# On some fixed-point theorems for *ψ*-contraction on metric space involving a graph

- Mahpeyker Öztürk
^{1}Email author and - Ekber Girgin
^{1}

**2014**:39

https://doi.org/10.1186/1029-242X-2014-39

© Öztürk and Girgin; licensee Springer. 2014

**Received: **2 October 2013

**Accepted: **2 January 2014

**Published: **24 January 2014

## Abstract

In this paper, we introduce the $(G,\psi )$-contraction and the $(G,\psi )$-graphic contraction in a metric space by using a graph. We explain some conditions for a mapping which is a $(G,\psi )$-contraction to have a unique fixed point and also we give conditions as regards the existence of a fixed point for $(G,\psi )$-graphic contraction by applying the connectivity of the graph in both cases. Moreover, we give examples to show that our results are a substantial improvement of some known results in the literature.

**MSC:**47H10, 54H25.

## Keywords

*ψ*-type contraction

## 1 Introduction

The metric fixed-point theory has been researched extensively in the past two decades such as in a metric space endowed with a partial ordering, and many results appeared giving sufficient conditions for a mapping to be a Picard operator. For these concepts have been given two main theorems, which are the Banach Contraction Principle and the Knaster-Tarski Theorem [1].

Recently Jachymski [2] and Gwóźdź-Lukawska and Jachymski [3] have given an interesting concept in fixed-point theory with some general structures by using the context of metric spaces endowed with a graph. Jachymski [2] has proved some generalizations of the Banach Contraction Principle to mappings on a metric space endowed with a graph and also has presented its applications to the Kelisky-Rivlin Theorem on iterates of the Bernstein operators on the space $C[0,1]$. Afterwards different contractions have been studied by various authors. In [4] the contraction principle for set-valued mappings, in [5–7] Kannan type, Reich type contractions, and *φ*-contractions have been investigated, respectively. Some new fixed-point results for graphic contractions on a complete metric space with a graph have been presented in [8]; also they gave a particular case of almost contractions.

In this paper, motivated by the work of Jachymski [2] and Petruşel [8], we introduce new contractions for the mappings on complete metric space and prove some fixed-point theorems. Our results generalize and unify some results by the above-mentioned authors.

## 2 Basic facts and definitions

Let $(X,d)$ be a metric space and Δ denote the diagonal of the Cartesian product $X\times X$. Let *G* be a directed graph such that the set $V(G)$ of its vertices coincides with *X*, and the set $E(G)$ of its edges contains all loops; that is, $E(G)\supseteq \mathrm{\Delta}$. Assume that *G* has no parallel edges, so one can identify *G* with the pair $(V(G),E(G))$.

*G*is denoted by ${G}^{-1}$ and this is a graph obtained from

*G*by reversing the direction of the edges. Hence

*G*by omitting the direction of the edges. Indeed, it is more convenient to treat $\tilde{G}$ as a directed graph for which the set of its edges is symmetric, and under this convention, we have

*G*is a graph

*H*such that $V(H)\subseteq V(G)$ and $E(H)\subseteq E(G)$. Let

*x*and

*y*be vertices in a graph

*G*. A path from

*x*to

*y*of length

*N*($N\in \mathbf{N}\cup \{0\}$) is a sequence ${({x}_{i})}_{i=0}^{N}$ of $N+1$ distinct vertices such that ${x}_{0}=x$, ${x}_{N}=y$ and $({x}_{i-1},{x}_{i})\in E(G)$ for $i=1,\dots ,N$. The number edges in

*G*forming the path is called the length of the path. A graph

*G*is connected if there is a path between any two vertices. If a graph

*G*is not connected then it is called disconnected and its different paths are called the components of

*G*. Every component of

*G*is a subgraph of it. Furthermore,

*G*is weakly connected if $\tilde{G}$ is connected. Let ${G}_{x}$ be the component of

*G*which consists of all edges and vertices contained in some path in

*G*beginning at

*x*. Suppose that

*G*is such that $E(G)$ is symmetric; then $V(G)={[x]}_{G}$ where ${[x]}_{G}$ denotes the equivalence class of relations ℜ defined on $V(G)$ by the rule

Some basic notations related to connectivity of graphs can be found in [9].

the set of all fixed points of *f*.

