take $\alpha +\gamma >0$; consequently, we have

$$p(x_{2n+2},x_{2n+3})<p(x_{2n+1},x_{2n+2})$$

for all $n\in \mathbb{N}$. Hence,

$$p^{\delta }(x_n,x_{n+1})\le \varphi (p^{\delta }(x_{n-1},x_n))<p^{\delta }(x_{n-1},x_n)$$

(2.1)

for all $n\in \mathbb{N}$. Therefore, the decreasing sequence of positive real numbers $\{p^{\delta }(x_n,x_{n+1})\}$ converges to some $c\ge 0$. If we assume that $c>0$, then from (2.1) we deduce that

$$0<c\le \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (p^{\delta }(x_{n-1},x_n))\le \varphi (c)<c,$$

a contradiction. So, $c=0$, that is, ${lim}_{n\to \mathrm{\infty }}{p}^{\delta }(x_n,x_{n+1})=0$ and so we have ${lim}_{n\to \mathrm{\infty }}p(x_n,x_{n+1})=0$. Also,

$$p^{\delta }(x_n,x_{n+1})\le \varphi (p^{\delta }(x_{n-1},x_n))\le \cdots \le {\varphi }^n(p^{\delta }(x_0,x_1)).$$

(2.2)

Now, for $m,n\in \mathbb{N}$ with $m>n$,

$$\begin{array}{rcl}{p}^{\delta }(x_n,x_m)& \le & p^{\delta }(x_n,x_{n+1})+p^{\delta }(x_{n+1},x_{n+2})+\cdots +p^{\delta }(x_{m-1},x_m)\\ -p^{\delta }(x_{n+1},x_{n+1})-p^{\delta }(x_{n+2},x_{n+2})-\cdots -p^{\delta }(x_{m-1},x_{m-1})\\ \le & {\varphi }^n(p^{\delta }(x_0,x_1))+{\varphi }^{n+1}(p^{\delta }(x_0,x_1))+\cdots +{\varphi }^{m-1}(p^{\delta }(x_0,x_1))\end{array}$$

implies that $p^{\delta }(x_n,x_m)$ converges to 0 as $n,m\to \mathrm{\infty }$. That is, ${lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)=0$. Since $(X,p)$ is complete, following similar arguments to those given in Theorem 2.1 of [9], there exists a $u\in X$ such that ${lim}_{n,m\to \mathrm{\infty }}p(x_n,x_m)={lim}_{n\to \mathrm{\infty }}p(x_n,u)=p(u,u)=0$. By the given hypothesis, $(x_{2n},u)\in E(G)$ for all $n\in \mathbb{N}$. We claim that the weight assigned to the edge $(u,gu)$ is zero. If not, then as f and g form a power graphic contraction, so we have

$$\begin{array}{rcl}{p}^{\delta }(x_{2n+1},u)& =& p^{\delta }(f{x}_{2n},gu)\\ \le & \varphi (p^{\alpha }(x_{2n},u)p^{\beta }(x_{2n},f{x}_{2n})p^{\gamma }(u,gu))\\ =& \varphi (p^{\alpha }(x_{2n},u)p^{\beta }(x_{2n},x_{2n+1})p^{\gamma }(u,gu)).\end{array}$$

(2.3)

We deduce, by taking upper limit as $n\to \mathrm{\infty }$ in (2.3), that

$$\begin{array}{rcl}{p}^{\delta }(u,gu)& \le & \underset{n\to \mathrm{\infty }}{lim\hspace{0.17em}sup}\varphi (p^{\alpha }(x_{2n},u)p^{\beta }(x_{2n},x_{2n+1})p^{\gamma }(u,gu))\\ \le & \varphi (p^{\alpha }(u,u)p^{\beta }(u,u)p^{\gamma }(u,gu))\\ \le & \varphi (p^{\alpha +\beta +\gamma }(u,gu))\\ <& p^{\delta }(u,gu),\end{array}$$

a contradiction. Hence, $u=gu$ and $u\in F(f)\cap F(g)$ by (i).

