Fixed point theorems for $\alpha$-integral type G-contraction mappings

Abstract

In this paper, we introduce the notion of $\alpha$-integral type G-contraction mappings to generalize the notions of Banach G-contraction and integral G-contraction mappings. We also prove some fixed point theorems for $\alpha$-integral type G-contraction mappings. By providing some example, we show that our results are real generalization of several results in literature.

1 Introduction and preliminaries

Development of fixed point theory on a metric space endowed with graph has a lot of activities in last few year. Jachymski [1] introduced the notion of Banach G-contraction. Later on various authors proved many fixed point theorems for single-valued and multi-valued mappings on a metric space endowed with a graph, see, for example [2–10]. Recently, Asl et al. [3] defined a graph-metric space and proved fixed point theorems on it. In this paper we introduce the notion of $\alpha$-integral type G-contraction to generalize the notions of Banach G-contraction and integral G-contraction. We also prove the fixed point theorems for such mappings and state some illustrative examples to claim that our results properly generalizes some results in literature.

Let (X, d) be a metric space and G be an undirected graph such that the set V(G) of its vertices coincides with X and the set E(G) of its edges contains all loops in V(G). Throughout this paper, we assume that G has no parallel edges. We also denote this space by $G_d$ and call it a graph-metric space [3]. A mapping $T:G_d\rightarrow G_d$ is said to be G-continuous if for given sequence $\{x_{n}\}$ such that $x_{n}\rightarrow x$ as $n\rightarrow \infty$, where $x\in G_d$ and $(x_{n},x_{n+1})\in E(G)$ for all $n \in {\mathbb {N}}$, we have $Tx_{n}\rightarrow Tx$ as $n\rightarrow \infty$.

In 2008, Jachymski [1] introduced the notion of Banach G-contraction mappings as follows:

Definition 1.1

[1] Let (X, d) be a metric space endowed with graph G. A mapping $T:X\rightarrow X$ is called a Banach G-contraction if T preserves the edges of G, i.e.,

$$\begin{aligned} \forall x,y\in X\quad ((x,y)\in E(G) \Rightarrow (Tx,Ty)\in E(G)) \end{aligned}$$

(1)

and

$$\begin{aligned} \exists c\in (0,1)\ \forall x,y\in X\quad ((x,y)\in E(G) \Rightarrow d(Tx,Ty)\le cd(x,y)). \end{aligned}$$

We denote by $\Phi$ the set of all Lebesgue integrable mappings $\phi :[0,\infty )\rightarrow [0,\infty )$ which are summable on each compact subset of $[0,\infty )$ and for each $\epsilon >0$, we have

$$\begin{aligned} \int _{0}^{\epsilon }\phi (t)~dt>0. \end{aligned}$$

(2)

In 2013, Samreen and Kamran [9] extended the notion of Banach G-contraction in the following way:

Definition 1.2

[9] Let (X, d) be a metric space endowed with graph G. A mapping $T:X\rightarrow X$ is called an integral G-contraction if T preserves the edges of G, i.e.,

$$\begin{aligned} \forall x,y\in X\quad ((x,y)\in E(G) \Rightarrow (Tx,Ty)\in E(G)) \end{aligned}$$

(3)

and

$$\begin{aligned} \forall x,y\in X\quad \left( (x,y)\in E(G) \Rightarrow \int _{0}^{d(Tx,Ty)}\phi (t)~dt\le c\int _{0}^{d(x,y)}\phi (t)~dt\right) , \end{aligned}$$

where $c\in (0,1)$ and $\phi \in \Phi$.

Recently, Kamran and Ali [11] introduced the notion of $\alpha$-subadmissible mappings in the following way:

Definition 1.3

Let $\alpha :G_d\times G_d\rightarrow [0,\infty )$ be a mapping. A mapping $T:G_d\rightarrow G_d$ is said to be $\alpha$-subadmissible if

• (i) for $x,y\in G_d$, $\alpha (x,y)\ge 1\Rightarrow \alpha (Tx,Ty)\ge 1$;

• (ii) for $x\in G_d$, $\alpha (T^{n}x,T^{n+1}x)\ge 1$ for all integers $n\ge 0$ implies $\alpha (T^{m}x,T^{n}x)\ge 1$ for all integers $m > n\ge 0$.