**Definition 1** [2]

*G*-contraction or simply

*G*-contraction if

*f*preserves edges of

*G*;

*f*decreases weights of edges of

*G*: for all $x,y\in X$ there exists $\alpha \in (0,1)$ such that

**Definition 2** [8]

*G*-graphic contraction

- (i)if
*f*preserves edges of*G*;$(x,y)\in E(G)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}(fx,fy)\in E(G),$(3)

- (ii)there exists $\alpha \in (0,1)$ such that$(x,y)\in E(G)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}d(fx,{f}^{2}x)\le \alpha d(x,fx),$(4)

for all $x,y\in {X}_{f}$.

**Definition 3** [2]

**Definition 4** [2]

*G*-continuous if for all $x,y\in X$ and any sequence ${({k}_{n})}_{n\in \mathbf{N}}$ of positive integers,

Now, we give a definition of the class Ψ which is used in several well-known papers to obtain some fixed-point results [10–13].

**Definition 5**Let us define the class $\mathrm{\Psi}=\{\psi :{\mathbf{R}}^{+}\to {\mathbf{R}}^{+}\mid \psi \text{is nondecreasing}\}$ which satisfies the following conditions:

- (i)
$\psi (\omega )=0$ if and only if $\omega =0$;

- (ii)
for every $({\omega}_{n})\in {\mathbf{R}}^{+}$, $\psi ({\omega}_{n})\to 0$ if and only if ${\omega}_{n}\to 0$;

- (iii)
for every ${\omega}_{1},{\omega}_{2}\in {\mathbf{R}}^{+}$, $\psi ({\omega}_{1}+{\omega}_{2})\le \psi ({\omega}_{1})+\psi ({\omega}_{2})$.

## 3 $(G,\psi )$-Contraction and related fixed-point theorems

We establish some fixed-point theorems in metric space with a graph by defining the $(G,\psi )$-contraction.

**Definition 6**We say that a mapping $f:X\to X$ is a $(G,\psi )$-contraction if the following hold;

- (i)
*f*preserves edges of*G*,*i.e.*$((x,y)\in E(G)\Rightarrow (fx,fy)\in E(G))$, $\mathrm{\forall}x,y\in X$; - (ii)
*f*decreases the weight of edges of*G*, that is, there exists $c\in (0,1)$ such that$(x,y)\in E(G)\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\psi (d(fx,fy))\le c\psi (d(x,y)),$

for all $x,y\in X$.

**Lemma 1** *If* $f:X\to X$ *is a* $(G,\psi )$-*contraction*, *then* *f* *is both a* $({G}^{-1},\psi )$-*contraction and a* $(\tilde{G},\psi )$-*contraction*.

*Proof* The proof can be obtained by the symmetry of *d* and the definition of the $(\tilde{G},\psi )$-contraction. □

**Lemma 2**

*Let*$f:X\to X$

*be a*$(G,\psi )$-

*contraction with constant*$c\in (0,1)$;

*for a given*$x\in X$

*and*$y\in {[x]}_{\tilde{G}}$,

*there exists*$r(x,y)\ge 0$

*such that*

*Proof*Let $x\in X$ and $y\in {[x]}_{\tilde{G}}$. Then there is a path ${({x}_{i})}_{i=0}^{N}$ in $\tilde{G}$ from

*x*to

*y*, which means ${x}_{0}=x$, ${x}_{N}=y$, and $({x}_{i-1},{x}_{i})\in E(\tilde{G})$ for $i=1,2,\dots ,N$. By Lemma 1,

*f*is a $(\tilde{G},\psi )$-contraction. With an easy induction, we have $({f}^{n}{x}_{i-1},{f}^{n}{x}_{i})\in E(\tilde{G})$ and

for all $n\in \mathbf{N}$ and $i=1,2,\dots ,N$.

