# A fixed point theorem for cyclic generalized contractions in metric spaces

- Maryam A Alghamdi
^{1}, - Adrian Petruşel
^{2}and - Naseer Shahzad
^{3}Email author

**2012**:122

https://doi.org/10.1186/1687-1812-2012-122

© Alghamdi et al.; licensee Springer 2012

**Received: **13 March 2012

**Accepted: **4 July 2012

**Published: **23 July 2012

## Abstract

In this paper, we extend a recent result of V. Pata (J. Fixed Point Theory Appl. 10:299-305, 2011) in the frame of a cyclic representation of a complete metric space.

## 1 Introduction

One of the fundamental result in fixed point theory is the Banach contraction principle. It has various non-trivial applications in many branches of pure and applied sciences (see, for instance, [2, 7, 14] and references cited therein).

*f*is a contraction if there exists $\lambda \in [0,1)$ such that, for all $x,y\in X$,

*f*is a contraction and $(X,d)$ is complete, then

*f*is a Picard operator. This result has been extended to other important classes of maps. Recently, Pata [8] proved that if $(X,d)$ is a complete metric space and $f:X\to X$ is an operator such that there exists fixed constants $\gamma \ge 0$, $\alpha \ge 1$ and $\beta \in [0,\alpha ]$ such that, for every $\epsilon \in [0,1]$ and every $x,y\in X$,

(where $\psi :[0,1]\to [0,\mathrm{\infty})$ is an increasing function vanishing with continuity at zero and $\parallel x\parallel :=d(x,{x}_{0})$, with arbitrary ${x}_{0}\in X$), then *f* has a unique fixed point in *X*.

**Remark 1.1** (see [8])

**Remark 1.2** (see [8])

has a unique fixed point ${x}^{\ast}=1$, but fails to be a contraction on any neighborhood both of 1 and of ∞.

Kirk, Srinivasan and Veeramani [6] obtained an extension of Banach’s fixed point theorem for mappings satisfying cyclical contractive conditions. Some generalizations of the results given in [6], using the setting of so-called fixed point structures, are presented in I. A. Rus [12]. In [10], Păcurar and Rus established a fixed point theorem for cyclic *φ*-contractions and they further discussed fixed point theory in metric spaces. In [3], Karapinar proved a fixed point theorem for cyclic weak *φ*-contraction mappings. Some other recent results concerning this topic are given in [1, 4, 5, 9, 11].

In the present paper, we obtain a fixed point theorem for a generalized contraction in the sense of the assumption (1.2), defined on a cyclic representation of a complete metric space.

## 2 Main results

We need first to recall a known concept.

**Definition 2.1** ([3])

*X*be a nonempty set,

*m*be a positive integer and $f:X\to X$ an operator. Then, we say that ${\bigcup}_{i=1}^{m}{A}_{i}$ is a cyclic representation of

*X*with respect to

*f*if:

- (i)
$X={\bigcup}_{i=1}^{m}{A}_{i}$, where ${A}_{i}$ are nonempty sets for each $i\in \{1,\dots ,m\}$;

- (ii)
$f({A}_{1})\subset {A}_{2},\dots ,f({A}_{m-1})\subset {A}_{m},f({A}_{m})\subset {A}_{1}$.

Our main result is as follows.

**Theorem 2.2**

*Let*$(X,d)$

*be a complete metric space*,

*m*

*be a positive integer*, ${A}_{1},\dots ,{A}_{m}$

*be closed nonempty subsets of*

*X*, $Y:={\bigcup}_{i=1}^{m}{A}_{i}$, $\psi :[0,1]\to [0,\mathrm{\infty})$

*be an increasing function vanishing with continuity at zero*,

*and*$f:Y\to Y$

*be an operator*.

*Assume that*:

- 1.
${\bigcup}_{i=1}^{m}{A}_{i}$

*is a cyclic representation of**Y**with respect to**f*; - 2.
*For every*$\epsilon \in [0,1]$, $x\in {A}_{i}$,*and*$y\in {A}_{i+1}$ ($i\in \{1,\dots ,m\}$,*where*${A}_{m+1}={A}_{1}$),*we have*$d(f(x),f(y))\le (1-\epsilon )d(x,y)+\gamma {\epsilon}^{\alpha}\psi (\epsilon ){[1+\parallel x\parallel +\parallel y\parallel ]}^{\beta},$(2.1)

*where* $\gamma \ge 0$, $\alpha \ge 1$ *and* $\beta \in [0,\alpha ]$ *are fixed constants*.

*Then*,

*we have the following conclusions*:

- (i)
*f**is a Picard operator*,*i*.*e*.,*f**has a unique fixed point*${x}^{\ast}\in {\bigcap}_{i=1}^{m}{A}_{i}$*and the Picard iteration sequence*${\{{f}^{n}(x)\}}_{n\in \mathbb{N}}$*converges to*${x}^{\ast}$,*for any initial point*$x\in Y$; - (ii)
*the following estimates hold*:$\begin{array}{c}d({x}_{n},{x}^{\ast})\le \parallel {x}^{\ast}\parallel ,\phantom{\rule{1em}{0ex}}n\ge 2;\hfill \\ d({x}_{n},{x}_{1})\le 2\parallel {x}^{\ast}\parallel ,\phantom{\rule{1em}{0ex}}n\ge 2.\hfill \end{array}$

*Proof*(i) For convenience of notation, if $j>m$, define ${A}_{j}={A}_{i}$ where $i=jmodm$ and $1\le i\le m$. Let ${x}_{1}\in {A}_{1}$. Starting from ${x}_{1}$, let ${\{{x}_{n}\}}_{n\ge 1}$ be the Picard iteration defined by the sequence

*n*. By (2.1), we have

Therefore, the sequence $({c}_{n})$ is bounded.

