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Fixed point theory for cyclic (ϕ - ψ)-contractions

Open Access
Research

Abstract

In this article, the concept of cyclic (ϕ - ψ)-contraction and a fixed point theorem for this type of mappings in the context of complete metric spaces have been presented. The results of this study extend some fixed point theorems in literature.

2000 Mathematics Subject Classification: 47H10;46T99 54H25.

Keywords

cyclic (ϕ - ψ)-contraction fixed point theory. 

1. Introduction and preliminaries

One of the most important results used in nonlinear analysis is the well-known Banach's contraction principle. Generalization of the above principle has been a heavily investigated branch research. Particularly, in [1] the authors introduced the following definition.

Definition 1. Let X be a nonempty set, m a positive integer, and T: XX a mapping. X = i = 1 m A i Open image in new window is said to be a cyclic representation of X with respect to T if
  1. (i)

    A i , i = 1, 2, ..., m are nonempty sets.

     
  2. (ii)

    T(A 1) ⊂ A 2, ..., T (A m-1) ⊂ A m , T (Am) ⊂ A 1.

     

Recently, fixed point theorems for operators T defined on a complete metric space X with a cyclic representation of X with respect to T have appeared in the literature (see, e.g., [2, 3, 4, 5]). Now, we present the main result of [5]. Previously, we need the following definition.

Definition 2. Let (X, d) be a metric space, m a positive integer A1, A2, ..., A m nonempty closed subsets of X and X = i = 1 m A i Open image in new window. An operator T: XX is said to be a cyclic weak ϕ-contraction if
  1. (i)

    X = i = 1 m A i Open image in new window is a cyclic representation of X with respect to T.

     
  2. (ii)

    d(Tx, Ty) ≤ d(x, y) - ϕ(d(x, y)), for any XA i , yA i+1, i = 1, 2, ..., m, where A m+1= A 1 and ϕ: [0, ∞) → [0, ∞) is a nondecreasing and continuous function satisfying ϕ(t) > 0 for t ∈ (0, ∞) and ϕ(0) = 0

     

Remark 3. For convenience, we denote by F the class of functions ϕ: [0, ∞) → [0, ∞) nondecreasing and continuous satisfying ϕ(t) > 0 for t ∈ (0, ∞) and ϕ(0) = 0.

The main result of [5] is the following.

Theorem 4. [[5], Theorem 6] Let (X, d) be a complete metric space, m a positive integer, A1, A2, ..., A m nonempty closed subsets of X and X = i = 1 m A i Open image in new window. Let T: XX be a cyclic weak ϕ-contraction with ϕF . Then T has a unique fixed point z i = 1 m A i Open image in new window.

The main purpose of this article is to present a generalization of Theorem 4.

2. Main results

First, we present the following definition.

Definition 5. Let (X, d) be a metric space, m a positive integer, A1, A2, ..., A m nonempty subsets of X and X = i = 1 m A i Open image in new window. An operator T: XX is a cyclic weak (ϕ - ψ)-contraction if
  1. (i)

    X = i = 1 m A i Open image in new window is a cyclic representation of X with respect to T,

     
  2. (ii)

    ϕ(d(Tx, Ty)) ≤ ϕ(d(x, y)) - ψ(d(x, y)), for any XA i , yA i+1, i = 1, 2, ..., m, where A m+1= A 1 and ϕ, ψF .

     

Our main result is the following.

Theorem 6. Let (X, d) be a complete metric space, m a positive integer, A1, A2, ..., A m nonempty subsets of X and X = i = 1 m A i Open image in new window. Let T: XX be a cyclic (ϕ - ψ)-contraction. Then T has a unique fixed point z i = 1 m A i Open image in new window.