By $\Xi$, we mean the family of functions $\xi :[0,\infty )\rightarrow [0,\infty )$ such that $\xi$ is nondecreasing, upper semicontinuous and $\lim _{n\rightarrow \infty }\xi ^{n}(t)=0$ for each $t\ge 0$. Note that if $\xi \in \Xi$, then $\xi (t)<t$ for each $t>0$, and $\xi (0)=0$.

2 Main results

We begin this section with the following definition.

Definition 2.1

A mapping $T:G_d\rightarrow G_d$ is said to be an $\alpha$-integral type G-contraction mapping if there exist three functions $\alpha :G_d\times G_d\rightarrow [0,\infty )$, $\phi \in \Phi$ and $\xi \in \Xi$ such that

$$\begin{aligned} \alpha (x,y)\int _{0}^{d(Tx,Ty)}\phi (t)~dt\le \xi \left( \int _{0}^{d(x,y)}\phi (t)~dt \right) , \end{aligned}$$

(4)

for each $(x,y)\in E(G)$.

Remark 2.2

Let (X, d) be a metric space. Define graph G by $V(G)=X$ and $E(G)=X\times X$, $\alpha (x,y)=1$ for each $x,y \in X$ and $\xi (t)=ct$ for all $t\ge 0$, where $c\in [0,1)$. Then (4) reduces to

$$\begin{aligned} \int _{0}^{d(Tx,Ty)}\phi (t)~dt\le c\int _{0}^{d(x,y)}\phi (t)~dt, \end{aligned}$$

for each $x,y\in X$, where $\phi \in \Phi$, which is the contractive condition considered by Branciari [12].

Note that every integral G-contraction mapping is an $\alpha$-integral type G-contraction mapping. The following example shows that the converse is not true in general.

Example 2.3

Let $X=[0,\infty )$ with the usual metric d and X endowed with graph G is defined by $V(G)=X$ and $E(G)=\{(x,y):x\ge y\}$. Define $T:G_d\rightarrow G_d$ and $\alpha :G_d\times G_d\rightarrow [0,\infty )$ by

$$\begin{aligned} Tx= {\left\{ \begin{array}{ll} \frac{x}{4}&{}\text{ if } x\ge 4, \\ x^{2}&{}\text{ if } 0\le x<4, \end{array}\right. } \end{aligned}$$

(5)

and

$$\begin{aligned} \alpha (x,y)={\left\{ \begin{array}{ll} 1&{}\text{ if } x,y\ge 4,\\ 0&{}\text{ otherwise }. \end{array}\right. } \end{aligned}$$

(6)

Take $\xi (t)=\frac{t}{2}$ for each $t\ge 0$ and $\phi (t)={\left\{ \begin{array}{ll} 0&{}\text{ if } t=0,\\ \frac{t^{-1/2}}{2}&{}\text{ if } t>0.\end{array}\right. }$ If $x,y\ge 4$, then we have

$$\begin{aligned} \alpha (x,y)\int _{0}^{d(Tx,Ty)}\phi (t)~dt=\sqrt{\frac{|x-y|}{4}}= \frac{1}{2}\int _{0}^{d(x,y)}\phi (t)~dt, \end{aligned}$$

and for otherwise, we have

$$\begin{aligned} \alpha (x,y)\int _{0}^{d(Tx,Ty)}\phi (t)~dt=0\le \frac{1}{2} \int _{0}^{d(x,y)}\phi (t)~dt. \end{aligned}$$

Thus, (4) holds for each $(x,y)\in E(G)$. Therefore, T is an $\alpha$-integral type G-contraction mapping. But T is not an integral G-contraction, since $(4,2)\in E(G)\nRightarrow (T4,T2)\in E(G)$.

Theorem 2.4

Let $G_d$ be a complete graph-metric space and $T:G_d\rightarrow G_d$ be an $\alpha$-integral type G-contraction mapping satisfying the following assumptions : 

• (i) T is $\alpha$-subadmissible; 

• (ii) there exists $x_{0}\in G_d$ such that $\alpha (x_{0},Tx_{0})\ge 1;$

• (iii) if $x,y\in G_d$ and $\alpha (x,y)\ge 1,$ then $(x,y)\in E(G);$

• (iv) T is G-continuous; 

Then T has a fixed point.