So it qualifies to set $r(x,y):={\sum}_{i=1}^{N}\psi (d({x}_{i-1},{x}_{i}))$. □

**Lemma 3** *Let* $(X,d)$ *be a complete metric space endowed with a graph* *G* *and* $f:X\to X$ *be a* $(G,\psi )$-*contraction for which there exists* ${x}_{0}\in X$ *such that* $f{x}_{0}\in {[{x}_{0}]}_{\tilde{G}}$. *Let* ${\tilde{G}}_{{x}_{0}}$ *be the component of* $\tilde{G}$ *containing* ${x}_{0}$. *Then* ${[{x}_{0}]}_{\tilde{G}}$ *is* *f*-*invariant and* $f{{|}_{[x]}}_{\tilde{G}}$ *is a* $({\tilde{G}}_{{x}_{0}},\psi )$-*contraction*. *Furthermore*, $x,y\in {[{x}_{0}]}_{\tilde{G}}$, *and the sequences* ${({f}^{n}x)}_{n\in \mathbf{N}}$ *and* ${({f}^{n}y)}_{n\in \mathbf{N}}$ *are Cauchy equivalent*.

*Proof* The proof of this lemma can obtained by using similar arguments as given in [7]. So we omit the proof. □

The following result shows that there is a close relation between convergence of an iteration sequence which can be obtained by using a $(G,\psi )$-contraction mapping and connectivity of the graph.

**Theorem 1**

*Let*$(X,d)$

*be a metric space endowed with a graph*

*G*

*and*$f:X\to X$

*be a*$(G,\psi )$-

*contraction*,

*then the following statements are equivalent*:

- (i)
*G**is weakly connected*; - (ii)
*for given*$x,y\in X$,*the sequences*${({f}^{n}x)}_{n\in \mathbf{N}}$*and*${({f}^{n}y)}_{n\in \mathbf{N}}$*are Cauchy equivalent*; - (iii)
$cardF(f)\le 1$.

*Proof*(i) ⇒ (ii) Let

*f*be a $(G,\psi )$-contraction and $x,y\in X$. By hypothesis, ${[x]}_{\tilde{G}}=X$, so $fx\in {[x]}_{\tilde{G}}$. By Lemma 2, we get

- (ii)
⇒ (iii) Let

*f*be a $(G,\psi )$-contraction and $x,y\in F(f)$. By (ii), ${({f}^{n}x)}_{n\in \mathbf{N}}$ and ${({f}^{n}y)}_{n\in \mathbf{N}}$ are equivalent, which yields $x=y$. - (iii)⇒ (ii) Suppose, to the contrary,
*G*is not weakly connected, that is, $\tilde{G}$ is disconnected. Let ${x}_{0}\in X$. Then the sets ${[{x}_{0}]}_{\tilde{G}}$ and $X-{[{x}_{0}]}_{\tilde{G}}$ both are nonempty. Let ${y}_{0}\in X-{[{x}_{0}]}_{\tilde{G}}$ and define$fx=\{\begin{array}{ll}{x}_{0},& \text{if}x\in {[{x}_{0}]}_{\tilde{G}},\\ {y}_{0},& \text{if}x\in X-{[{x}_{0}]}_{\tilde{G}}.\end{array}$

Obviously, $F(f)=\{{x}_{0},{y}_{0}\}$. We show *f* is a $(G,\psi )$-contraction. Let $(x,y)\in E(G)$. Then ${[x]}_{\tilde{G}}={[y]}_{\tilde{G}}$, so either $x,y\in {[{x}_{0}]}_{\tilde{G}}$ or $x,y\in X-{[{x}_{0}]}_{\tilde{G}}$. Hence in both cases $fx=fy$, so $(fx,fy)\in E(G)$ as $E(G)\supseteq \mathrm{\Delta}$, and $\psi (d(fx,fy))=0$. Thereby, *f* is a $(G,\psi )$-contraction having two fixed points which violates the assumption. □

The following result is an easy consequence of Theorem 1.

