Open Access

Fixed point theorems for F-expanding mappings

Fixed Point Theory and Applications20172017:9

https://doi.org/10.1186/s13663-017-0602-3

Received: 24 October 2016

Accepted: 10 May 2017

Published: 19 May 2017

Abstract

Recently, Wardowski (Fixed Point Theory Appl. 2012:94, 2012) introduced a new concept of F-contraction and proved a fixed point theorem which generalizes the Banach contraction principle. Following this direction of research, in this paper, we present some new fixed point results for F-expanding mappings, especially on a complete G-metric space.

Keywords

fixed point F-contraction map F-expanding map G-metric space

MSC

47H1054H25

1 Introduction

Let \((X,d)\) be a metric space. A mapping \(T:X\rightarrow X\) is said to be expanding if
$$ \forall_{x,y\in X}\quad d(Tx,Ty)\geqslant \lambda d(x,y),\quad \mbox{where}\quad \lambda >1. $$
(1)

The condition \(\lambda >1\) is important, the function \(T:\mathbb{R} \rightarrow \mathbb{R}\) defined by \(Tx=x+e^{x}\) satisfies the condition \(\vert Tx-Ty \vert \geqslant \vert x-y \vert \) for all \(x,y\in \mathbb{R}\), and T has no fixed point.

For an expanding map, the following result is well known.

Theorem 1.1

Let \((X,d)\) be a complete metric space, and let \(T:X\rightarrow X\) be surjective and expanding. Then T is bijective and has a unique fixed point.

It follows from the Banach contraction principle and the following very simple observation.

Lemma 1.2

If \(T:X\rightarrow X\) is surjective, then there exists a mapping \(T^{*}:X\rightarrow X\) such that \(T\circ T^{*}\) is the identity map on X.

Proof

For any point \(x\in X\), let \(y_{x}\in X\) be any point such that \(Ty_{x}=x\). Let \(T^{*}x=y_{x}\) for all \(x\in X\). Then \((T\circ T^{*})(x)=T(T^{*}x)=Ty_{x}=x\) for all \(x\in X\). □

In the present paper, we introduce a new type of expanding mappings.

Definition 1.3

Let \(\mathcal{F}\) be the family of all function \(F:(0,+\infty )\rightarrow \mathbb{R}\) such that
(F1): 

F is strictly increasing, i.e., for all \(\alpha , \beta \in (0,+\infty )\), if \(\alpha <\beta \), then \(F(\alpha )< F( \beta )\);

(F2): 
for each sequence \(\{\alpha_{n}\}\subset (0,+\infty )\), the following holds:
$$ \lim_{n\rightarrow \infty }\alpha_{n}=0\quad \mbox{if and only if} \quad \lim_{n\rightarrow \infty }F(\alpha_{n})=- \infty ; $$
(F3): 

there exists \(k\in (0,1)\) such that \(\lim_{\alpha \rightarrow 0_{+}}\alpha^{k}F(\alpha )=0\).

Definition 1.4

Let \((X,d)\) be a metric space. A mapping \(T:X\rightarrow X\) is called F-expanding if there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y\in X\),
$$ d(x,y)>0\quad \Rightarrow \quad F\bigl(d(Tx,Ty)\bigr)\geqslant F\bigl(d(x,y) \bigr)+t. $$
(2)

When we consider in (2) the different types of the mapping \(F\in \mathcal{F}\), then we obtain a variety of expanding mappings.

Example 1.5

Let \(F_{1}(\alpha )=\ln \alpha \). It is clear that \(F_{1}\) satisfies (F1), (F2), (F3) for any \(k\in (0,1)\). Each mapping \(T:X\rightarrow X\) satisfying (2) is an \(F_{1}\)-expanding map such that
$$ d(Tx,Ty)\geqslant e^{t}d(x,y)\quad\mbox{for all}\quad x,y\in X,~d(x,y)>0. $$
It is clear that for \(x,y\in X\) such that \(x=y\), the inequality \(d(Tx,Ty)\geqslant e^{t}d(x,y)\) also holds.

