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

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 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 $(V(G),E(G))$.

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

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

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

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

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 $(x_{i-1},x_i)\in E(G)$ 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

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

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

$$F(f)=\{x\in X:x=fx\}$$

the set of all fixed points of f.

Definition 1 [2]

A mapping $f:X\to X$ is a Banach G-contraction or simply G-contraction if f preserves edges of G;

$$(x,y)\in E(G)\Rightarrow (fx,fy)\in E(G),$$

(1)

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

$$(x,y)\in E(G)\Rightarrow d(fx,fy)<\le \alpha d(x,y).$$

(2)

Definition 2 [8]

The mapping $f:X\to X$ is a G-graphic contraction

• (i) if f preserves edges of G;

  $$(x,y)\in E(G)\Rightarrow (fx,fy)\in E(G),$$

  (3)

for all $x,y\in X$;

• (ii) there exists $\alpha \in (0,1)$ such that

  $$(x,y)\in E(G)\Rightarrow d(fx,f^{2}x)\le \alpha d(x,fx),$$

  (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,

$$f^{k_n}x\to y\text{implies}f\left(f^{k_n}x\right)\to fy\text{as }n\to \mathrm{\infty }.$$

Definition 4 [2]

A mapping $f:X\to X$ is called orbitally G-continuous if for all $x,y\in X$ and any sequence ${(k_n)}_{n\in \mathbf{N}}$ of positive integers,

$$f^{k_n}x\to y,(f^{k_n}x,f^{k_{n+1}}x)\in E(G)\text{imply}f\left(f^{k_n}x\right)\to fy\text{as }n\to \mathrm{\infty }.$$

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

  $$(x,y)\in E(G)\Rightarrow \psi (d(fx,fy))\le c\psi (d(x,y)),$$

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

$$\psi \left(d(f^{n}x,f^{n}y)\right)\le c^{n}r(x,y).$$

(5)

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

$$\begin{array}{rl}\psi \left(d(f^n{x}_{i-1},f^n{x}_i)\right)& \le c\psi \left(d(f^{n-1}{x}_{i-1},f^{n-1}{x}_i)\right)\\ \le c\left(c\psi \left(d(f^{n-2}{x}_{i-1},f^{n-2}{x}_i)\right)\right)\le \cdots \le c^n\psi (d(x_{i-1},x_i))\end{array}$$

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

Hence using the triangle inequality, we get

$$\psi \left(d(f^{n}x,f^{n}y)\right)\le \sum _{i=1}^N\psi \left(d(f^n{x}_{i-1},f^n{x}_i)\right)\le c^n\sum _{i=1}^N\psi (d(x_{i-1},x_i)).$$

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

$$\psi \left(d(f^{n}x,f^{n+1}x)\right)\le c^{n}r(x,fx)$$

for all $n\in \mathbf{N}$. Hence

$$\sum _{n=0}^{\mathrm{\infty }}\psi \left(d(f^{n}x,f^{n+1}x)\right)<\mathrm{\infty }$$

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

$$E(G)=\{(0,0)\}\cup \{(0,1)\}\cup \{(x,y)\in (0,1]\times (0,1]:x\ge y\},$$

and $f:X\to X$ as follows:

$$fx=\{\begin{array}{ll}\frac{x}{3},& \text{if }x\in (0,1],\\ \frac{1}{2},& \text{if }x=0.\end{array}$$

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:

$$\begin{array}{r}\mathit{\text{for any }}{(x_n)}_{n\in \mathbf{N}}\mathit{\text{ in }}X,\mathit{\text{ if }}{x}_n\to x\mathit{\text{ and }}(x_n,x_{n+1})\in E(G)\mathit{\text{ for }}n\in \mathbf{N},\\ \mathit{\text{then there is a subsequence }}{(x_{k_n})}_{n\in \mathbf{N}}\mathit{\text{ with }}(x_{k_n},x)\in E(G)\mathit{\text{ for }}n\in \mathbf{N}.\end{array}$$

(6)

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

$$(f^{n}x,f^{n+1}x)\in E(G)$$

(7)

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

$$\psi \left(d(f^{k_{n+1}}x,f{x}^{\ast })\right)\le c\psi \left(d(f^{k_n}x,x^{\ast })\right),$$

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),

$$\underset{n\to \mathrm{\infty }}{lim}{f}^n{x}_2\in {[x_1]}_{\tilde{G}}\cap F(f)=\{x_1\},$$

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

$$\psi \left(d(f^{n}x,f^{n+1}x)\right)\le c^{n}r(x),$$

(8)

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

$$\psi \left(d(f^{n}x,f^{n+1}x)\right)\le c^{n}r(x),$$

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.

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 $x_{N-1}=x$. Also let ${(y_i)}_{i=0}^M$ be a path in $\tilde{G}$ from $x_0$ to $f{x}_0$. Then we realize

$$(y_0,y_1,\dots ,y_M,f{x}_1,f{x}_2,\dots ,f{x}_{N-1}=fx,f{x}_N=fy)$$

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:

$$\begin{array}{r}\mathit{\text{for any }}{(x_n)}_{n\in \mathbf{N}}\mathit{\text{ in }}X,\mathit{\text{ if }}{x}_n\to x\mathit{\text{ and }}(x_n,x_{n+1})\in E(G)\\ (\mathit{\text{or}},\mathit{\text{respectively}},(x_{n+1},x_n)\in E(G))\\ \mathit{\text{ for all }}n\in \mathbf{N},\mathit{\text{ then there is a subsequence }}{(x_{k_n})}_{n\in \mathbf{N}}\mathit{\text{ with }}(x_{k_n},x)\in E(G)\\ (\mathit{\text{or}},\mathit{\text{respectively}},(x,x_{k_n})\in E(G))\mathit{\text{ for all }}n\in \mathbf{N}.\end{array}$$

(9)

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

$$E(G)=\{(0,0)\}\cup \{(0,x):x\ge 1/2\}\cup \{(x,y):x,y\in (0,1]\},$$

and $f:X\to X$,

$$fx=\{\begin{array}{ll}\frac{x}{2},& \text{if }x\in (0,1];\\ \frac{1}{2},& \text{if }x=0.\end{array}$$

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

$$E(G)=\{(0,0)\}\cup \{(0,1)\}\cup \{(x,y)\in (0,1]\times (0,1]:x\ge y\},$$

and $f:X\to X$ as follows:

$$fx=\{\begin{array}{ll}\frac{x}{2},& \text{if }x\in (0,1],\\ \frac{3}{4},& \text{if }x=0.\end{array}$$

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

$$\psi \left(d(f0,f\frac{1}{2})\right)\le c\psi \left(d(0,\frac{1}{2})\right)\Rightarrow \frac{1}{4}\le c\frac{1}{4},$$

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].