Proof

Starting from $x_{0}\in G_d$ in (ii). Define a sequence $\{x_{n}\}$ in $G_d$ such that $x_{n+1}=Tx_{n}$ for each $n\in {\mathbb {N}}\cup \{0\}$. Since T is $\alpha$-subadmissible, by induction we have

$$\begin{aligned} \alpha (x_{n},x_{n+1})\ge 1\quad \text {for each }n\in {\mathbb {N}}\cup \{0\}. \end{aligned}$$

(7)

From (4), we have

$$\begin{aligned} \int _{0}^{d(x_{n},x_{n+1})}\phi (t)~dt= & {} \int _{0}^{d(Tx_{n-1},Tx_{n})}\phi (t)~dt\nonumber \\\le & {} \alpha (x_{n-1},x_{n})\int _{0}^{d(Tx_{n-1},Tx_{n})}\phi (t)~dt\nonumber \\\le & {} \xi \left( \int _{0}^{d(x_{n-1},x_{n})}\phi (t)~dt\right) , \end{aligned}$$

for all $n\in {\mathbb {N}}$. By induction, we have

$$\begin{aligned} \int _{0}^{d(x_{n},x_{n+1})}\phi (t)~dt\le \xi ^{n} \left( \int _{0}^{d(x_{0},x_{1})}\phi (t)~dt\right) , \end{aligned}$$

(8)

for all $n\in {\mathbb {N}}$. Letting $n\rightarrow \infty$ in (8), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{0}^{d(x_{n},x_{n+1})}\phi (t)~dt=0. \end{aligned}$$

(9)

From (2) and (9), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }d(x_{n},x_{n+1})=0. \end{aligned}$$

(10)

Now, using a similar argument as given by authors in [13, 14], we will show that $\{x_{n}\}$ is a Cauchy sequence in $G_d$. Assume on contrary that $\{x_{n}\}$ is not a Cauchy sequence. Then there exists $\epsilon >0$ for which we can find two sequences of positive integers $\{n_{k}\}$ and $\{m_{k}\}$ such that for each $k\in {\mathbb {N}}$, we have

$$\begin{aligned} n_{k}>m_{k}>k,\quad d(x_{n_{k}},x_{m_{k}})\ge \epsilon ,\quad d(x_{n_{k}-1},x_{m_{k}})< \epsilon . \end{aligned}$$

(11)

By using triangular inequality and (11), we have

$$\begin{aligned} \epsilon \le d(x_{n_{k}},x_{m_{k}})\le & {} d(x_{n_{k}},x_{n_{k}-1})+ d(x_{n_{k}-1},x_{m_{k}}) \nonumber \\< & {} d(x_{n_{k}},x_{n_{k}-1})+\epsilon . \end{aligned}$$

Letting $k\rightarrow \infty$ in above inequality and using (10), we get

$$\begin{aligned} \lim _{k\rightarrow \infty }d(x_{n_{k}},x_{m_{k}})=\epsilon . \end{aligned}$$

(12)

Again by using triangular inequality, we have

$$\begin{aligned} d(x_{n_{k}-1},x_{m_{k}-1}) \le d(x_{n_{k}-1},x_{n_{k}})+ d(x_{n_{k}},x_{m_{k}})+d(x_{m_{k}},x_{m_{k}-1}) \end{aligned}$$

and

$$\begin{aligned} d(x_{n_{k}},x_{m_{k}}) \le d(x_{n_{k}},x_{n_{k}-1})+ d(x_{n_{k}-1},x_{m_{k}-1})+d(x_{m_{k}-1},x_{m_{k}}) \end{aligned}$$

for all $k\in {\mathbb {N}}$. Using (10) and (12) in above two inequalities, we get

$$\begin{aligned} \lim _{k\rightarrow \infty }d(x_{n_{k}-1},x_{m_{k}-1})=\epsilon . \end{aligned}$$

(13)

As T is $\alpha$-subadmissible, from (7), we have $\alpha (x_{n_{k}-1},x_{m_{k}-1})\ge 1$ for all $k\in {\mathbb {N}}$. Then by Condition (iii) we get $(x_{n_{k}-1},x_{m_{k}-1})\in E(G)$ for all $k\in {\mathbb {N}}$. From (4), we have

$$\begin{aligned} \int _{0}^{d(x_{n_{k}},x_{m_{k}})}\phi (t)~dt= & {} \int _{0}^{d(Tx_{n_{k}-1},Tx_{m_{k}-1})}\phi (t)~dt\nonumber \\\le & {} \alpha (x_{n_{k}-1},x_{m_{k}-1})\int _{0}^{d(Tx_{n_{k}-1}, Tx_{m_{k}-1})}\phi (t)~dt\nonumber \\\le & {} \xi \left( \int _{0}^{d(x_{n_{k}-1},x_{m_{k}-1})} \phi (t)~dt\right) \end{aligned}$$