Example 1.6

If \(F_{2}(\alpha )=\ln \alpha +\alpha \), \(\alpha >0\), then \(F_{1}\) satisfies (F1), (F2) and (F3), and condition (2) is of the form
$$ d(Tx,Ty)e^{d(Tx,Ty)-d(x,y)}\geqslant e^{t}d(x,y)\quad\mbox{for all}\quad x,y \in X. $$

Example 1.7

Consider \(F_{3}(\alpha )=\ln (\alpha^{2}+\alpha )\), \(\alpha >0\). \(F_{3}\) satisfies (F1), (F2) and (F3), and for \(F_{3}\)-expanding T, the following condition holds:
$$ d(Tx,Ty)\cdot \frac{d(Tx,Ty)+1}{d(x,y)+1}\geqslant e^{t}d(x,y) \quad\mbox{for all}\quad x,y\in X. $$

Example 1.8

Consider \(F_{4}(\alpha )=\arctan (-\frac{1}{ \alpha })\), \(\alpha >0\). \(F_{4}\) satisfies (F1), (F2) and (F3), and for \(F_{4}\)-expanding T, the following condition holds:
$$ d(Tx,Ty)\geqslant \biggl[\frac{1+\frac{\tan t }{d(x,y)}}{1-\tan t\cdot d(x,y)} \biggr]d(x,y)\quad\mbox{for some}\quad 0< t< \frac{\pi }{2}. $$
Here, we have obtained a special type of nonlinear expanding map \(d(Tx,Ty)\geqslant \varphi (d(x, y))d(x,y)\).

Other functions belonging to \(\mathcal{F}\) are, for example, \(F(\alpha )=\ln (\alpha^{n})\), \(n\in \mathbb{N}\), \(\alpha >0\); \(F(\alpha )=\ln (\arctan \alpha )\), \(\alpha >0\).

Now we recall the following.

Definition 1.9

Let \((X,d)\) be a metric space. A mapping \(T:X\rightarrow X\) is an F-contraction on X if there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y\in X\),
$$ d(Tx,Ty)>0\quad \Rightarrow \quad t+F\bigl(d(Tx,Ty)\bigr)\leqslant F\bigl(d(x,y) \bigr). $$
(3)

For such mappings, Wardowski [1] proved the following theorem.

Theorem 1.10

Let \((X,d)\) be a complete metric space and \(T:X\rightarrow X\) be an F-contraction. Then T has a unique fixed point \(u\in X\) and for every \(x\in X\), a sequence \(\{x_{n}=T ^{n}x\}\) is convergent to u.

2 The result

In this section, we give some fixed point theorem for F-expanding maps.

Theorem 2.1

Let \((X,d)\) be a complete metric space and \(T:X\rightarrow X\) be surjective and F-expanding. Then T has a unique fixed point.

Proof

From Lemma 1.2, there exists a mapping \(T^{*}:X\rightarrow X\) such that \(T\circ T^{*}\) is the identity mapping on X. Let \(x,y\in X\) be arbitrary points such that \(x\neq y\), and let \(z=T^{*}x\) and \(w=T^{*}y\) (obviously, \(z\neq w\)). By using (2) applied to z and w, we have
$$ F\bigl(d(Tz,Tw)\bigr)\geqslant F\bigl(d(z,w)\bigr)+t. $$
Since \(Tz=T(T^{*}x)=x\) and \(Tw=T(T^{*}y)=y\), then
$$ F\bigl(d(x,y)\bigr)\geqslant F\bigl(d\bigl(T^{*}x,T^{*}y \bigr)\bigr)+t, $$
so \(T^{*}:X\rightarrow X\) is an F-contraction. By Theorem 1.10, \(T^{*}\) has a unique fixed point \(u\in X\). In particular, u is also a fixed point of T because \(T^{*}u=u\) implies that \(Tu=T(T^{*}u)=u\).
Let us observe that T has at most one fixed point. If \(u,v\in X\) and \(Tu=u\neq v=Tv\), then we would get the contradiction
$$\begin{aligned}& F\bigl(d(Tu,Tv)\bigr)\geqslant F\bigl(d(u,v)\bigr)+t, \\& 0=F\bigl(d(Tu,Tv)\bigr)-F\bigl(d(u,v)\bigr)\geqslant t>0, \end{aligned}$$
so the fixed point of T is unique. □

Remark 2.2

If T is not surjective, the previous result is false. For example, let \(X=[0,\infty )\) endowed with the metric \(d(x,y)=\vert x-y \vert \) for all \(x,y\in X\), and let \(T:X\rightarrow X\) be defined by \(Tx=2x+1\) for all \(x\in X\). Then T satisfies the condition \(d(Tx,Ty)\geqslant 2d(x,y)\) for all \(x,y\in X\) and T is fixed point free.

