Fixed Points for Contractive Mappings of Integral Type Involving ω -Distance and Applications

In this paper, we use ω -distance to prove the existence, uniqueness, and iterative approximations of ﬁ xed points for a few contractive mappings of integral type in complete metric spaces. The proved results are used to investigate the solvability of certain nonlinear integral equations. Four examples are given.

Kada et al. [8] introduced the concept of ω-distance in metric spaces and proved a few fixed point theorems for some contractive mappings by using ω-distance. It is clear that the results in [8] extended the Caristi's fixed point theorem, Ekeland's ε-variational's principle, and the nonconvex minimization theorem. The researchers in [3, 5-7, 9, 10, 17] got several fixed point results for certain contractive mappings with respect to ω-distance.
In this paper, we prove the existence, uniqueness, and iterative approximations of fixed points for several kinds of mappings, which satisfy some contractive conditions of integral type with respect to ω-distance in complete metric spaces. We also construct four illustrative examples and give applications of the obtained results in nonlinear Fredholm and Volterra integral equations, respectively. Our results generalize or differ from the corresponding fixed point theorems in [1,14,15].

Preliminaries
Let ℕ denotes the set of all positive integers, R = ð−∞, + ∞Þ, R + = ½0,+∞Þ, N 0 = N ∪ f0g and Recall that a self-mapping f in a metric space X is called orbitally continuous if lim n→∞ f n x = u implies lim n→∞ f n+1 x = f u for each f f n xg n∈N 0 ⊆ X and u∈X.

Fixed Point Results with respect to ω -Distance
In this section, using ω-distance, we give four fixed point theorems for the contractive mappings (2) Here, ðw, ψÞ ∈ Φ 1 × Φ 4 . Then, f possesses a unique fixed point u ∈ X such that pðu, uÞ = 0, lim n→∞ p f n x 0 , u ð Þ= 0 and lim n→∞ f n x 0 = u for each x 0 ∈ X: Proof. Firstly, we claim the existence of fixed points of f in X. Put x 0 ∈ X and x n = f n x 0 for each n ∈ N 0 . Now, we need to think over two situations as follows: Case 2. x n 0 = x n 0 −1 for some n 0 ∈ N. Clearly, x n 0 −1 is a fixed point of f and lim n→∞ f n x 0 = x n 0 −1 . Suppose that pðx n 0 −1 , x n 0 −1 Þ > 0. Making use of (2) and ðw, ψÞ ∈ Φ 1 × Φ 4 , we obtain that which is ridiculous. Hence, pðx n 0 −1 , x n 0 −1 Þ = 0, which means that Case 3. x n ≠ x n−1 for all n ∈ N. Suppose that (2), (6), and ðw, ψÞ ∈ Φ 1 × Φ 4 ensure that that is, The above equation and w ∈ Φ 1 give that Combining (6), (9), and ðp 1 Þ, we know that that is, Because of Lemma 1 in [8], (6), and (11), we deduce that x n 0 = x n 0 +1 , which is contradictive, and, hence, By means of ðw, ψÞ ∈ Φ 1 × Φ 4 , (2) and (12), we have which together with (12) and w ∈ Φ 1 ensures that We see from (14) that fpðx n , x n+1 Þg n∈N 0 is a positive and strictly decreasing sequence. It follows that Now, we claim that c = 0. Otherwise, c > 0. In view of Lemma 2.1 in [12], (2), (15), and ðw, ψÞ ∈ Φ 1 × Φ 4 , we see that 2 Journal of Function Spaces It is ridiculous. Therefore, c = 0. Consequently, In the same way, we have Now, we proceed to show that lim n,m→∞ Suppose that there exists a real number ε > 0 such that for every k ∈ N, there exist mðkÞ, nðkÞ ∈ N satisfying For each k ∈ N, mðkÞ denotes the least integer exceeding nðkÞ and satisfying (20). Obviously, On account of ðp 1 Þ and (21), we obtain that Letting k tend to infinity in (22) and taking advantage of In light of Lemma 2.1 in [12], (2), (23), and ðw, ψÞ ∈ Φ 1 × Φ 4 , we deduct that It is ridiculous. Of course, (19) is true. Assume that ε > 0 and δ denotes the real number appearing in (3) of [8]. By means of (19), we infer that there is N ∈ N satisfying which ensures that So fx n g n∈N 0 is a Cauchy sequence. Completeness of X means that lim n→∞ x n = u for some u ∈ X: According to (19), we are aware of the fact that for any which together with (27) gives that It follows that Taking account of (2), (30), ðw, ψÞ ∈ Φ 1 × Φ 4 , and Lemma 2.1 in [12], we infer that 3 Journal of Function Spaces It follows that Lemma 2.2 in [12] and the above equation give that We get from (1) in [8] and (17) that Clearly, Applying (30), (35), and Lemma 1 in [8], we gain that u = f u.

Theorem 4.
Let p be a ω-distance in a complete metric space ðX, dÞ and let f : X → X satisfy that Here, ðw, ψÞ ∈ Φ 2 × Φ 4 . Then, f possesses a unique fixed point u ∈ X such that pðu, uÞ = 0, Proof. Firstly, we show that f possesses fixed points in X. Let x 0 ∈X and x n = f n x 0 for each n ∈ N 0 . Now, we divide the proof into two steps.

Four Examples
Now, we give four examples to explain the fixed point results obtained in Section 3.

Journal of Function Spaces
If there is ψ ∈ Φ 3 satisfying the conditions of Theorem 1 in [15], we know that which is absurd. If there are c ∈ ð0, 1Þ and w ∈ Φ 1 satisfying the conditions of Theorem 2.1 in [1], we attain that 0 < Example 16. Let X = R + , dðx, yÞ = |x − y | and pðx, yÞ = x + y , ∀x, y ∈ X. Let f : X → X, w and ψ : R + → R + be defined by, respectively, and w r ð Þ = 2r, ψ r ð Þ = r 2 ,∀r ∈ R + : Evidently, p is a ω-distance in X and ðw, ψÞ ∈ Φ 2 × Φ 4 . Let x, y ∈ X. For the sake of verifying (38), we take into account the following four possible cases: That is to say, (38) is true. Therefore, the conditions of Theorem 4 are fulfilled. Consequently, Theorem 4 means that f possesses a unique fixed point in X. However, we cannot use Theorem 2.1 in [14] to show the existence of fixed points for the mapping f in X. Or else, there is ðw, ψÞ ∈ Φ 1