*j*, $0\le j\le m-1$, such that $(n+p)-n+j=1modm$,

*i.e.*, $p+j=1modm$. Now, let

*p*be fixed, $j=0$ and let

*n*

This shows that $\{{x}_{n}\}$ is a Cauchy sequence in the complete metric space $(Y,d)$ and, thus, it is convergent to a point $y\in Y={\bigcup}_{i=1}^{m}{A}_{i}$. The case $j\ne 0$ similar.

*y*. Since each ${A}_{i}$ is closed for $i\in \{1,\dots ,m\}$, we get that $y\in {\bigcap}_{i=1}^{m}{A}_{i}$. Then ${\bigcap}_{i=1}^{m}{A}_{i}\ne \mathrm{\varnothing}$ and we can consider the restriction

which satisfies the conditions of Theorem 1 in [8], since ${\bigcap}_{i=1}^{m}{A}_{i}$ is also closed and complete. From this result, it follows that *g* has a unique fixed point, say ${x}^{\ast}\in {\bigcap}_{i=1}^{m}{A}_{i}$.

*g*has a unique fixed point, say $z\in {\bigcap}_{i=1}^{m}{A}_{i}$. Since ${x}^{\ast}$, $z\in {\bigcap}_{i=1}^{m}{A}_{i}$, we have ${x}^{\ast}$, $z\in {A}_{i}$ for all $i\in \{1,\dots ,m\}$ and, hence, $d({x}^{\ast},z)$ and $d(f({x}^{\ast}),f(z))$ are well defined. We can write (2.1) in the form

is valid for every $\epsilon \in [0,1]$, which implies $d({x}^{\ast},z)=0$. Thus, ${x}^{\ast}$ is the unique fixed point of *f* for any initial value $x\in Y$.

*f*, we obtain

*i.e.*, the Picard iteration converges to the unique fixed point of

*f*for any initial point ${x}_{1}\in Y$.

- (ii)Since ${x}^{\ast}$ is a fixed point and ${x}^{\ast}\in {\bigcap}_{i=1}^{m}{A}_{i}$, we obtain that$d({x}_{n},{x}^{\ast})=d(f({x}_{n-1}),f\left({x}^{\ast}\right))\le d({x}_{n-1},{x}^{\ast})\le \cdots \le d({x}_{1},{x}^{\ast})=\parallel {x}^{\ast}\parallel .$(2.4)

□

In view of Remark 1.1, we immediately obtain the following corollary.

**Corollary 2.3** (Kirk, Srinivasan, Veeramani [2], Theorem 1.3])

*Let*$(X,d)$

*be a complete metric space*,

*m*

*be a positive integer*, ${A}_{1},\dots ,{A}_{m}$

*be closed nonempty subsets of*

*X*, $Y:={\bigcup}_{i=1}^{m}{A}_{i}$

*and*$f:Y\to Y$

*be an operator*.

*Assume that*:

- (i)
${\bigcup}_{i=1}^{m}{A}_{i}$

*is a cyclic representation of**Y**with respect to**f*; - (ii)
*there exists*$\lambda \in [0,1)$*such that*,*for any*$x\in {A}_{i}$, $y\in {A}_{i+1}$,*where*${A}_{m+1}={A}_{1}$,*we have*$d(f(x),f(y))\le \lambda d(x,y).$

*Then* *f* *has a unique fixed point* ${x}^{\ast}\in {\bigcap}_{i=1}^{m}{A}_{i}$.

*f*satisfies (1.2) with constants

*α*,

*β*,

*γ*and function

*ψ*, and if $\parallel f(x)\parallel \le \parallel x\parallel $ for each $x\in X$, then its

*m*-iterate ${f}^{m}$ also satisfies the condition (1.2) with constants

*α*,

*β*,

*mγ*and function

*ψ*. Indeed, let us suppose that

*f*satisfies (1.2) with constants

*α*,

*β*,

*γ*. Then, for every $\epsilon \in [0,1]$, we have

Notice also that if ${\bigcup}_{i=1}^{m}{A}_{i}$ is a cyclic representation of *X* with respect to *f*, then each ${A}_{i}$ ($i\in \{1,2,\dots ,m\}$) is an invariant set with respect to ${f}^{m}$. Using these two remarks, we get the following periodic point theorem.

**Theorem 2.4**

*Let*$(X,d)$

*be a complete metric space*,

*m*

*be a positive integer*, ${A}_{1},\dots ,{A}_{m}$

*be nonempty subsets of*

*X*, $Y:={\bigcup}_{i=1}^{m}{A}_{i}$, $\psi :[0,1]\to [0,\mathrm{\infty})$

*be an increasing function vanishing with continuity at zero and*$f:Y\to Y$

*be an operator such that*$\parallel f(x)\parallel \le \parallel x\parallel $

*for each*$x\in Y$.

*Assume that*:

- 1.
${\bigcup}_{i=1}^{m}{A}_{i}$

*is a cyclic representation of**Y**with respect to**f*. - 2.
*There exists*${i}_{0}\in \{1,\dots ,m\}$*such that*${A}_{{i}_{0}}$*is closed*. - 3.

*where* $\gamma \ge 0$, $\alpha \ge 1$ *and* $\beta \in [0,\alpha ]$ *are fixed constants*.

*Then*, ${f}^{m}$ *has a fixed point*.

*Proof* Notice that, by the above considerations, ${f}^{m}$ is a self mapping on ${A}_{{i}_{0}}$ and it satisfies the condition (1.2) with constants *α*, *β*, *mγ* and function *ψ*. Thus, by Theorem 1 in [8] we get the conclusion. □

## Notes

## Declarations

## Authors’ Affiliations

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