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

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.

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: X → X a mapping. X = ∪ i = 1 m A i is said to be a cyclic representation of X with respect to T if

(i) A_i, i = 1, 2, ..., m are nonempty sets.

(ii) T(A₁) ⊂ A₂, ..., T (A_m-1) ⊂ A_m, T (Am) ⊂ A₁.

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., [2-5]). 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 . An operator T: X → X is said to be a cyclic weak ϕ-contraction if

(i) X = ∪ i = 1 m A i is a cyclic representation of X with respect to T.

(ii) d(Tx, Ty) ≤ d(x, y) - ϕ(d(x, y)), for any X ∈ A_i, y ∈ A_i+1, i = 1, 2, ..., m, where A_m+1 = A₁ 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 . Let T: X → X be a cyclic weak ϕ-contraction with ϕ ∈ F. Then T has a unique fixed point z ∈ ∩ i = 1 m A i .

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

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 . An operator T: X → X is a cyclic weak (ϕ - ψ)-contraction if

(i) X = ∪ i = 1 m A i is a cyclic representation of X with respect to T,

(ii) ϕ(d(Tx, Ty)) ≤ ϕ(d(x, y)) - ψ(d(x, y)), for any X ∈ A_i, y ∈ A_i+1, i = 1, 2, ..., m, where A_m+1 = A₁ and ϕ, ψ ∈ F.

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 . Let T: X → X be a cyclic (ϕ - ψ)-contraction. Then T has a unique fixed point z ∈ ∩ i = 1 m A i .

Proof. Take x₀ ∈ X and consider the sequence given by

If there exists n₀ ∈ ℕ such that x n 0 + 1 = x n 0 then, since x n 0 + 1 = T x n 0 = x n 0 , the part of existence of the fixed point is proved. Suppose that x_n+1 ≠ x_n for any n = 0, 1, 2, .... Then, since X = ∪ i = 1 m A i , for any n > 0 there exists i_n ∈ {1, 2, ..., m} such that x n - 1 ∈ A i n and x n ∈ A i n + 1 . Since T is a cyclic (ϕ - ψ)-contraction, we have

From 2.1 and taking into account that ϕ is nondecreasing we obtain

Thus {d(x_n, x_n+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 . Taking n → ∞ in (2.1) and using the continuity of ϕ and ψ, we have

and, therefore, ψ(γ) = 0. Since ψ ∈ F, γ = 0, that is,

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, q ≥ n 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

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 . Therefore, d ( x q n , x p n - m ) ≤ ε . Using the triangular inequality

Letting n → ∞ in the last inequality and taking into account that lim_n→∞ d(x_n, x_n+1) = 0, we obtain

Again, by the triangular inequality

Letting n → ∞ in (2.4) and taking into account that lim n → ∞ d ( x n , x n + 1 ) = 0 and (2.4), we get

Since x q n and x p n lie in different adjacently labelled sets A_i and A_i+1 for certain 1 ≤ i ≤ m, using the fact that T is a cyclic (ϕ - ψ)-contraction, we have

Taking into account (2.4) and (2.6) and the continuity of ϕ and ψ, letting n → ∞ in the last inequality, we obtain

and consequently, ψ(ε) = 0. Since ψ ∈ F, 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, q ≥ n₀ with p - q ≡ 1(m)

Since lim n → ∞ d ( x n , x n + 1 ) = 0 we also find n₁ ∈ ℕ such that

for any n ≥ n₁.