3 Applications to G-metric spaces

In 2006 Mustafa and Sims (see [2] and the references therein) introduced the notion of a G-metric space and investigated the topology of such spaces. The G-metric space is as follows.

Definition 3.1

Let X be a nonempty set. A function \(G:X\times X\times X\rightarrow [0,\infty )\) satisfying the following axioms:
(\(G_{1}\)): 

\(G(x,y,z)=0\) if \(x=y=z\),

(\(G_{2}\)): 

\(G(x,x,y)>0\) for all \(x,y\in X\) with \(x\neq y\),

(\(G_{3}\)): 

\(G(x,x,y)\leqslant G(x,y,z)\) for all \(x,y,z\in X\) with \(z\neq y\),

(\(G_{4}\)): 

\(G(x,y,z)=G(x,z,y)=G(y,z,x)=\cdots \) (symmetry in all three variables),

(\(G_{5}\)): 

\(G(x,y,z)\leqslant G(x,a,a)+G(a,y,z)\) for all \(x,y,z,a\in X\),

is called a G-metric on X, and the pair \((X,G)\) is called a G-metric space.

Recently, Samet et al. [3] observed that some fixed point theorems in the context of G-metric spaces can be concluded from existence results in the setting of quasi-metric spaces. Especially, the following theorem is a simple consequence of Theorem 1.10.

Theorem 3.2

Let \((X,G)\) be a complete G-metric space, and let \(T:X\rightarrow X\) satisfy one of the following conditions:
  1. (a)
    T is an F-contraction of type I on a G-metric space X, i.e., there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y\in X\),
    $$ G(Tx,Ty,Ty)>0\quad \Rightarrow \quad t+F\bigl(G(Tx,Ty,Ty)\bigr)\leqslant F \bigl(G(x,y,y)\bigr); $$
    (4)
     
  2. (b)
    T is an F-contraction of type II on a G-metric space X, i.e., there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y,z\in X\),
    $$ G(Tx,Ty,Tz)>0\quad \Rightarrow \quad t+F\bigl(G(Tx,Ty,Tz)\bigr)\leqslant F \bigl(G(x,y,z)\bigr). $$
    (5)
     
Then T has a unique fixed point \(u\in X\), and for any \(x\in X\), a sequence \(\{x_{n}=T^{n}x\}\) is G-convergent to u.

The previous ideas lead also to analogous fixed point theorems for F-expanding mappings on G-metric spaces.

Definition 3.3

A mapping \(T:X\rightarrow X\) from a G-metric space \((X,G)\) into itself is said to be
  1. (a)
    F-expanding of type I on a G-metric space X if there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y\in X\),
    $$ G(x,y,y)>0\quad \Rightarrow \quad F\bigl(G(Tx,Ty,Ty)\bigr)\geqslant F \bigl(G(x,y,y)\bigr)+t; $$
    (6)
     
  2. (b)
    F-expanding of type II on a G-metric space X if there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y,z\in X\),
    $$ G(x,y,z)>0\quad \Rightarrow \quad F\bigl(G(Tx,Ty,Tz)\bigr)\geqslant F \bigl(G(x,y,z)\bigr)+t. $$
    (7)
     

Theorem 3.4

Let \((X,G)\) be a complete G-metric space and \(T:X\rightarrow X\) be a surjective and F-expanding mapping of type I (or type II). Then T has a unique fixed point.

Proof

Let T be an F-expanding mapping of type I. From Lemma 1.2, there exists a mapping \(T^{*}:X\rightarrow X\) such that \(T\circ T^{*}\) is the identity mapping on X. Let \(x,y\in X\) be arbitrary points such that \(x\neq y\), and let \(\xi =T^{*}x\) and \(\eta =T^{*}y\). Obviously, \(\xi \neq \eta \) and \(G(\xi ,\eta ,\eta )>0\). By using (6) applied to ξ and η, we have
$$ F\bigl(G(T\xi ,T\eta ,T\eta )\bigr)\geqslant F\bigl(G(\xi ,\eta ,\eta )\bigr)+t. $$
Since \(T\xi =T(T^{*}x)=x\) and \(T\eta =T(T^{*}y)=y\), then
$$ F\bigl(G(x,y,y)\bigr)\geqslant F\bigl(G\bigl(T^{*}x,T^{*}y,T^{*}y \bigr)\bigr)+t, $$
so \(T^{*}\) is an F-contraction of type I on a G-metric space \((X,G)\). Theorem 3.2 guarantees that \(T^{*}\) has a unique fixed point \(u\in X\). The point u is also a fixed point of T because \(Tu=T(T^{*}u)=u\).
Now, we prove the uniqueness of the fixed point. Assume that v is another fixed point of T different from u: \(Tu=u\neq v=Tv\). This means \(G(u,v,v)>0\), so by (6)
$$ 0< t\leqslant F\bigl(G(Tu,Tv,Tv)\bigr)-F\bigl(G(u,v,v)\bigr)=0, $$
which is a contradiction, and hence \(u=v\).