Finally, to prove (iv), suppose the set $F(f)\cap F(g)$ is complete. We are to show that $F(f)\cap F(g)$ is a singleton. Assume on the contrary that there exist u and v such that $u,v\in F(f)\cap F(g)$ but $u\ne v$. As $(u,v)\in E(G)$ and f and g form a power graphic contraction, so

$$\begin{array}{rcl}0& <& p^{\delta }(u,v)=p^{\delta }(fu,fv)\\ \le & \varphi (p^{\alpha }(u,v)p^{\beta }(u,fu)p^{\gamma }(v,gv))\\ =& \varphi (p^{\alpha }(u,v)p^{\beta }(u,u)p^{\gamma }(v,v))\\ \le & \varphi (p^{\delta }(u,v)),\end{array}$$

a contradiction. Hence, $u=v$. Conversely, if $F(f)\cap F(g)$ is a singleton, then it follows that $F(f)\cap F(g)$ is complete. □

Corollary 2.2 Let $(X,p)$ be a complete partial metric space endowed with a directed graph G. If we replace (1.1) by

$$p^{\delta }(f^{s}x,g^{t}y)\le \varphi (p^{\alpha }(x,y)p^{\beta }(x,f^{s}x)p^{\gamma }(y,g^{t}y)),$$

(2.4)

where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in (0,\mathrm{\infty })$ and $s,t\in \mathbb{N}$, then the conclusions obtained in Theorem 2.1 remain true.

Proof It follows from Theorem 2.1, that $F(f^s)\cap F(g^t)$ is a singleton provided that $F(f^s)\cap F(g^t)$ is complete. Let $F(f^s)\cap F(g^t)=\{w\}$, then we have $f(w)=f(f^s(w))=f^{s+1}(w)=f^s(f(w))$, and $g(w)=g(g^t(w))=g^{t+1}(w)=g^t(g(w))$ implies that fw and gw are also in $F(f^s)\cap F(g^t)$. Since $F(f^s)\cap F(g^t)$ is a singleton, we deduce that $w=fw=gw$. Hence, $F(f)\cap F(g)$ is a singleton. □

The following remark shows that different choices of α, β and γ give a variety of power graphic contraction pairs of two mappings.

Remarks 2.3 Let $(X,p)$ be a complete partial metric space endowed with a directed graph G.

(R1) We may replace (1.1) with the following:

$$p^3(fx,gy)\le \varphi (p(x,y)p(x,fx)p(y,gy))$$

(2.5)

to obtain conclusions of Theorem 2.1. Indeed, taking $\alpha =\beta =\gamma =1$ in Theorem 2.1, one obtains (2.5).

(R2) If we replace (1.1) by one of the following condition:

(2.6)

(2.7)

(2.8)

then the conclusions obtained in Theorem 2.1 remain true. Note that

• (i) if we take $\alpha =\beta =1$ and $\gamma =0$ in (1.1), then we obtain (2.6),

• (ii) take $\alpha =\gamma =1$, $\beta =0$ in (1.1) to obtain (2.7),

• (iii) use $\beta =\gamma =1$, $\alpha =0$ in (1.1) and obtain (2.8).

(R3) Also, if we replace (1.1) by one of the following conditions:

(2.9)

(2.10)

(2.11)

then the conclusions obtained in Theorem 2.1 remain true. Note that

• (iv) take $\alpha =1$ and $\beta =\gamma =0$ in (1.1) to obtain (2.9),

• (v) to obtain (2.10), take $\beta =1$, $\alpha =\gamma =0$ in (1.1),

• (vi) if one takes $\gamma =1$, $\alpha =\beta =0$ in (1.1), then we obtain (2.11).

Remark 2.4 If we take $f=g$ in a power graphic contraction pair, then we obtain fixed point results for a power graphic contraction.