# 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.

## 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  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.

## 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.

## Applications to G-metric spaces

In 2006 Mustafa and Sims (see  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.  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

, 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 . 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 , 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 , 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 (, 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

, 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 , 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 , 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 (, 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

, 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.

## References

1. 1.

Wardowski, D: Fixed points of a new type of contractive mappings in complete metric spaces. Fixed Point Theory Appl. 2012, Article ID 94 (2012)

2. 2.

Agarwal, RP, Karapinar, E, O’Regan, D, Roldán-López-de-Hierro, AF: Fixed Point Theory in Metric Type Spaces. Springer, Switzerland (2015)

3. 3.

Samet, B, Vetro, C, Vetro, F: Remarks on G-metric spaces. Int. J. Anal. 2013, Article ID 917158 (2013)

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Correspondence to Jarosław Górnicki.

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