For F-expanding mappings of type II, it is necessary to take \(z=y\) and apply the proof for F-expanding mappings of type I. □

As a corollary of Theorem 3.4, taking \(F_{1}\in \mathcal{F}\), see Examples 1.5, we obtain the following.

Corollary 3.5

[2], Corollary 9.1.4

Let \((X,G)\) be a complete G-metric space and \(T:X\rightarrow X\) be surjective, and let there exist \(\lambda >1\) such that
$$ G(Tx,Ty,Ty)\geqslant \lambda G(x,y,y)\quad\textit{for all}\quad x,y\in X, $$
or
$$ G(Tx,Ty,Tz)\geqslant \lambda G(x,y,z)\quad\textit{for all}\quad x,y,z\in X. $$
Then T has a unique fixed point.

Remark 3.6

If T is not surjective, the previous results are false. Consider \(X=(-\infty ,-1]\cup [1,\infty )\) endowed with the G-metric \(G(x,y,z)=\vert x-y \vert +\vert x-z \vert +\vert y-z \vert \) for all \(x,y,z\in X\) and the mapping \(T:X\rightarrow X\) defined by \(Tx=-2x\). Then \(G(Tx,Ty,Tz)\geqslant 2G(x,y,z)\) for all \(x,y,z\in X\) and T has no fixed point.

Now, we will improve some results contained in the book [2]. We will use the following observation: if \(T:X\rightarrow X\) is a surjective mapping, based on each \(x_{0}\in X\), there exists a sequence \(\{x_{n}\}\) such that \(Tx_{n+1}=x_{n}\) for all \(n\geqslant 0\). Generally, a sequence \(\{x_{n}\}\) verifying the above condition is not necessarily unique.

Theorem 3.7

Let \((X,G)\) be a complete G-metric space, and let \(T:X\rightarrow X\) be a surjective mapping. Suppose that there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y\in X\),
$$ G(x,Tx,y)>0\quad \Rightarrow \quad F\bigl(G\bigl(Tx,T^{2}x,Ty\bigr) \bigr)\geqslant F\bigl(G(x,Tx,y)\bigr)+t. $$
(8)
Then T has a unique fixed point.

Proof

Let \(x_{0}\in X\) be arbitrary. Since T is surjective, there exists \(x_{1}\in X\) such that \(Tx_{1}=x_{0}\). By continuing this process, we can find a sequence \(\{x_{n}=Tx_{n+1}\}\) for all \(n=0,1,2,\ldots\) . If there exists \(n_{0}\in \mathbb{N}\cup \{0\}\) such that \(x_{n_{0}}=x_{n_{0}+1}\), then \(x_{n_{0}+1}\) is a fixed point of T.

