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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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DOI: https://doi.org/10.1186/1029242X201439
Keywords
 connected graph
 fixed point
 metric space
 ψtype contraction