**Corollary 1**

*Let*$(X,d)$

*be a complete metric space endowed with a graph*

*G*

*and*$f:X\to X$

*be a*$(G,\psi )$-

*contraction*,

*then the following statements are equivalent*:

- (i)
*G**is weakly connected*; - (ii)
*there is*${x}^{\ast}\in X$*such that*${lim}_{n\to \mathrm{\infty}}{f}^{n}x={x}^{\ast}$,*for all*$x\in X$.

Now, we give an example of *f* being a $(G,\psi )$-contraction and this example shows that we could not add that ${x}^{\ast}$ is a fixed point of *f* in Corollary 1.

**Example 1**Let $X=[0,1]$ be endowed with the usual metric. Take

Then *f* is a $(G,\psi )$-contraction where $\psi (\omega )=\frac{\omega}{\omega +1}$.

*Proof* It can be easily seen that *G* is a weakly connected graph and *f* is a $(G,\psi )$-contraction where $\psi (\omega )=\frac{\omega}{\omega +1}$. It is a fact that $({f}^{n}x)\to 0$, for all $x\in X$ but *f* has no fixed point. □

For any mapping which satisfies the condition of Corollary 1 to have a fixed point we need to add condition (6), which is given in the following theorem.

**Theorem 2**

*Let*$(X,d)$

*be a complete metric space and the triple*$(X,d,G)$

*have the following condition*:

*Let*$f:X\to X$

*be a*$(G,\psi )$-

*contraction*,

*and*${X}_{f}=\{x\in X:(x,fx)\in E(G)\}$.

*Then the following statements hold*.

- (i)
$cardF(f)=card\{{[x]}_{\tilde{G}}:x\in {X}_{f}\}$.

- (ii)
$F(f)\ne \mathrm{\varnothing}$

*iff*${X}_{f}\ne \mathrm{\varnothing}$. - (iii)
*f**has a unique fixed point iff there exists*${x}_{0}\in {X}_{f}$*such that*${X}_{f}\subseteq {[{x}_{0}]}_{\tilde{G}}$. - (iv)
*For any*$x\in {X}_{f}$, $f{{|}_{[x]}}_{\tilde{G}}$*is a Picard operator*. - (v)
*If*${X}_{f}\ne \mathrm{\varnothing}$*and**G**is weakly connected*,*then**f**is a Picard operator*. - (vi)
*If*${X}^{\prime}:=\bigcup \{{[x]}_{\tilde{G}}:x\in {X}_{f}\}$ ,*then*$f{|}_{{X}^{\prime}}$*is a weakly Picard operator*. - (vii)
*If*$f\subseteq E(G)$,*then**f**is a weakly Picard operator*.

*Proof*Initially, we prove the items (iv) and (v). Take $x\in {X}_{f}$ and then $fx\in {[x]}_{\tilde{G}}$, so by Lemma 3, if $y\in {[x]}_{\tilde{G}}$, then ${({f}^{n}x)}_{n\in \mathbf{N}}$ and ${({f}^{n}y)}_{n\in \mathbf{N}}$ are Cauchy equivalent. Since

*X*is complete, ${({f}^{n}x)}_{n\in \mathbf{N}}$ converges to some ${x}^{\ast}\in X$. It is obvious that ${lim}_{n\to \mathrm{\infty}}{f}^{n}y={x}^{\ast}$. Then by using induction we get

*G*and also in $\tilde{G}$ from

*x*to ${x}^{\ast}$, and this means that ${x}^{\ast}\in {[x]}_{\tilde{G}}$. Since

*f*is a $(G,\psi )$-contraction we have

for all $n\in \mathbf{N}$. By taking the limit as $n\to \mathrm{\infty}$, we deduce $f{x}^{\ast}={x}^{\ast}$. Thereby, $f{{|}_{[x]}}_{\tilde{G}}$ is a Picard operator. Also, we conclude that *f* is a Picard operator, when ${[x]}_{\tilde{G}}=X$, since there is weakly connectedness of *G*.