Proof. Take x0X and consider the sequence given by
x n + 1 = T x n , n = 0 , 1 , 2 , . Open image in new window
If there exists n0 ∈ ℕ such that x n 0 + 1 = x n 0 Open image in new window then, since x n 0 + 1 = T x n 0 = x n 0 Open image in new window, the part of existence of the fixed point is proved. Suppose that xn+1x n for any n = 0, 1, 2, .... Then, since X = i = 1 m A i Open image in new window, for any n > 0 there exists i n ∈ {1, 2, ..., m} such that x n - 1 A i n Open image in new window and x n A i n + 1 Open image in new window. Since T is a cyclic (ϕ - ψ)-contraction, we have
ϕ ( d ( x n , x n + 1 ) ) = ϕ ( d ( T x n - 1 , T x n ) ) ϕ ( d ( x n - 1 , x n ) ) - ψ ( d ( x n - 1 , x n ) ) ϕ ( d ( x n - 1 , x n ) ) Open image in new window
(2.1)
From 2.1 and taking into account that ϕ is nondecreasing we obtain
d ( x n , x n + 1 ) d ( x n - 1 , x n ) for any  n = 1 , 2 , Open image in new window
Thus {d(x n , xn+1)} is a nondecreasing sequence of nonnegative real numbers. Consequently, there exists γ ≥ 0 such that lim n d ( x n , x n + 1 ) = γ Open image in new window. Taking n → ∞ in (2.1) and using the continuity of ϕ and ψ, we have
ϕ ( γ ) ϕ ( γ ) - ψ ( γ ) ϕ ( γ ) , Open image in new window
and, therefore, ψ(γ) = 0. Since ψF, γ = 0, that is,
lim n d ( x n , x n + 1 ) = 0 . Open image in new window
(2.2)

In the sequel, we will prove that {x n } is a Cauchy sequence.

First, we prove the following claim.

Claim: For every ε > 0 there exists n ∈ ℕ such that if p, qn with p - q ≡ 1(m) then d(x p , x q ) < ε.

In fact, suppose the contrary case. This means that there exists ε > 0 such that for any n ∈ ℕ we can find p n > q n n with p n - q n ≡ 1(m) satisfying
d ( x q n , x p n ) ε . Open image in new window
(2.3)
Now, we take n > 2m. Then, corresponding to q n n use can choose p n in such a way that it is the smallest integer with p n > q n satisfying p n - q n ≡ 1(m) and d ( x q n , x p n ) ε Open image in new window. Therefore, d ( x q n , x p n - m ) ε Open image in new window. Using the triangular inequality
ε d ( x q n , x p n ) d ( x q n , x p n - m ) + i = 1 m d ( x p n - i , x p n - i + 1 ) < ε + i = 1 m d ( x p n - i , x p n - i + 1 ) . Open image in new window
Letting n → ∞ in the last inequality and taking into account that limn→∞d(x n , xn+1) = 0, we obtain
lim n d ( x q n , x p n ) = ε Open image in new window
(2.4)
Again, by the triangular inequality
ε d ( x q n , x p n ) d ( x q n , x q n + 1 ) + d ( x q n + 1 , x p n + 1 ) + d ( x p n + 1 , x p n ) d ( x q n , x q n + 1 ) + d ( x q n + 1 , x q n ) + d ( x q n , x p n ) + d ( x p n , x p n + 1 ) + d ( x p n + 1 , x p n ) = 2 d ( x q n , x q n + 1 ) + d ( x q n , x p n ) + 2 d ( x p n , x p n + 1 ) Open image in new window
(2.5)
Letting n → ∞ in (2.4) and taking into account that lim n d ( x n , x n + 1 ) = 0 Open image in new window and (2.4), we get
lim n d ( x q n + 1 , x p n + 1 ) = ε . Open image in new window
(2.6)
Since x q n Open image in new window and x p n Open image in new window lie in different adjacently labelled sets A i and Ai+1for certain 1 ≤ im, using the fact that T is a cyclic (ϕ - ψ)-contraction, we have
ϕ ( d ( x q n + 1 , x p n + 1 ) ) = ϕ ( d ( T x q n , T x q n ) ϕ ( d ( x q n , x p n ) ) - ψ ( d ( x q n , x p n ) ) ϕ ( d ( x q n , x p n ) ) Open image in new window
Taking into account (2.4) and (2.6) and the continuity of ϕ and ψ, letting n → ∞ in the last inequality, we obtain
ϕ ( ε ) ϕ ( ε ) - ψ ( ε ) ϕ ( ε ) Open image in new window

and consequently, ψ(ε) = 0. Since ψF, then ε = 0 which is contradiction.

Therefore, our claim is proved.