Now assume that \(x_{n}\neq x_{n+1}\) for all \(n\geqslant 0\). Then \(G(x_{n+1},x_{n},x_{n})>0\) for all \(n\geqslant 0\), and from (8) with \(x=x_{n+1}\) and \(y=x_{n}\), we have, for all \(n\geqslant 1\),
$$\begin{aligned} F\bigl(G(x_{n},x_{n-1},x_{n-1})\bigr)&=F\bigl(G \bigl(Tx_{n+1},T^{2}x_{n+1},Tx_{n}\bigr) \bigr) \\ &\geqslant F\bigl(G(x_{n+1},Tx_{n+1},x_{n})\bigr)+t = F\bigl(G(x_{n+1},x_{n},x_{n})\bigr)+t, \end{aligned}$$
and hence
$$ t+F\bigl(G(x_{n+1},x_{n},x_{n})\bigr)\leqslant F \bigl(G(x_{n},x_{n-1},x_{n-1})\bigr). $$
(9)
Using (9), the following holds for every \(n\geqslant 1\):
$$\begin{aligned} F\bigl(G(x_{n+1},x_{n},x_{n})\bigr)&\leqslant F \bigl(G(x_{n},x_{n-1},x_{n-1})\bigr)-t \\ &\leqslant F\bigl(G(x_{n-1},x_{n-2},x_{n-2})\bigr)-2t \leqslant \cdots \leqslant F\bigl(G(x_{1},x_{0},x_{0}) \bigr)-nt. \end{aligned}$$
(10)
From (10) we obtain
$$ \lim_{n\rightarrow \infty } F\bigl(G(x_{n+1},x_{n},x_{n}) \bigr)=-\infty, $$
which together with (F2) gives
$$ \lim_{n\rightarrow \infty }G(x_{n+1},x_{n},x_{n})=0. $$
(11)
From (F3) there exists \(k\in (0,1)\) such that
$$ \lim_{n\rightarrow \infty }\bigl[G(x_{n+1},x_{n},x_{n}) \bigr]^{k}F\bigl(G(x _{n+1},x_{n},x_{n}) \bigr)=0. $$
(12)
By (10), the following holds for all \(n\geqslant 1\):
$$\begin{aligned}& \bigl[G(x_{n+1},x_{n},x_{n})\bigr]^{k}F \bigl(G(x_{n+1},x_{n},x_{n})\bigr)- \bigl[G(x_{n+1},x _{n},x_{n})\bigr]^{k}F \bigl(G(x_{1},x_{0},x_{0})\bigr) \\& \quad \leqslant \bigl[G(x_{n+1},x_{n},x_{n}) \bigr]^{k}\bigl(F\bigl(G(x_{1},x_{0},x_{0}) \bigr)-nt\bigr) \\& \quad \quad {}-\bigl[G(x_{n+1},x_{n},x_{n}) \bigr]^{k}F\bigl(G(x_{1},x_{0},x_{0}) \bigr)= -\bigl[G(x_{n+1},x _{n},x_{n}) \bigr]^{k}\cdot nt\leqslant 0. \end{aligned}$$
(13)
Letting \(n\rightarrow \infty \) in (13) and using (11), (12), we obtain
$$ \lim_{n\rightarrow \infty } \bigl[G(x_{n+1},x_{n},x_{n}) \bigr]^{k}\cdot n=0. $$
(14)
Now, let us observe that from (14) there exists \(n_{1}\geqslant 1\) such that
$$ \bigl[G(x_{n+1},x_{n},x_{n})\bigr]^{k} \cdot n\leqslant 1\quad\mbox{for all}\quad n \geqslant n_{1}. $$
Consequently, we have
$$ G(x_{n+1},x_{n},x_{n})\leqslant \frac{1}{n^{1/k}} \quad\mbox{for all}\quad n \geqslant n_{1}. $$
Since the series \(\sum_{i=1}^{\infty }\frac{1}{i^{1/k}}\) converges, for any \(\varepsilon >0\), there exists \(n_{2}\geqslant 1\) such that \(\sum_{i=n_{2}}^{\infty }\frac{1}{i^{1/k}}<\varepsilon \). In order to show that \(\{x_{n}\}\) is a Cauchy sequence, we consider \(m>n>\max \{n_{1},n_{2}\}\). From [2], Lemma 3.1.2(4), we get
$$\begin{aligned} G(x_{m},x_{n},x_{n})&\leqslant \sum _{j=n}^{m-1}G(x_{j+1},x_{j},x _{j})\leqslant \sum_{j=n}^{\infty }G(x_{j+1},x_{j},x_{j}) \\ &\leqslant \sum_{j=n}^{\infty }\frac{1}{j^{1/k}}\leqslant \sum_{j=n_{2}}^{\infty }\frac{1}{j^{1/k}}< \varepsilon . \end{aligned}$$
Therefore by [2], Lemma 3.2.2 and axiom \((G_{4})\), \(\{x_{n}\}\) is a Cauchy in a G-metric space \((X,G)\). From the completeness of \((X,G)\), there exists \(u\in X\) such that \(\{x_{n}\} \rightarrow u\). As T is surjective, there exists \(w\in X\) such that \(u=Tw\). From (8) with \(x=x_{n+1}\) and \(y=w\), we have, for all \(n\geqslant 1\),
$$\begin{aligned} F\bigl(G(x_{n},x_{n-1},u)\bigr)&=F\bigl(G\bigl(Tx_{n+1},T^{2}x_{n+1},Tw \bigr)\bigr) \\ &\geqslant F\bigl(G(x_{n+1},Tx_{n+1},w)\bigr)+t=F \bigl(G(x_{n+1},x_{n},w)\bigr)+t, \end{aligned}$$
and hence
$$ F\bigl(G(x_{n},x_{n-1},u)\bigr) > F\bigl(G(x_{n+1},x_{n},w) \bigr). $$
(15)
By (F1) from (15), we have
$$ G(x_{n},x_{n-1},u) > G(x_{n+1},x_{n},w) \quad\mbox{for all}\quad n\geqslant 1. $$
(16)
Using the fact that the function G is continuous on each variable ([2], Theorem 3.2.2), taking the limit as \(n\rightarrow \infty \) in the above inequality, we get
$$ G(u,u,w)=\lim_{n\rightarrow \infty }G(x_{n},x_{n-1},u)=0, $$
that is, \(u=w\). Then u is a fixed point of T because \(u=Tw=Tu\).
To prove uniqueness, suppose that \(u,v\in X\) are two fixed points. If \(Tu=u\neq v=Tv\), then \(G(u,u,v)>0\). So, by (8),
$$\begin{aligned} F\bigl(G(u,u,v)\bigr)&=F\bigl(G\bigl(Tu,T^{2}u,Tv\bigr)\bigr) \\ &\geqslant F\bigl(G(u, Tu,v)\bigr)+t=F\bigl(G(u,u,v)\bigr)+t, \end{aligned}$$
which is a contradiction, because \(t>0\). Hence, \(u=v\). □

