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On some fixedpoint theorems for ψcontraction on metric space involving a graph
Journal of Inequalities and Applications volume 2014, Article number: 39 (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.
1 Introduction
The metric fixedpoint 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 KnasterTarski Theorem [1].
Recently Jachymski [2] and GwóźdźLukawska and Jachymski [3] have given an interesting concept in fixedpoint 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 KeliskyRivlin 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 setvalued mappings, in [5–7] Kannan type, Reich type contractions, and φcontractions have been investigated, respectively. Some new fixedpoint 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 fixedpoint theorems. Our results generalize and unify some results by the abovementioned 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 $(V(G),E(G))$.
The conversion of a graph G is denoted by ${G}^{1}$ and this is a graph obtained from G by reversing the direction of the edges. Hence
By $\tilde{G}$ we denote the undirected graph obtained from 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
A subgraph of a graph 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 $({x}_{i1},{x}_{i})\in E(G)$ 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].
If $f:X\to X$ is an operator, then we denote by
the set of all fixed points of f.
Definition 1 [2]
A mapping $f:X\to X$ is a Banach Gcontraction or simply Gcontraction if f preserves edges of G;
for all $x,y\in X$, and f decreases weights of edges of G: for all $x,y\in X$ there exists $\alpha \in (0,1)$ such that
Definition 2 [8]
The mapping $f:X\to X$ is a Ggraphic 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)
for all $x,y\in X$;

(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]
A mapping $f:X\to X$ is called orbitally continuous if for all $x,y\in X$ and any sequence ${({k}_{n})}_{n\in \mathbf{N}}$ of positive integers,
Definition 4 [2]
A mapping $f:X\to X$ is called orbitally Gcontinuous 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 wellknown papers to obtain some fixedpoint 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 fixedpoint theorems
We establish some fixedpoint 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 $({x}_{i1},{x}_{i})\in E(\tilde{G})$ for $i=1,2,\dots ,N$. By Lemma 1, f is a $(\tilde{G},\psi )$contraction. With an easy induction, we have $({f}^{n}{x}_{i1},{f}^{n}{x}_{i})\in E(\tilde{G})$ and
for all $n\in \mathbf{N}$ and $i=1,2,\dots ,N$.
Hence using the triangle inequality, we get
So it qualifies to set $r(x,y):={\sum}_{i=1}^{N}\psi (d({x}_{i1},{x}_{i}))$. □
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 finvariant 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
for all $n\in \mathbf{N}$. Hence
and if we use a standard argument, then ${({f}^{n}x)}_{n\in \mathbf{N}}$ is obtained as a Cauchy sequence. Since also $y\in {[x]}_{\tilde{G}}$, Lemma 2 leads to $\psi (d({f}^{n}x,{f}^{n}y))\le {c}^{n}r(x,y)$. Therefore, ${({f}^{n}x)}_{n\in \mathbf{N}}$ and ${({f}^{n}y)}_{n\in \mathbf{N}}$ are equivalent. Clearly, because ${({f}^{n}x)}_{n\in \mathbf{N}}$ is a Cauchy sequence, so is ${({f}^{n}y)}_{n\in \mathbf{N}}$.

(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
and $f:X\to X$ as follows:
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
for all $n\in \mathbf{N}$, since $(x,fx)\in E(G)$. By (6), there is a subsequence ${({f}^{{k}_{n}}x)}_{n\in \mathbf{N}}$ such that $({f}^{{k}_{n}}x,{x}^{\ast})\in E(G)$ for all $n\in \mathbf{N}$. If we use (7), we conclude that $(x,fx,{f}^{2}x,\dots ,{f}^{{k}_{1}},{x}^{\ast})$ is a path in 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).
Let us define a mapping to prove (i): $\rho (x)={[x]}_{\tilde{G}}$ for all $x\in F(f)$. It is sufficient to show that $\rho :F(f)\to C=\{{[x]}_{\tilde{G}}:x\in {X}_{f}\}$ is a bijection. Because $E(G)\supseteq \mathrm{\Delta}$, we deduce $F(f)\subseteq {X}_{f}$ and then $\rho (F(f))\subseteq C$. Beside, if $x\in {X}_{f}$, then by (iv), ${lim}_{n\to \mathrm{\infty}}{f}^{n}x\in {[x]}_{\tilde{G}}\cap F(f)$, which implies $\rho ({lim}_{n\to \mathrm{\infty}}{f}^{n}x)={[x]}_{\tilde{G}}$ and so 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 fixedpoint 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 selfmapping 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 $({x}_{i1},{x}_{i})\in E(\tilde{G})$ for $i=1,2,\dots ,N$. Since f is a $(G,\psi )$graphic contraction we get $(f{x}_{i1},f{x}_{i})\in E(\tilde{G})$ 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.
Take $(x,y)\in E({\tilde{G}}_{{x}_{0}})$; then there is a path ${({x}_{i})}_{i=0}^{N}$ in $\tilde{G}$ from ${x}_{0}$ to y such that ${x}_{N1}=x$. 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 Gcontinuous. 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 Gcontinuous, 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 Gcontinuous in order to obtain statements which are given in Theorem 3.
Example 2 Let $X=[0,1]$ be endowed with the usual metric. Consider
and $f:X\to X$,
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 Gcontinuous. 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 Gcontinuous 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
and $f:X\to X$ as follows:
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].
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Öztürk, M., Girgin, E. On some fixedpoint theorems for ψcontraction on metric space involving a graph. J Inequal Appl 2014, 39 (2014). https://doi.org/10.1186/1029242X201439
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Keywords
 connected graph
 fixed point
 metric space
 ψtype contraction