(vi) is obvious from (iv). For proof of (vii), if $f\subseteq E(G)$ then ${X}_{f}=X$ and so ${X}^{\prime}=X$ holds. Thus *f* is a weakly Picard operator because of (vi).

*ρ*is a surjective mapping. We show that

*f*is injective. Take ${x}_{1},{x}_{2}\in F(f)$ which are such that $\rho ({x}_{1})=\rho ({x}_{2})\Rightarrow {[{x}_{1}]}_{\tilde{G}}={[{x}_{2}]}_{\tilde{G}}$, then ${x}_{2}\in {[{x}_{1}]}_{\tilde{G}}$ and so, by (i),

which gives ${x}_{1}={x}_{2}$. Thus, *f* is injective and this is the desired result. Finally, one can see that (ii) and (iii) are easy consequences of (i). □

**Corollary 2**

*Let*$(X,d)$

*be complete metric space and*$(X,d,G)$

*obey condition*(6).

*The following are equivalent*:

- (i)
*G**is weakly connected*; - (ii)
*every*$(G,\psi )$-*contraction*$f:X\to X$*such that*$({x}_{0},f{x}_{0})\in E(G)$,*for some*${x}_{0}\in X$,*is a Picard operator*; - (iii)
*for any*$(G,\psi )$-*contraction*, $cardF(f)\le 1$.

*Proof*(i) ⇒ (ii): This can be obtained directly from Theorem 2(v).

- (ii)
⇒ (iii): Let $f:X\to X$ be a $(G,\psi )$-contraction. If ${X}_{f}$ is empty, so is $F(f)$, because $F(f)$ is a subset of ${X}_{f}$. If ${X}_{f}$ is nonempty, then by (ii), $F(f)$ is singleton. In these two cases, $cardF(f)\le 1$.

- (iii)
⇒ (i): This implication follows from Theorem 1. □

**Remark 1** In the above results by taking $\psi (\omega )=\omega $, we obtain Corollary 3.2, which is given in [2].

## 4 $(G,\psi )$-Graphic contraction and fixed-point theorems

Now, we define $(G,\psi )$-graphic contraction and give some results and examples.

**Definition 7**Let $(X,d)$ be a metric space and

*G*be a graph. The mapping $f:X\to X$ is called a $(G,\psi )$-graphic contraction if the following conditions hold:

- (i)
$(x,y)\in E(G)$ implies $(fx,fy)\in E(G)$ (

*f*is edge preserving); - (ii)there exists a $\psi \in \mathrm{\Psi}$ with constants $c\in [0,1)$ such that$\psi \left(d(fx,{f}^{2}x)\right)\le c\psi (d(x,fx))$

for all $x\in {X}^{f}$, where ${X}^{f}:=\{x\in X:(x,fx)\in E(G)\text{or}(fx,x)\in E(G)\}$.

Firstly, we give the following lemmas which can be proved as in the above section.

**Lemma 4** *If* $f:X\to X$ *is a* $(G,\psi )$-*graphic contraction*, *then* *f* *is both a* $({G}^{-1},\psi )$-*graphic contraction and a* $(\tilde{G},\psi )$-*graphic contraction*.

**Lemma 5**

*Let*$f:X\to X$

*be a*$(G,\psi )$-

*graphic contraction with constant*$c\in [0,1)$.

*Then*,

*given*$x\in {X}^{f}$,

*there exists*$r(x)\ge 0$

*such that*

*for all* $n\in \mathbf{N}$, *where* $r(x):=\psi (d(x,fx))$.