Taking \(F_{1}\in \mathcal{F}\), see Example 1.5, we obtain the following.

Corollary 3.8

[2], Theorem 9.1.2

Let \((X,G)\) be a complete G-metric space and \(T:X\rightarrow X\) be a surjective mapping. Suppose that there exists \(\lambda >1\) such that
$$ G\bigl(Tx,T^{2}x,Ty\bigr)\geqslant \lambda G(x,Tx,y)\quad\textit{for all}\quad x,y \in X. $$
Then T has a unique fixed point.

Next result does not guarantee the uniqueness of the fixed point.

Theorem 3.9

Let \((X,G)\) be a complete G-metric space, and let \(T:X\rightarrow X\) be a surjective mapping. Suppose that there exist \(F\in \mathcal{F}\) and \(t>0\) such that for all \(x,y\in X\),
$$ G\bigl(x,Tx,T^{2}x\bigr)>0\quad \Rightarrow \quad F\bigl(G \bigl(Tx,Ty,T^{2}y\bigr)\bigr)\geqslant F\bigl(G\bigl(x,Tx,T ^{2}x\bigr)\bigr)+t. $$
(17)
Then T has a fixed point.

Proof

Let \(x_{0}\in X\) be arbitrary. Since T is surjective, there exists \(x_{1}\in X\) such that \(x_{0}=Tx_{1}\). By continuing this process, we can find a sequence \(\{x_{n}=Tx_{n+1}\}\) for all \(n\geqslant 0\). If there exists \(n_{0}\geqslant 0\) such that \(x_{n_{0}}=x_{n_{0}+1}\), then \(x_{n_{0}+1}\) is a fixed point of T.