In the sequel, we will prove that (X, d) is a Cauchy sequence. Fix ε > 0. By the claim, we find n0 ∈ ℕ such that if p, qn0 with p - q ≡ 1(m)
d ( x p , x q ) ε 2 Open image in new window
(2.7)
Since lim n d ( x n , x n + 1 ) = 0 Open image in new window we also find n1 ∈ ℕ such that
d ( x n , x n + 1 ) ε 2 m Open image in new window
(2.8)

for any nn1.

Suppose that r, s ≥ max{n0, n1} and s > r. Then there exists k ∈ {1, 2, ..., m} such that s - rk(m). Therefore, s - r + j ≡ 1(m) for j = m - k + 1. So, we have d(x r , x s ) ≤ d(x r , xs+j)+ d(xs+j, xs+j-1)+ ⋯ + d(xs+1, x s ). By (2.7) and (2.8) and from the last inequality, we get
d ( x r , x s ) ε 2 + j ε 2 m ε 2 + m ε 2 m = ε Open image in new window

This proves that (x n ) is a Cauchy sequence. Since X is a complete metric space, there exists xX such that limn→∞x n = x. In what follows, we prove that x is a fixed point of T. In fact, since lim n x n = x Open image in new window and, as X = i = 1 m A i Open image in new window is a cyclic representation of X with respect to T, the sequence (x n ) has infinite terms in each A i for i ∈ {∈ 1, 2, ..., m}.

Suppose that xA i , TxAi+1and we take a subsequence x n k Open image in new window of (x n ) with x n k A i - 1 Open image in new window (the existence of this subsequence is guaranteed by the above-mentioned comment). Using the contractive condition, we can obtain
ϕ ( d ( x n k + 1 , T x ) ) = ϕ ( d ( T x n k , T x ) ) ϕ ( d ( T x n k , x ) ) - ψ ( d ( x n k , x ) ) ϕ ( d ( x n k , x ) ) Open image in new window
and since x n k x Open image in new window and ϕ and ψ belong to F, letting k → ∞ in the last inequality, we have
ϕ ( d ( x , T x ) ) ϕ ( d ( x , x ) ) = ϕ ( 0 ) = 0 Open image in new window

or, equivalently, ϕ(d(x, Tx)) = 0. Since ϕF, then d(x, Tx) = 0 and, therefore, x is a fixed point of T.

Finally, to prove the uniqueness of the fixed point, we have y, zX with y and z fixed points of T. The cyclic character of T and the fact that y, zX are fixed points of T, imply that y , z i = 1 m A i Open image in new window. Using the contractive condition we obtain
ϕ ( d ( y , z ) ) = ϕ ( d ( T y , T x ) ) ϕ ( d ( y , z ) ) - ψ ( d ( y , z ) ) ϕ ( d ( y , z ) ) Open image in new window
and from the last inequality
ψ ( d ( y , z ) ) = 0 Open image in new window

Since ψF, d(y, z) = 0 and, consequently, y = z. This finishes the proof.

In the sequel, we will show that Theorem 6 extends some recent results.

If in Theorem 6 we take as ϕ the identity mapping on [0, ∞) (which we denote by Id[0, ∞)), we obtain the following corollary.

Corollary 7. Let (X, d) be a complete metric space m a positive integer, A1, A2, ..., A m nonempty subsets of X and X = i = 1 m A i Open image in new window. Let T: XX be a cyclic (Id[0, ∞) - ψ) contraction. Then T has a unique fixed point z i = 1 m A i Open image in new window.

Corollary 7 is a generalization of the main result of [5] (see [[5], Theorem 6]) because we do not impose that the sets A i are closed.

If in Theorem 6 we consider ϕ = Id[0, ∞) and ψ = (1 - k)Id[0, ∞) for k ∈ [0, 1) (obviously, ϕ, ψF), we have the following corollary.

Corollary 8. Let (X, d) be a complete metric space m a positive integer, A1, A2, ..., A m nonempty subsets of X and X = i = 1 m A i Open image in new window. Let T: XX be a cyclic (Id[0, ∞) - (1 - k)Id[0, ∞)) contraction, where k ∈ [0, 1). Then T has a unique fixed point z i = 1 m A i Open image in new window.

Corollary 8 is Theorem 1.3 of [1].

The following corollary gives us a fixed point theorem with a contractive condition of integral type for cyclic contractions.

