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# Correction: A fixed point theorem for cyclic generalized contractions in metric spaces. Fixed Point Theory and Applications 2012, 2012:122

Fixed Point Theory and Applications20132013:39

https://doi.org/10.1186/1687-1812-2013-39

• Received: 31 January 2013
• Accepted: 10 February 2013
• Published:

## Abstract

The purpose of this short note is to present some corrections and clarifications concerning the proof of the main result given in the above mentioned paper.

## Keywords

• Point Theorem
• Initial Point
• Differential Geometry
• Fixed Point Theorem
• Point Theory

Concerning the text and the proof of the main result (Theorem 2.2), we would like to do the following corrections and clarifications:

(1) In Theorem 2.2, an additional hypothesis is needed, namely:

3. There exists ${y}_{1}\in Y$ such that
(2) On page 3, the fact that the sequence ${\left({c}_{n}\right)}_{n\in \mathbb{N}}$ is bounded follows now from the above mentioned hypothesis, using the following remark: by 3., for each $x\in Y$, we have that
Indeed, let $n\in {\mathbb{N}}^{\ast }$ and suppose that ${y}_{1}\in {A}_{1}$. Let $x\in {A}_{l}$ (where $l\in \left\{1,\dots ,m\right\}$). Let us choose ${u}_{2}\in {A}_{2},\phantom{\rule{0.2em}{0ex}}\dots \phantom{\rule{0.2em}{0ex}},{u}_{l-1}\in {A}_{l-1}$. Then
$\begin{array}{rcl}d\left(x,{f}^{n}\left(x\right)\right)& \le & d\left(x,{y}_{1}\right)+d\left({y}_{1},{f}^{n}\left({y}_{1}\right)\right)+d\left({f}^{n}\left({y}_{1}\right),{f}^{n}\left(x\right)\right)\\ \le & d\left(x,{y}_{1}\right)+d\left({y}_{1},{f}^{n}\left({y}_{1}\right)\right)+d\left({f}^{n}\left({y}_{1}\right),{f}^{n}\left({u}_{2}\right)\right)+\cdots +d\left({f}^{n}\left({u}_{l-1}\right),{f}^{n}\left(x\right)\right)\le \cdots \\ \le & d\left(x,{y}_{1}\right)+d\left({y}_{1},{f}^{n}\left({y}_{1}\right)\right)+d\left({y}_{1},{u}_{2}\right)+\cdots +d\left({u}_{l-1},x\right)<+\mathrm{\infty }.\end{array}$

(3) On page 5, to prove that the Picard iteration converges to ${x}^{\ast }$, we have to do as follows.

Let us show now that the Picard iteration converges to ${x}^{\ast }$ for any initial point ${x}_{1}$. We know that f has a unique fixed point (denoted by ${x}^{\ast }$) and the sequence ${\left({x}_{n}\right)}_{n\in \mathbb{N}}$ converges to a certain $y\in {\bigcap }_{i=1}^{m}{A}_{i}$. We will show that y is also a fixed point of f. For this purpose, we have

This shows that y is a fixed point of f and, thus, $y={x}^{\ast }$.

## Authors’ Affiliations

(1)
Department of Mathematics, Sciences Faculty for Girls, King Abdulaziz University, P.O. Box 4087, Jeddah, 21491, Saudi Arabia
(2)
Department of Mathematics, Babeş-Bolyai University, Kogǎlniceanu Street No. 1, Cluj-Napoca, 400084, Romania
(3)
Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21859, Saudi Arabia 