**Lemma 6** *Suppose that* $f:X\to X$ *is a* $(G,\psi )$-*graphic contraction*. *Then for each* $x\in {X}^{f}$, *there exists* ${x}^{\ast}\in X$ *such that the sequence* ${({f}^{n}x)}_{n\in \mathbf{N}}$ *converges to* ${x}^{\ast}$ *as* $n\to \mathrm{\infty}$.

*Proof*Take an arbitrary element

*x*in ${X}^{f}$. By Lemma 5, we obtain

for all $n\in \mathbf{N}$. Therefore, ${\sum}_{n=0}^{\mathrm{\infty}}\psi (d({f}^{n}x,{f}^{n+1}x))<\mathrm{\infty}$ and so $\psi (d({f}^{n}x,{f}^{n+1}x))\to 0$; consequently using the property of *ψ* we have $d({f}^{n}x,{f}^{n+1}x)\to 0$. Then we say that ${({f}^{n}x)}_{n\in \mathbf{N}}$ is a Cauchy sequence. By the completeness of *X*, there exists ${x}^{\ast}\in X$ such that ${({f}^{n}x)}_{n\in \mathbf{N}}$ converges as $n\to \mathrm{\infty}$. □

**Lemma 7** *The self*-*mapping* *f* *is a* $(G,\psi )$-*graphic contraction for which there exists* ${x}_{0}\in X$ *such that* $f{x}_{0}\in {[{x}_{0}]}_{\tilde{G}}$. *Then the set* ${[{x}_{0}]}_{\tilde{G}}$ *invariant with respect to* *f* *and* $f{|}_{{[{x}_{0}]}_{\tilde{G}}}$ *is a* $({\tilde{G}}_{{x}_{0}},\psi )$-*graphic contraction*, *where* ${\tilde{G}}_{{x}_{0}}$ *is the component of* $\tilde{G}$ *containing* ${x}_{0}$.

*Proof* Let *x* be an element in ${[{x}_{0}]}_{\tilde{G}}$. Then there exist ${({x}_{i})}_{i=0}^{N}$ in $\tilde{G}$ from ${x}_{0}$ to *x*, *i.e.*, ${x}_{N}=x$ and $({x}_{i-1},{x}_{i})\in E(\tilde{G})$ for $i=1,2,\dots ,N$. Since *f* is a $(G,\psi )$-graphic contraction we get $(f{x}_{i-1},f{x}_{i})\in E(\tilde{G})$ for $i=1,2,\dots ,N$. So we have a path from $f{x}_{0}$ to *fx*. Therefore $fx\in {[f{x}_{0}]}_{\tilde{G}}={[{x}_{0}]}_{\tilde{G}}$ since $f{x}_{0}\in {[{x}_{0}]}_{\tilde{G}}$. Consequently ${[{x}_{0}]}_{\tilde{G}}$ is invariant with respect to *f*.

*y*such that ${x}_{N-1}=x$. Also let ${({y}_{i})}_{i=0}^{M}$ be a path in $\tilde{G}$ from ${x}_{0}$ to $f{x}_{0}$. Then we realize

is a path in $\tilde{G}$ from ${x}_{0}$ to *fy* such that $(fx,fy)\in E({\tilde{G}}_{{x}_{0}})$. Furthermore, *f* is a $({\tilde{G}}_{{x}_{0}},\psi )$-graphic contraction because $E({\tilde{G}}_{{x}_{0}})\subseteq E(\tilde{G})$ and *f* is a $(\tilde{G},\psi )$-graphic contraction. □

**Theorem 3**

*Let*$(X,d)$

*be a complete metric space and let the triple*$(X,d,G)$

*have the following condition*:

*Let*$f:X\to X$

*be a*$(G,\psi )$-

*graphic contraction and*

*f*

*is orbitally*

*G*-

*continuous*.