Now, assume that \(x_{n}\neq x_{n+1}\) for all \(n\geqslant 0\). From (17) with \(x=x_{n+1}\) and \(y=x_{n}\), we have \(G(x_{n+1},Tx_{n+1},T ^{2}x_{n+1})=G(x_{n+1},x_{n},x_{n-1})>0\) and
$$\begin{aligned} F\bigl(G(x_{n},x_{n-1},x_{n-2})\bigr)&=F\bigl(G \bigl(Tx_{n+1},Tx_{n},T^{2}x_{n}\bigr) \bigr) \\ &\geqslant F\bigl(G\bigl(x_{n+1},Tx_{n+1},T^{2}x_{n+1} \bigr)\bigr)+t=F\bigl(G(x_{n+1},x_{n},x _{n-1}) \bigr)+t, \end{aligned}$$
and hence
$$\begin{aligned} F\bigl(G(x_{n+1},x_{n},x_{n-1})\bigr)&\leqslant F \bigl(G(x_{n},x_{n-1},x_{n-2})\bigr)-t \\ &\leqslant F\bigl(G(x_{n-1},x_{n-2},x_{n-3})\bigr)-2t \\ &\leqslant \cdots \leqslant F\bigl(G(x_{2},x_{1},x_{0}) \bigr)-(n-1)t. \end{aligned}$$
(18)
From (18), we obtain
$$ \lim_{n\rightarrow \infty }F\bigl(G(x_{n+1},x_{n},x_{n-1}) \bigr) =- \infty , $$
which together with (F2) gives
$$ \lim_{n\rightarrow \infty }G(x_{n+1},x_{n},x_{n-1})=0. $$
Mimicking the proof of Theorem 3.7, we obtain
$$ \lim_{n\rightarrow \infty } \bigl[G(x_{n+1},x_{n},x_{n-1}) \bigr]^{k} \cdot (n-1)=0; $$
and consequently, there exists \(n_{1}\geqslant 1\) such that
$$ G(x_{n+1},x_{n},x_{n-1})\leqslant \frac{1}{(n-1)^{1/k}} \quad \mbox{for all}\quad n>n_{1}. $$
Since the series \(\sum_{i=1}^{\infty }\frac{1}{i^{1/k}}\) converges, for any \(\varepsilon >0\), there exists \(n_{2}\geqslant 1\) such that \(\sum_{i=n_{2}}^{\infty }\frac{1}{i^{1/k}}<\varepsilon \). In order to show that \(\{x_{n}\}\) is a Cauchy sequence, we consider \(m>n>\max \{n_{1},n_{2}\}\). From [2], Lemma 3.1.2(4) and axioms (\(G_{3}\)), \((G_{4})\), we get
$$\begin{aligned} G(x_{m},x_{n},x_{n})&\leqslant \sum _{j=n}^{m-1}G(x_{j+1},x_{j},x _{j})\leqslant \sum_{j=n}^{\infty }G(x_{j+1},x_{j},x_{j}) \\ &\leqslant \sum_{j=n}^{\infty }G(x_{j+1},x_{j},x_{j-1}) \leqslant \sum_{j=n}^{\infty }\frac{1}{j^{1/k}} \leqslant \sum_{j=n_{2}}^{\infty }\frac{1}{j^{1/k}}< \varepsilon . \end{aligned}$$
Therefore, by [2], Lemma 3.2.2, \(\{x_{n}\}\) is a Cauchy in a G-metric space \((X,G)\). From the completeness of \((X,G)\), there exists \(u\in X\) such that \(\{x_{n}\}\rightarrow u\). As T is surjective, there exists \(w\in X\) such that \(u=Tw\). From (17) with \(x=w\) and \(y=x_{n+1}\), we have
$$ F\bigl(G(u,x_{n},x_{n-1})\bigr)=F\bigl(G\bigl(Tw,Tx_{n+1},T^{2}x_{n+1} \bigr)\bigr)\geqslant F\bigl(G\bigl(w,Tw,T ^{2}w\bigr)\bigr)+t, $$
so
$$ F\bigl(G\bigl(w,Tw,T^{2}w\bigr)\bigr) \leqslant F\bigl(G(u,x_{n},x_{n-1}) \bigr)-t< F\bigl(G(u,x_{n},x_{n-1})\bigr). $$
Using (F1), we have
$$ G\bigl(w,Tw,T^{2}w\bigr) < G(u,x_{n},x_{n-1})\quad \mbox{for all}\quad n\geqslant 1. $$
Using the fact that the function G is continuous on each variable ([2], Theorem 3.2.2), taking the limit as \(n\rightarrow \infty \) in the above inequality, we get
$$ G\bigl(w,Tw,T^{2}w\bigr) =\lim_{n\rightarrow \infty } G(u,x_{n},x_{n-1})=0, $$
that is, \(w=Tw=T^{2}w\). Hence, \(u=Tu\). □

Taking \(F_{1}\in \mathcal{F}\), see Examples 1.5, we obtain the following.

Corollary 3.10

[2], Theorem 9.1.3

Let \((X,G)\) be a complete G-metric space and \(T:X\rightarrow X\) be a surjective mapping. Suppose that there exists \(\lambda >1\) such that
$$ G\bigl(Tx,Ty,T^{2}y\bigr)\geqslant \lambda G\bigl(x,Tx,T^{2}x \bigr)\quad\textit{for all}\quad x,y \in X. $$
Then T has, at least, a fixed point.

Declarations

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Authors’ Affiliations

(1)
Department of Mathematics and Applied Physics, Rzeszów University of Technology

References

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