Implicit-Relation-Type Cyclic Contractive Mappings and Applications to Integral Equations

and Applied Analysis 3 relations on metric spaces have been used in many articles for details see 14–19 and references cited therein . In this section, we define a suitable implicit function involving six real nonnegative arguments to prove our results, that was given in 20 . Let R denote the nonnegative real numbers and let T be the set of all continuous functions T : R6 → R satisfying the following conditions: T1: T t1, . . . , t6 is non-increasing in variables t2, . . . , t6; T2: there exists a right continuous function f : R → R , f 0 0, f t < t for t > 0, such that for u ≥ 0, T u, v, u, v, 0, u v ≤ 0 2.1 or T u, v, 0, 0, v, v ≤ 0 2.2 implies u ≤ f v ; T3: T u, 0, u, 0, 0, u > 0, T u, u, 0, 0, u, u > 0, for all u > 0. Example 2.1. T t1, . . . , t6 t1 − αmax{t2, t3, t4} − 1 − α at5 bt6 , where 0 ≤ α < 1, 0 ≤ a < 1/2, 0 ≤ b < 1/2. Example 2.2. T t1, . . . , t6 t1 − kmax{t2, t3, t4, 1/2 t5 t6 }, where k ∈ 0, 1 . Example 2.3. T t1, . . . , t6 t1 − φ max{t2, t3, t4, 1/2 t5 t6 } , where φ : R → R is right continuous and φ 0 0, φ t < t for t > 0. Example 2.4. T t1, . . . , t6 t1 − t1 at2 bt3 ct4 − dt5t6, where a > 0, b, c, d ≥ 0, a b c < 1 and a d < 1. We need the following lemma for the proof of our theorems. Lemma 2.5 see 21 . Let f : R → R be a right continuous function such that f t < t for every t > 0. Then limn→∞f t 0, where f denotes the n times repeated composition of f with itself. Next, we introduce a new notion of cyclic contractive mapping and establish a new results for such mappings. Definition 2.6. Let X, d be a metric space. Let p be a positive integer, let A1, A2, . . . , Ap be nonempty subsets ofX, and Y ⋃p i 1 Ai. An operator F : Y → Y is called an implicit relation type cyclic contractive mapping if ∗ Y ⋃p i 1 Ai is a cyclic representation of Y with respect to F; ∗∗ for any x, y ∈ Ai ×Ai 1, i 1, 2, . . . , p with Ap 1 A1 , T ( d ( Fx,Fy ) , d ( x, y ) , d x,Fx , d ( y,Fy ) , d ( x,Fy ) , d ( y,Fx )) ≤ 0, 2.3 for some T ∈ T. Using Example 2.2, we present an example of an implicit relation type cyclic contractive mapping. 4 Abstract and Applied Analysis Example 2.7. Let X 0, 1 with the usual metric. Suppose A1 0, 1/2 , A2 1/2, 1 , and A3 A1; note that X ⋃2 i 1 Ai. Define F : X → X such that Fx ⎧ ⎨ ⎩ 1 2 , x ∈ 0, 1 , 0, x 1. 2.4 Clearly,A1 andA2 are closed subsets ofX. Moreover,F Ai ⊂ Ai 1 for i 1, 2, so that ⋃2 i 1 Ai is a cyclic representation of Xwith respect to F. Furthermore, if T : R 6 → R is given by T t1, t2, t3, t4, t5, t6 t1 − 3 4 max { t2, t3, t4, t5 t6 2 } , 2.5 then T ∈ T. We will show that implicit relation type cyclic contractive conditions are verified. We will distinguish the following cases: 1 x ∈ A1, y ∈ A2. i When x ∈ 0, 1/2 and y ∈ 1/2, 1 , we deduce d Fx,Fy 0 and inequality 2.3 is trivially satisfied. ii When x ∈ 0, 1/2 and y 1, we deduce d Fx,Fy 1/2 and t2 |x − 1|, t3 ∣∣∣x − 1 2 ∣∣∣, t4 1, t5 x, t6 1 2 , 2.6 then T t1, t2, t3, t4, t5, t6 1/2 − 3/4. Inequality 2.3 holds as it reduces to 1/2 < 3/4. 