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

This proves that (x_n) is a Cauchy sequence. Since X is a complete metric space, there exists x ∈ X such that lim_n→∞ x_n = x. In what follows, we prove that x is a fixed point of T. In fact, since lim n → ∞ x n = x and, as X = ∪ i = 1 m A i 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 x ∈ A_i, Tx ∈ A_i+1 and we take a subsequence x n k of (x_n) with x n k ∈ A i - 1 (the existence of this subsequence is guaranteed by the above-mentioned comment). Using the contractive condition, we can obtain

and since x n k → x and ϕ and ψ belong to F, letting k → ∞ in the last inequality, we have

or, equivalently, ϕ(d(x, Tx)) = 0. Since ϕ ∈ F, 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, z ∈ X with y and z fixed points of T. The cyclic character of T and the fact that y, z ∈ X are fixed points of T, imply that y , z ∈ ∩ i = 1 m A i . Using the contractive condition we obtain

and from the last inequality

Since ψ ∈ F, 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 . Let T: X → X be a cyclic (Id_{[0, ∞)} - ψ) contraction. Then T has a unique fixed point z ∈ ∩ i = 1 m A i .

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, ψ ∈ F), 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 . Let T: X → X 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 .

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 . Let T: X → X be an operator such that

(i) X = ∪ i = 1 m A i is a cyclic representation of X with respect to T .

(ii) There exists k ∈ [0, 1) such that

for any X ∈ A_i, y ∈ A_i+1, i = 1, 2, ..., m where A_m+1 = A₁, and ρ: [0, ∞) → [0, ∞) is a Lebesgue-integrable mapping satisfying ∫ 0 ε ρ ( t ) d t for ε > 0.

Then T has unique fixed point z ∈ ∩ i = 1 m A i .

Proof. It is easily proved that the function ϕ: [0, ∞) → [0, ∞) given by φ ( t ) = ∫ 0 t p ( s ) d s satisfies that ϕ ∈ F. 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: X → X a mapping such that for x, y ∈ X,

where ρ: [0, ∞) → [0, ∞) is a Lebesgue-integrable mapping satisfying ∫ 0 ε ρ ( t ) d t for 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: X → X an operator such that for x, y ∈ X,

where ϕ, ψ ∈ F. Then T has a unique fixed point.

This result appears in [7].

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 ∈ ℕ * ∪ { 0 } 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:

Obviously, X = A₁ ∪ A₂. Let T: X → X be the mapping defined by

It is easily seen that X = A₁ ∪ A₂ is a cyclic representation of X with respect to T. Now we consider the function ρ: [0, ∞) → [0, ∞) defined by

It is easily proved that ∫ 0 t ρ ( s ) d s = t 1 ∕ t for 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 for ε > 0. We take m, n ∈ ℕ* with m ≥ n and we will prove

Since ∫ 0 t ρ ( s ) d s = t 1 ∕ t for t ≤ 1 and, as diam(X) ≤ 1, the last inequality can be written as

or equivalently,

or equivalently,

or equivalently,

In order to prove that this last inequality is true, notice that

and, therefore,

On the other hand, from

we obtain

and, thus,

Since n + m + 1 m - n ≥ 1 ,

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

Now we take x = 1 n , n ∈ ℕ* and y = 0. In this case, condition (ii) of Corollary 9 for k = 1 2 has the form

The last inequality is true since

and, then,

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 ) 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 and φ: [0, ∞) → [0, ∞) a (c)-comparison function. Let T: X → X be an operator and we assume that

(i) X = ∪ i = 1 m A i is a cyclic representation of X with respect to T.

(ii) d(Tx, Ty) ≤ φ(d(x, y)), for any X ∈ A_i and y ∈ A_i+1, where A_m+1 = A.

Then T has a unique fixed point z ∈ ∩ i = 1 m A i .

Now, the contractive condition of Theorem 6 can be written as

for any x ∈ A_i, y ∈ A_i+1, where A_m+1 = A₁, and ϕ, φ ∈ F.

Particularly, if we take ϕ = Id_{[0, ∞)} and φ ( t ) = t 2 1 + t , it is easily seen that ϕ, φ ∈ F. On the other hand,

and

Moreover, for every t ∈ (0, ∞), ∑ k = 0 ∞ ( ϕ - φ ) ( k ) ( t ) 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.

The authors declare that they have no competing interests.

The authors have contributed in obtaining the new results presented in this article. All authors read and approved the final manuscript.

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

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