Different common fixed point theorems of integral type for pairs of subcompatible mappings

Abstract. In this paper, a general common fixed point theorem for two pairs of subcompatible mappings satisfying integral type implicit relations is obtained in a metric space. Our result improves several results especially the result of Pathak et al. [6]. Also, another common fixed point theorem of Greguš type for four mappings satisfying a contractive condition of integral type in a metric space using the concept of subcompatibility is established which generalizes the result of Djoudi and Aliouche [1] and others. Again a third common fixed point theorem for two pairs of near-contractive subcompatible mappings is given which enlarges the result of Mbarki [5] and references therein.


Introduction
Let (X , d) be a metric space and let f , g be two mappings from X into itself.f and g commute if f gx = gf x for all x ∈ X .
This commutativity was weakened in 1982 by Sessa [7] with the notion of weakly commuting mappings.f and g above are weakly commuting if d(f gx, gf x) ≤ d(gx, f x) for all x in X .
Later on, Jungck [3] enlarged the class of commuting and weakly commuting mappings by compatible mappings which asserts that the above mappings f and g are compatible if lim n→∞ d(f gx n , gf x n ) = 0 whenever {x n } is a sequence in X such that lim n→∞ f x n = lim n→∞ gx n = t for some t ∈ X .This concept was further improved by Jungck [4] with the notion of weakly compatible mappings.f and g are weakly compatible if f t = gt for some t ∈ X implies that f gt = gf t.
Recently in 2007, Pathak et al. [6] stated and proved a general common fixed point theorem of integral type for two pairs of weakly compatible mappings satisfying integral type implicit relations in a symmetric space.
Our aim here is to improve and extend the result of [6] by using the new concept of mappings called subcompatibility which enlarges the concept of weakly compatible mappings.
We introduce the notion of subcompatible mappings as follows: Let f and g be two self-mappings of a metric space (X , d). f and g are subcompatible if and only if there exists a sequence {x n } in X such that lim n→∞ f x n = lim n→∞ gx n = t for some t ∈ X and lim n→∞ d(f gx n , gf x n ) = 0.It is clear to see that weakly compatible mappings are subcompatible, however the implication is not reversible.
Example 1.1.Let X = [0, ∞) with the usual metric d.Define f , g : X → X as follows Let {x n } be a sequence in X defined by x n = 4+ 1 n for n ∈ N * = {1, 2, ...}.Then, we have and Hence, f and g are subcompatible mappings.
Thus, f and g are not weakly compatible.
For our first main result we need the following implicit relations.

Implicit relations
Let R + be the set of all nonnegative real numbers, Ψ be the family of all ψ : R + → R Lebesgue-integrable and summable mappings and Φ be the set of all real continuous functions ϕ : R 6 + → R satisfying the following conditions: (ϕ 1 ) for all u, v ≥ 0, if , where k ∈ (0, 1) and ψ(t) = t.Then ϕ is continuous and ψ is a Lebesgue-integrable mapping which is summable.We have for u > 0.