2 x ∈ A2, y ∈ A1. i When x ∈ 1/2, 1 and y ∈ 0, 1/2 , we deduce d Fx,Fy 0 and inequality 2.3 is trivially satisfied. ii When x ∈ 1 and y 0, 1/2 , we deduce d Fx,Fy 1/2 and t2 ∣1 − y ∣, t3 1, t4 ∣∣∣y − 1 2 ∣∣∣, t5 1 2 , t6 y. 2.7 Then T t1, t2, t3, t4, t5, t6 1/2 − 3/4. Inequality 2.3 holds as it reduces to 1/2 < 3/4. Hence, F is an implicit relation type cyclic contractive mapping. 3. Main Result Our main result is the following. Abstract and Applied Analysis 5 Theorem 3.1. Let X, d be a complete metric space, p ∈ N, A1, A2, . . . , Ap nonempty closed subsets of X, and Y ⋃p i 1 Ai. Suppose F : Y → Y is an implicit relation type cyclic contractive mapping, for some T ∈ T. Then F has a unique fixed point. Moreover, the fixed point of F belongs to ⋂p i 1 Ai. Proof. Let x0 ∈ A1 such a point exists since A1 / ∅ . Define the sequence {xn} in X by xn 1 Fxn, n 0, 1, 2, . . . . 3.1and Applied Analysis 5 Theorem 3.1. Let X, d be a complete metric space, p ∈ N, A1, A2, . . . , Ap nonempty closed subsets of X, and Y ⋃p i 1 Ai. Suppose F : Y → Y is an implicit relation type cyclic contractive mapping, for some T ∈ T. Then F has a unique fixed point. Moreover, the fixed point of F belongs to ⋂p i 1 Ai. Proof. Let x0 ∈ A1 such a point exists since A1 / ∅ . Define the sequence {xn} in X by xn 1 Fxn, n 0, 1, 2, . . . . 3.1 We will prove that lim n→∞ d xn, xn 1 0. 3.2 If for some k, we have xk 1 xk, then 3.2 follows immediately. So, we can suppose that d xn, xn 1 > 0 for all n. From the condition ∗ , we observe that for all n, there exists i i n ∈ {1, 2, . . . , p} such that xn, xn 1 ∈ Ai ×Ai 1. Then, from the condition ∗∗ , we have T d Fxn,Fxn−1 , d xn, xn−1 , d xn,Fxn , d xn−1,Fxn−1 , d xn,Fxn−1 , d xn−1,Fxn ≤ 0 3.3 and so T d xn 1, xn , d xn, xn−1 , d xn, xn 1 , d xn−1, xn , 0, d xn−1, xn 1 ≤ 0. 3.4 Now using T1, we have T d xn 1, xn , d xn, xn−1 , d xn, xn 1 , d xn−1, xn , 0, d xn−1, xn d xn, xn 1 ≤ 0 3.5 and from T2, there exists a right continuous function f : R → R , f 0 0, f t < t, for t > 0, such that for all n ∈ {1, 2, . . .}, d xn 1, xn ≤ f d xn, xn−1 . 3.6 If we continue this procedure, we can have d xn 1, xn ≤ f d x1, x0 3.7 and so from Lemma 2.5, lim n→∞ d xn 1, xn 0. 3.8 Next we show that {xn} is a Cauchy sequence. Suppose it is not true. Then we can find a δ > 0 and two sequences of integers {m k }, {n k }, n k > m k ≥ k with rk d ( xm k , xn k ) ≥ δ for k ∈ {1, 2, . . .}. 3.9 6 Abstract and Applied Analysis We may also assume d ( xm k , xn k −1 ) < δ 3.10 by choosing n k to be the smallest number exceedingm k for which 3.9 holds. Now 3.7 , 3.9 , and 3.10 imply δ ≤ rk ≤ d ( xm k , xn k −1 ) d ( xn k −1, xn k ) < δ f k −1 d x0, x1 3.11 and so lim k→∞ rk δ. 3.12 On the other hand, for all k, there exists j k ∈ {1, . . . , p} such that n k −m k j k ≡ 1 p . Then xm k −j k for k large enough, m k > j k and xn k lie in different adjacently labelled sets Ai and Ai 1 for certain i ∈ {1, . . . , p}. Using the triangle inequality, we get ∣d ( xm k −j k , xn k ) − d ( xn k , xm k )∣ ≤ d ( xm k −j k , xm k )