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A fixed point theorem for cyclic generalized contractions in metric spaces
Fixed Point Theory and Applications volume 2012, Article number: 122 (2012)
Abstract
In this paper, we extend a recent result of V. Pata (J. Fixed Point Theory Appl. 10:299305, 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 nontrivial applications in many branches of pure and applied sciences (see, for instance, [2, 7, 14] and references cited therein).
Let (X,d) be a metric space and f:X\to X be an operator. We say that f is a contraction if there exists \lambda \in [0,1) such that, for all x,y\in X,
In terms of Picard operator theory (see [13]), Banach contraction principle asserts that if 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])
The condition (1.2) is weaker than the contraction condition (1.1). In fact, if
then it can be verified that, for every x,y\in X, we have
where
Remark 1.2 (see [8])
The function f:[1,\mathrm{\infty})\to [1,\mathrm{\infty}) defined as
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 socalled fixed point structures, are presented in I. A. Rus [12]. In [10], Păcurar and Rus established a fixed point theorem for cyclic φcontractions and they further discussed fixed point theory in metric spaces. In [3], Karapinar proved a fixed point theorem for cyclic weak φcontraction 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])
Let 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}_{m1})\subset {A}_{m},f({A}_{m})\subset {A}_{1}.
Let (X,d) be a complete metric space. Selecting an arbitrary {x}_{1}\in X, we denote
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
and set {c}_{n}=\parallel {x}_{n}\parallel. Assume {x}_{n}\ne {x}_{n+1} for all n. By (2.1), we have
First, we prove that the sequence {({c}_{n})}_{n\in {\mathbb{N}}^{\ast}} is bounded. By (2.2) we get that
Since {x}_{1}\in {A}_{1} and {x}_{n}\in {A}_{n}, from (2.1), we obtain that
where {c}_{1}=\parallel {x}_{1}\parallel =d({x}_{1},{x}_{1})=0, \beta \le \alpha, and for some a,b>0. Thus,
If there is a subsequence {({c}_{{n}_{k}})}_{k\in {\mathbb{N}}^{\ast}}\to \mathrm{\infty}, the choice \epsilon ={\epsilon}_{k}=\frac{(1+b)}{{c}_{{n}_{k}}} leads to the contradiction
Therefore, the sequence ({c}_{n}) is bounded.
From (2.2) we obtain that the sequence \{d({x}_{n},{x}_{n+1})\} is nonincreasing and then it is convergent to the real number
Now we show that r=0. Assume that r>0. Let {x}_{n}\in {A}_{n} and {x}_{n+1}\in {A}_{n+1}. By (2.1), we have
for some K>0. Letting n\to \mathrm{\infty}, we obtain
which implies r=0. This leads to a contradiction, therefore
For p\ge 1, suppose there exists j, 0\le j\le m1, such that (n+p)n+j=1modm, i.e., p+j=1modm. Now, let p be fixed, j=0 and let
So, we have
Since p=1modm, {x}_{n} and {x}_{n+p} lie in different sets {A}_{i} and {A}_{i+1}, for some 1\le i\le m. Then by (2.1) we have
where C=sup\gamma {(1+2{c}_{n})}^{\beta}<\mathrm{\infty}. Choosing for each n
the relation (2.3) becomes
Since {q}_{0}=0, it follows that
Consequently,
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.
On the other hand, the sequence \{{x}_{n}\} has an infinite number of terms in each {A}_{i}, for every i\in \{1,\dots ,m\}. Since (Y,d) is complete, in each {A}_{i}, i\in \{1,\dots ,m\} we can construct a subsequence of \{{x}_{n}\} which converges to 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}.
We claim now that for any initial value x\in Y, we get the same limit point {x}^{\ast}\in {\bigcap}_{i=1}^{m}{A}_{i}. Indeed, for x\in Y={\bigcup}_{i=1}^{m}{A}_{i}, by repeating the above process, the corresponding iterative sequence yields that 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
for some K>0. Suppose that \epsilon =0. Then we have
If equality occurs, the relation
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.
To prove that the Picard iteration converges to {x}^{\ast}, let us consider {x}_{1}\in Y={\bigcup}_{i=1}^{m}{A}_{i}. Then there exists {i}_{0}\in \{1,\dots ,m\} such that {x}_{n}\in {A}_{{i}_{0}}. As {x}^{\ast}\in {\bigcap}_{i=1}^{m}{A}_{i} it follows that {x}^{\ast}\in {A}_{{i}_{0}+1} as well. By the continuity of f, we obtain
Letting n\to \mathrm{\infty}, it follows that ({x}_{n})\to {x}^{\ast}, 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}_{n1}),f\left({x}^{\ast}\right))\le d({x}_{n1},{x}^{\ast})\le \cdots \le d({x}_{1},{x}^{\ast})=\parallel {x}^{\ast}\parallel .(2.4)
By (2.4), it follows that
□
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}.
Finally, we will prove a periodic point theorem. For this purpose, notice first that if 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 miterate {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
Thus, we immediately get that, for m\in \mathbb{N} with m\ge 2, 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.
For every \epsilon \in [0,1] and each x,y\in {A}_{{i}_{0}}, we have
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. □
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An erratum to this article is available at http://dx.doi.org/10.1186/16871812201339.
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Alghamdi, M.A., Petruşel, A. & Shahzad, N. A fixed point theorem for cyclic generalized contractions in metric spaces. Fixed Point Theory Appl 2012, 122 (2012). https://doi.org/10.1186/168718122012122
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DOI: https://doi.org/10.1186/168718122012122
Keywords
 Point Theorem
 Fixed Point Theorem
 Nonempty Subset
 Fixed Point Theory
 Unique Fixed Point