Fixed point theorems for F-expanding mappings

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.

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:

• (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)

• (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

• (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)

• (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.