Main results
Theorem 3.1.Let f , g, h and k be four mappings of a metric space (X , d) into itself such that (1) ϕ(d(f x,gy),d(hx,ky),d(f x,hx),d(gy,ky),d(ky,f x),d(hx,gy)) 0 ψ(t) d t ≤ 0, for all x, y in X , where ϕ ∈ Φ and ψ ∈ Ψ. Suppose that (f, h) and (g, k) are subcompatible and h and k are continuous, then, f , g, h and k have a unique common fixed point.
Proof.Since the pairs (f, h) and (g, k) are subcompatible, then, there exist two sequences {x n } and {y n } in X such that lim Since ϕ is continuous, we obtain at infinity Since h is continuous, then h 2 x n → ht, hf x n → ht.Also we have Since f and h are subcompatible, taking the limit as n → ∞ in the above inequality we have lim n→∞ f hx n = ht.The use of condition (1) gives At infinity we obtain which contradicts (ϕ 2 ).Hence ht = t.Again using (1) we get Taking the limit as n → ∞, we get Using condition (1) we have When n tends to infinity, we get If gt = t, using inequality (1) we have which implies d(t, gt) = 0 by using condition (ϕ a ).Thus, gt = t.
For the uniqueness of common fixed point t, let z = t be another common fixed point of f , g, h and k.Then using (1) we obtain ϕ(d(f t,gz),d(ht,kz),d(f t,ht),d(gz,kz),d(kz,f t),d(ht,gz)) which is a contradiction of (ϕ 2 ).Therefore z = t.Corollary 3.1.Let (X , d) be a metric space and let f and h be two mappings from X into itself satisfying the condition for all x, y in X , where ϕ ∈ Φ and ψ ∈ Ψ.If h is continuous and the pair (f, h) is subcompatible, then, f and h have a unique common fixed point.Corollary 3.2.Let (X , d) be a metric space and let f , g and h be three self-mappings of X such that Now, let F be the family of mappings F : R + → R + such that each F is upper semi-continuous and F (t) < t for all t > 0 and let Ω be the family of ω : R + → R + such that every ω is a Lebesgue-integrable mapping which is summable and 0 ω(t) d t > 0 for each > 0.
In their paper [1], Djoudi and Aliouche proved a common fixed point theorem of Greguš type for four mappings satisfying a contractive condition of integral type in a metric space using the concept of weak compatibility.
Our objective here is to improve, extend and generalize the result of [1] by using the notion of subcompatibility.
Theorem 3.3.Let f , g, h and k be mappings from a metric space (X , d) into itself satisfying inequality for all x, y in X , where 0 < a < 1, p is an integer such that p ≥ 1, F ∈ F and ω ∈ Ω.If h and k are continuous and the pairs (f, h) and (g, k) are subcompatible, then, f , g, h and k have a unique common fixed point.
Proof.Since the pair (f, h) as well as (g, k) is subcompatible, then, there are two sequenses {x n } and {y n } in X such that lim First, we prove that z = t.If t = z, using inequality (2) we get Letting n → ∞, we obtain Since h is continuous, then we have h 2 x n → ht, hf x n → ht.Also, we have As f and h are subcompatible, letting n tends to infinity in the above inequality, we obtain lim n→∞ f hx n = ht.The use of condition (2) gives Different common fixed point theorems of integral type for . . .
We obtain at infinity Again by inequality (2) we have At infinity we obtain We get at infinity This contradiction implies that kt = t.Suppose that gt = t, the use of inequality (2) gives which is a contradiction.Hence gt = t.Therefore t = z is a common fixed point of both f , g, h and k.Suppose that f , g, h and k have another common fixed point z = t.Then, by inequality (2) we get This contradiction implies that z = t.
If f = g and h = k in Theorem 3.3, we get the next result: Corollary 3.3.Let f and h be two self-mappings of a metric space (X , d) such that for all x, y in X , where 0 < a < 1, p is an integer such that p ≥ 1, F ∈ F and ω ∈ Ω.If h is continuous and the pair (f, h) is subcompatible, then, f and h have a unique common fixed point.
If we let in Theorem 3.3 h = k, then we get the following corollary: Corollary 3.4.Let f , g and h be three self-mappings of a metric space (X , d) such that for all x, y in X , where 0 < a < 1, p is an integer such that p ≥ 1, F ∈ F and ω ∈ Ω.If h is continuous and the pairs (f, h) and (g, h) are subcompatible, then, f , g and h have a unique common fixed point.
The next result is a generalization of Theorem 3.3.We end our work by establishing another result which improves, extends and generalizes especially the result of [5].
Theorem 3.5.Let (X , d) be a metric space, f , g, h and k be mappings from X into itself and be an upper semi-continuous function of [0, ∞) into itself such that (t) = 0 if and only if t = 0 and satisfying inequality If the pairs (f, h) and (g, k) are subcompatible and h and k are continuous, then, f , g, h and k have a unique common fixed point.
Proof.Since the pairs (f, h) and (g, k) are subcompatible, then, there exist two sequences {x n } and {y n } in X such that lim First, we prove that z = t.Suppose that (d(t, z)) > 0, using inequality (3) we get Taking the limit as n → ∞, we obtain which is a contradiction.Hence (d(t, z)) = 0 which implies that d(t, z) = 0, thus t = z.
Since h is continuous, then, we have h 2 x n → ht, hf x n → ht.Also, we have d(f hx n , ht) ≤ d(f hx n , hf x n ) + d(hf x n , ht).As f and h are subcompatible, letting n tends to infinity in the above inequality, we obtain lim Letting n → ∞ we obtain  This contradiction implies that (d(ht, t)) = 0 and hence ht = t.
n→∞ hx n = lim n→∞ f x n = t for some t ∈ X and lim n→∞ d(f hx n , hf x n ) = 0; lim n→∞ gy n = lim n→∞ ky n = z for some z ∈ X and lim n→∞ d(gky n , kgy n ) = 0.
n→∞ f x n = lim n→∞ hx n = t for some t ∈ X and lim n→∞ d(f hx n , hf x n ) = 0; lim n→∞ gy n = lim n→∞ ky n = z for some z ∈ X and lim n→∞ d(gky n , kgy n ) = 0.
Now, since k is continuous, then, we have k 2 y n → kt and kgy n → kt and d(gky n , kt) ≤ d(gky n , kgy n ) + d(kgy n , kt).Since the pair (g, k) is subcompatible, we get at infinity lim d t p < d(f t,t) 0 ω(t) d t p ,which is a contradiction.Hence f t = t.n→∞ gky n = kt.