Corollary 9. Let (X, d) be a complete metric space, m a positive integer, A1, A2, ..., A m nonempty closed subsets of X and X = i = 1 m A i Open image in new window. Let T: XX be an operator such that
  1. (i)

    X = i = 1 m A i Open image in new window is a cyclic representation of X with respect to T .

     
  2. (ii)
    There exists k ∈ [0, 1) such that
    0 d ( T x , T y ) ρ ( t ) d t k 0 d ( x , y ) ρ ( t ) d t Open image in new window
     

for any XA i , yAi+1, i = 1, 2, ..., m where Am+1= A1, and ρ: [0, ∞) → [0, ∞) is a Lebesgue-integrable mapping satisfying 0 ε ρ ( t ) d t Open image in new window for ε> 0.

Then T has unique fixed point z i = 1 m A i Open image in new window.

Proof. It is easily proved that the function ϕ: [0, ∞) → [0, ∞) given by φ ( t ) = 0 t p ( s ) d s Open image in new window satisfies that ϕF. Therefore, Corollary 9 is obtained from Theorem 6, taking as ϕ the above-defined function and as ψ the function ψ(t) = (1 - k)ϕ(t).

If in Corollary 9, we take A i = X for i = 1, 2, ..., m we obtain the following result.

Corollary 10. Let (X, d) be a complete metric space and T: XX a mapping such that for x, yX,
0 d ( T x , T y ) ρ t d t k 0 d ( x , y ) ρ t d t Open image in new window

where ρ: [0, ∞) → [0, ∞) is a Lebesgue-integrable mapping satisfying 0 ε ρ ( t ) d t Open image in new windowfor ε > 0 and the constant k ∈ [0, 1). Then T has a unique fixed point.

Notice that this is the main result of [6]. If in Theorem 6 we put A i = X for i = 1, 2, ..., m we have the result.

Corollary 11. Let (X, d) be a complete metric space and T: XX an operator such that for x, yX,
ϕ ( d ( T x , T y ) ) ϕ ( d ( x , y ) ) - ψ ( d ( x , y ) ) , Open image in new window

where ϕ, ψF . Then T has a unique fixed point.

This result appears in [7].

3. Example and remark

In this section, we present an example which illustrates our results. Throughout the article, we let ℕ* = ℕ\{0}.

Example 12. Consider X = 1 n : n * { 0 } Open image in new window with the metric induced by the usual distance in ℝ, i.e., d(x, y) = |x - y|. Since X is a closed subset of ℝ, it is a complete metric space. We consider the following subsets of X:
A 1 = { 1 n : n o d d } { 0 } Open image in new window
A 2 = { 1 n : n e v e n } { 0 } Open image in new window
Obviously, X = A1A2. Let T: XX be the mapping defined by
T x = 1 n + 1 i f x = 1 n 0 i f x = 0 Open image in new window
It is easily seen that X = A1A2is a cyclic representation of X with respect to T. Now we consider the function ρ: [0, ∞) → [0, ∞) defined by
ρ ( t ) = 0 i f t = 0 t 1 t - 2 [ 1 - ln t ] i f 0 < t < e 0 i f t e Open image in new window

It is easily proved that 0 t ρ ( s ) d s = t 1 t Open image in new windowfor t ≤ 1.

In what follows, we prove that T satisfies condition (ii) of Corollary 9.