(14)

for all $k\in {\mathbb {N}}$. Using properties of $\xi$ and let $k\rightarrow \infty$ in (14), we have

$$\begin{aligned} \int _{0}^{\epsilon }\phi (t)~dt\le \xi \left( \int _{0}^{\epsilon }\phi (t)~dt\right) < \int _{0}^{\epsilon }\phi (t)~dt. \end{aligned}$$

A contradiction to our assumption. Hence $\{x_{n}\}$ is a Cauchy sequence in $G_d$. Since $G_d$ is complete, there exists $x^{*}\in G_d$ such that $x_{n}\rightarrow x^{*}$ as $n\rightarrow \infty$. G-continuity of T implies $x_{n+1}=Tx_{n}\rightarrow Tx^{*}$ as $n\rightarrow \infty$. By uniqueness of limit, we have $Tx^{*}=x^{*}$. $\square$

Theorem 2.5

Let $G_d$ be a complete graph-metric space and $T:G_d\rightarrow G_d$ be an $\alpha$-integral type G-contraction mapping satisfying the following assumptions : 

• (i) T is $\alpha$-subadmissible; 

• (ii) there exists $x_{0}\in G_d$ such that $\alpha (x_{0},Tx_{0})\ge 1;$

• (iii) if $x,y\in G_d$ and $\alpha (x,y)\ge 1,$ then $(x,y)\in E(G);$

• (iv) if $\{x_{n}\}$ is a sequence in $G_d$ such that $x_{n}\rightarrow x$ as $n\rightarrow \infty$ and $\alpha (x_{n},x_{n+1})\ge 1$ for each $n\in \mathbb {N}\cup \{0\}$, then $\alpha (x_{n},x)\ge 1$ for each $n\in \mathbb {N}\cup \{0\}$.

Then T has a fixed point.

Proof

According to the proof of Theorem 2.4, we know that $\{x_{n}\}$ be a Cauchy sequence in $G_d$. Since $G_d$ is complete, there exists $x^{*}\in G_d$ such that $x_{n}\rightarrow x^{*}$ as $n\rightarrow \infty$. From (7) and Conditions (iv) and (iii), we have

$$\begin{aligned} \alpha (x_{n},x^{*})\ge 1 \text { and }(x_{n},x^{*})\in E(G) \quad \text {for each }n\in {\mathbb {N}}\cup \{0\}. \end{aligned}$$

(15)

By the triangular inequality, we have

$$\begin{aligned} |d(Tx^{*},x^{*})-d(Tx_{n},x^{*})|\le d(Tx^{*},Tx_{n}) \end{aligned}$$

(16)

for all $n\in {\mathbb {N}} \cup \{0\}$. Thus, we have

$$\begin{aligned} \int _{0}^{|d(Tx^{*},x^{*})-d(Tx_{n},x^{*})|}\phi (t)~dt\le & {} \int _{0}^{d(Tx^{*},Tx_{n})}\phi (t)~dt\nonumber \\\le & {} \alpha (x_{n},x^{*})\int _{0}^{d(Tx^{*},Tx_{n})}\phi (t)~dt\nonumber \\\le & {} \xi \left( \int _{0}^{d(x^{*},x_{n})}\phi (t)~dt\right) \end{aligned}$$

(17)

for all $n\in {\mathbb {N}} \cup \{0\}$. Letting $n\rightarrow \infty$ in (17), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\int _{0}^{|d(Tx^{*},x^{*})-d(Tx_{n},x^{*})|}\phi (t)~dt=0. \end{aligned}$$

(18)

From (18), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }|d(Tx^{*},x^{*})-d(Tx_{n},x^{*})|=0. \end{aligned}$$

(19)

This implies that

$$\begin{aligned} d(Tx^{*},x^{*})=\lim _{n\rightarrow \infty }d(Tx_{n},x^{*})= \lim _{n\rightarrow \infty }d(x_{n+1},x^{*})=0. \end{aligned}$$

Hence $Tx^{*} = x^{*}$. $\square$

We use the following condition to obtain the uniqueness of the fixed point.

(A) For each fixed points x and y of T, we have $\alpha (x,y)\ge 1$.