*Then the following statements hold*:

- (i)
$F(f)\ne \mathrm{\varnothing}$

*if and only if*${X}^{f}\ne \mathrm{\varnothing}$. - (ii)
*If*${X}^{f}\ne \mathrm{\varnothing}$*and**G**is weakly connected*,*then**f**is a weakly Picard operator*. - (iii)
*For any*$x\in {X}^{f}$,*we see that*$f{{|}_{[x]}}_{\tilde{G}}$*is a weakly Picard operator*.

*Proof* We begin with the statement (iii). Let $x\in {X}^{f}$; by Lemma 6, there exists ${x}^{\ast}\in X$ such that ${lim}_{n\to \mathrm{\infty}}{f}^{n}x={x}^{\ast}$. Since $x\in {X}^{f}$, then ${f}^{n}x\in {X}^{f}$ for every $n\in \mathbf{N}$. Now assume that $(x,fx)\in E(G)$. (A similar deduction can be made if $(fx,x)\in E(G)$.) By condition (9), there is a subsequence ${({f}^{{k}_{n}}x)}_{n\in \mathbf{N}}$ of ${({f}^{n}x)}_{n\in \mathbf{N}}$ such that $({f}^{{k}_{n}}x,{x}^{\ast})\in E(G)$ for each $n\in \mathbf{N}$. A path in *G* can be formed by using the points $x,fx,\dots ,{f}^{{k}_{1}}x,{x}^{\ast}$ and hence ${x}^{\ast}\in {[x]}_{\tilde{G}}$. Since *f* is orbitally *G*-continuous, we see that ${x}^{\ast}$ is a fixed point for $f{{|}_{[x]}}_{\tilde{G}}$.

To prove (i), using (iii) we have $F(f)\ne \mathrm{\varnothing}$ if ${X}^{f}\ne \mathrm{\varnothing}$. Suppose that $F(f)\ne \mathrm{\varnothing}$. By using the assumption that $\mathrm{\Delta}\subseteq E(G)$, we immediately obtain ${X}^{f}\ne \mathrm{\varnothing}$. Hence (i) holds.

For proving (ii) let $x\in {X}^{f}$. If we use weak connectivity of *G*, we have $X={[x]}_{\tilde{G}}$ and by applying (iii) we obtain the desired result. □

The next example illustrates that *f* must be orbitally *G*-continuous in order to obtain statements which are given in Theorem 3.

**Example 2**Let $X=[0,1]$ be endowed with the usual metric. Consider

Then *G* is weakly connected, ${X}^{f}$ is nonempty and *f* is a $(G,\psi )$-graphic contraction where $\psi (\omega )=\frac{\omega}{3}$, but it is not orbitally *G*-continuous. Thus, *f* does not have a fixed point.

**Remark 2** In Theorem 3, by replacing the condition that the triple $(X,d,G)$ satisfies (9) and *f* is orbitally *G*-continuous with the mapping *f* is orbitally continuous, we have the above result, too.

The following example demonstrates that the $(G,\psi )$-graphic contraction is more general than the $(G,\psi )$-contraction.

**Example 3**Let $X=[0,1]$ be endowed with the usual metric. Take

Then *G* is weakly connected and ${X}^{f}$ is nonempty and *f* is a $(G,\psi )$-graphic contraction with $\psi (\omega )=\frac{\omega}{2}$ which is not a $(G,\psi )$-contraction.

*Proof*It is clear that

*G*is weakly connected, ${X}^{f}\ne \mathrm{\varnothing}$, and with simple calculations it can be easily seen that

*f*is a $(G,\psi )$-graphic contraction. Take

which is a contradiction since $c\in [0,1)$. Thus, *f* is not $(G,\psi )$-contraction. □

**Remark 3** In Theorem 3, if we take $\psi (\omega )=\omega $, then we get Theorem 2.1, which is given in [8].

## Declarations

### Acknowledgements

The authors are grateful to the reviewers for their careful reviews and useful comments.

## Authors’ Affiliations

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