In fact, notice that the function ρ(t) is a Lebesgue-integrable mapping satisfying 0 ε ρ ( t ) d t > 0 Open image in new windowfor ε > 0. We take m, n ∈ ℕ* with mn and we will prove
0 d ( T ( 1 n ) , T ( 1 m ) ) ρ ( s ) d s 1 2 0 d ( 1 n , 1 m ) ρ ( s ) d s Open image in new window
Since 0 t ρ ( s ) d s = t 1 t Open image in new windowfor t ≤ 1 and, as diam(X) ≤ 1, the last inequality can be written as
d ( T ( 1 n ) , T ( 1 m ) ) 1 d ( T ( 1 n ) , T ( 1 m ) ) 1 2 d 1 n , 1 m 1 d 1 n , 1 m Open image in new window
or equivalently,
1 n + 1 - 1 m + 1 1 1 n + 1 - 1 m + 1 1 2 1 n - 1 m 1 1 n - 1 m Open image in new window
or equivalently,
m - n ( n + 1 ) ( m + 1 ) ( n + 1 ) ( m + 1 ) m - n 1 2 m - n n m n m m - n Open image in new window
or equivalently,
m - n ( n + 1 ) ( m + 1 ) n + m + 1 m - n × n m ( n + 1 ) ( m + 1 ) n m m - n 1 2 Open image in new window
(3.1)
In order to prove that this last inequality is true, notice that
n m ( n + 1 ) ( m + 1 ) < 1 Open image in new window
(3.2)
and, therefore,
n m ( n + 1 ) ( m + 1 ) n m m - n 1 Open image in new window
On the other hand, from
m - n m . n m - n m + n Open image in new window
we obtain
2 ( m - n ) ( n + 1 ) ( m + 1 ) Open image in new window
and, thus,
m - n ( n + 1 ) ( m + 1 ) 1 2 . Open image in new window
Since n + m + 1 m - n 1 Open image in new window,
m - n ( n + 1 ) ( m + 1 ) n + m + 1 m - n 1 2 . Open image in new window
(3.3)

Finally, (3.2) and (3.3) give us (3.1).

Now we take x = 1 n Open image in new window, n ∈ ℕ* and y = 0. In this case, condition (ii) of Corollary 9 for k = 1 2 Open image in new window has the form
d T 1 n , T ( 0 ) 1 d ( T 1 n , T ( 0 ) ) = 1 n + 1 n + 1 1 2 d 1 n , 0 ) 1 d ( 1 n , 0 ) = 1 2 1 n n Open image in new window
The last inequality is true since
1 n + 1 n < 1 n n Open image in new window
and, then,
1 n + 1 n + 1 = 1 n + 1 n 1 n + 1 1 2 1 n + 1 n < 1 2 1 n n . Open image in new window

Consequently, since assumptions of Corollary 9 are satisfied, this corollary gives us the existence of a unique fixed point (which is obviously x = 0).

This example appears in [6].

Now, we connect our results with the ones appearing in [3]. Previously, we need the following definition.

Definition 13. A function φ: [0, ∞) → [0, ∞) is a (c)-comparison function if k = 0 φ k ( t ) Open image in new window converges for any t∈ [0, ∞). The main result of[3]is the following.

Theorem 14. Let (X, d) be a complete metric space, m a positive integer, A1, A2, ..., A m nonempty subsets of X, X = i = 1 m A i Open image in new windowand φ: [0, ∞) → [0, ∞) a (c)-comparison function. Let T: XX be an operator and we assume that
  1. (i)

    X = i = 1 m A i Open image in new window is a cyclic representation of X with respect to T.

     
  2. (ii)

    d(Tx, Ty) ≤ φ(d(x, y)), for any XA i and yA i+1, where A m+1= A.

     

Then T has a unique fixed point z i = 1 m A i Open image in new window.

Now, the contractive condition of Theorem 6 can be written as
ϕ ( d ( T x , T y ) ) ϕ ( d ( x , y ) ) - φ ( d ( x , y ) ) = ( ϕ - φ ) ( d ( x , y ) ) Open image in new window

for any xA i , yAi+1, where Am+1= A1, and ϕ, φF.

Particularly, if we take ϕ = Id[0, ∞) and φ ( t ) = t 2 1 + t Open image in new window, it is easily seen that ϕ, φF. On the other hand,
( ϕ - φ ) ( t ) = t - t 2 1 + t = t 1 + t Open image in new window
and
( ϕ - φ ) ( n ) ( t ) = t 1 + n t Open image in new window

Moreover, for every t ∈ (0, ∞), k = 0 ( ϕ - φ ) ( k ) ( t ) Open image in new window diverges. Therefore, ϕ - φ is not a (c)-comparison function. Consequently, our Theorem 6 can be applied to cases which cannot treated by Theorem 14.

Notes

Acknowledgements

KS was partially supported by the "Ministerio de Education y Ciencia", Project MTM 2007/65706.

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© Karapinar and Sadarangani; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  1. 1.Department of MathematicsAtilim UniversityIncek, AnkaraTurkey
  2. 2.Department of MathematicsUniversity of Las Palmas de Gran Canaria, Campus Universitario de TafiraLas Palmas de Gran CanariaSpain

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