Theorem 2.6

Adding the Condition $(\mathrm{A})$ to the hypotheses of Theorem 2.4 (resp. Theorem 2.5), we get the uniqueness of the fixed point of T.

Proof

Suppose that $x^{*}$ and $y^{*}$ are two distinct fixed points of T. From Condition (A), we get $\alpha (x^*,y^*)\ge 1$ and hence

$$\begin{aligned} \int _{0}^{d(x^{*},y^{*})}\phi (t)~dt\le & {} \alpha (x^{*},y^{*})\int _{0}^{d(Tx^{*},Ty^{*})}\phi (t)~dt\nonumber \\\le & {} \xi \left( \int _{0}^{d(x^{*},y^{*})}\phi (t)~dt\right) \nonumber \\< & {} \int _{0}^{d(x^{*},y^{*})}\phi (t)~dt. \end{aligned}$$

(20)

A contradiction to our assumption. Hence T has unique fixed point. $\square$

Example 2.7

Let $X={\mathbb {R}}$ with the usual metric d and X endowed with the graph G such that $V(G)=X$ and $E(G)=\{(x,y): x,y\ge -1\}$. Define $T:G_d\rightarrow G_d$ and $\alpha :G_d\times G_d\rightarrow [0,\infty )$ by

$$\begin{aligned} Tx= {\left\{ \begin{array}{ll} \frac{x}{2}&{}\text{ if } x\ge 0,\\ x&{}\text{ if } x<0, \end{array}\right. } \end{aligned}$$

(21)

and

$$\begin{aligned} \alpha (x,y)={\left\{ \begin{array}{ll} 1&{}\text{ if } x,y\ge 0,\\ 0&{}\text{ otherwise }.\end{array}\right. } \end{aligned}$$

(22)

Take $\xi (t)=\frac{t}{4}$ and $\phi (t)=t$ for each $t\ge 0$. If $x,y\ge 0$, then we have

$$\begin{aligned} \alpha (x,y)\int _{0}^{d(Tx,Ty)}\phi (t)~dt=\frac{1}{2} \left( \frac{|x-y|}{2}\right) ^{2}= \frac{1}{4}\int _{0}^{d(x,y)}\phi (t)~dt=\xi \left( \int _{0}^{d(x,y)}\phi (t)~dt\right) , \end{aligned}$$

and for otherwise, we have

$$\begin{aligned} \alpha (x,y)\int _{0}^{d(Tx,Ty)}\phi (t)~dt=0\le \frac{1}{4} \int _{0}^{d(x,y)}\phi (t)~dt=\xi \left( \int _{0}^{d(x,y)}\phi (t)~dt\right) . \end{aligned}$$

This shows that (4) holds for each $(x,y)\in E(G)$. Therefore, T is an $\alpha$-integral type G-contraction mapping. If $x,y\in G_d$ and $\alpha (x,y)=1$, then $x,y\ge 0$. By definition of T and $\alpha$, we have $\alpha (Tx,Ty)=1$. For each $x\in G_d$ such that $\alpha (T^{n}x,T^{n+1}x) \ge 1$ for all integers $n\ge 0$, we have $\alpha (T^{m}x,T^{n}x)=1$ for all integers $m > n\ge 0$. Hence T is $\alpha$-subadmissible. Also, we have $x_{0}=1\in G_d$ such that $\alpha (1,T1)=\alpha (1,1/2)=1$. Moreover, if $x,y\in G_d$ such that $\alpha (x,y) \ge 1$, then $x,y\ge 0$ which implies that $(x,y)\in E(G)$. Finally, it is easy to see that the continuity of T implies that T is G-continuous. Therefore, all the hypotheses of Theorem 2.4 hold. Hence T has a fixed point, that is, a point 0.

Note that T is neither a Banach G-contraction mapping nor an integral G-contraction mapping. Indeed, if we take $(x,y) = (-1,-1/2)\in E(G)$, we get

$$\begin{aligned} d(Tx,Ty)=d(T(-1),T(-1/2)) > cd(-1,-1/2)=cd(x,y) \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{d(Tx,Ty)}\phi (t)~dt = \int _{0}^{d(T(-1),T(-1/2))}\phi (t)~dt > c\int _{0}^{d(-1,-1/2)}\phi (t)~dt = c\int _{0}^{d(x,y)}\!\phi (t)~dt \end{aligned}$$

for all $c\in (0,1)$ and $\phi \in \Phi$.