A fixed point theorem involving rational expressions without using Picard iteration

: In this paper, we consider a certain fixed point theorem that contains some rational expressions. The main aim of this paper is to prove a fixed point theorem without using the Picard iteration.


Introduction and Preliminaries
Let T be a self-mapping on a Banach space (X , · ) and x ∈ X . The sequence {xp} is called Picard sequence if xp = Tx p− for n = , , · · · . In this case, the mapping T is said to be Picard operator.
In [1], the authors set out a di erent general principle for obtaining xed points. We recall the main results of their paper. Theorem 1. [1] Let T be self-mapping on a nonempty closed convex subset C of a Banach space X . If for some x ∈ C there exists a constant c, ≤ c < such that [2]- [6]. Among them, we recall the following essential result.
By Ψ we will denote the set of all (c)-comparison functions. Very recently, Karapınar ([7]) announced the following interesting result.

Theorem 4. [7]
On a nonempty closed convex subset C of a Banach space X , let T be a self-mapping. If for some x ∈ C, there exists a φ ∈ Ψ such that for all p su ciently large, then u is a xed point of T.

Main results
First, we demonstrate the following useful Lemma: Lemma 5. Let C be a nonempty closed convex subset of a Banach space X and the sequence {xp} in C. If for some x ∈ C there exists a function φ ∈ Ψ such that then the sequence {xp} converges to a point u ∈ C.
Proof. Let x ∈ C be such that the inequality (2.1) holds. Thus, recursively, from the inequality (2.1) and taking into account the monotony of the function φ, it follows that and by (c ), we have that lim p→∞ x p+ − xp = , so the sequence {xp} is asymptotically regular. Moreover, by the triangle inequality and taking (c ) into account, we get where Up = p j=n φ j ( x − x ) . Thereupon, {xp} is a Cauchy sequence and since, the subset C is closed, we deduce that {xp} converges to a point in u ∈ C. Theorem 6. Let C be a nonempty closed convex subset of a Banach space X , T be a self-mapping on C, x be arbitrary in C and the sequence {xn} in C de ned as (

2.4)
Then the sequence {xp} is convergent provided that there exists φ ∈ Φ such that Moreover if u = lim p→∞ xp and there exist the constants c , c ≥ , ≤ λ < , such that for p su ciently large, then u is a xed point of T.
Proof. Let x ∈ C and the sequence {xp} be de ned by (2.4). Thus, keeping in mind (2.10), by Lemma 5 it follows that {xp} is a convergent sequence. Let u ∈ C be the limit of xp. We claim that u is a xed point of the mapping T. Indeed, we can easily see that (2.4) can be rewritten as Txp − xp = (x p+ − xp). Therefore, But, since lim p→∞ xp = u, taking into account the uniqueness of the limit we get that lim p→∞ Txp = u. Now, letting the limit of (2.11) as p → ∞, we have which is a contradiction. Thereupon, u = Tu.
Adding a supplementary condition in Theorem 6, we can assure the uniqueness of the xed point.

Theorem 7.
If in Theorem 6 the constants c , c , λ ≥ are such that c + c + λ < then the xed point of the mapping T is unique.
Proof. Let {xp} be the sequence de ned by (2.4). We know by the previous proof, that {xp} is convergent to a point u ∈ C. More than that, lim p→∞ xp = lim p→∞ Txp = u and Tu = u.
Supposing than there exists a point v ∈ C such that Tv = v ≠ u, by (2.11) we have (2.8) Taking the limit as p → ∞ in (2.8) we get This is a contradiction, so that u = v.
Corollary 8. Let C be a nonempty closed convex subset of a Banach space X , T be a self-mapping on C, x be arbitrary in C and the sequence {xn} in C de ned as Then the sequence {xp} is convergent provided that there exists ≤ κ < such that  Proof. Put φ(t ) = κt , with ≤ κ < , in Theorem 6.
Corollary 9. Let C be a nonempty closed convex subset of a Banach space X , and T, G : C → C be two mappings such that Gx = x+Tx , for any x ∈ C. Supposing that there exists < Υ < such that for any x, y ∈ C. If there exist two positive real numbers a, b, with a < Υ and b < such that

13)
then there exists at least one point u ∈ C such that Tu = u.
Proof. First of all, by considering the de nition of the mapping G, we can easily obtain the following relations: (2.14) Let x be an arbitrary but xed point in C and the sequence {xp} de ned as follows: (2.15) Using this notation, the relations (2.14) become (2.16) Thus, In case that max xp − Tx p+ , Txp − Tx p+ = Txp −Tx p+ , by (2.12) and keeping in mind (2.16), we obtain (2.17) Of course, since in case that max xp − x p+ , x p+ − x p+ = x p+ , x p+ , we get a contradiction, it follows that In the second case, when max x p − Tx p+ , Tx p − Tx p+ = x p − Tx p+ , we have But, from (2.13), we get xp − Tx p+ ≤ a T xp − Tx p+ , and then Consequently, since aΥ < , we get Thereupon, if we denote κ = max Υ , aΥ , from (2.18) and (2.18) we have On the other hand, from the inequality (2.12), for x = xp and y = u, we get Moreover, choosing c = Υ , c = Υ and λ = Υ , for p su ciently large, we get (2.11) and then, by Theorem 6 it follows that u is a xed point of T.
Corollary 10. Let C be a nonempty closed convex subset of a Banach space X , and T : C → C be a mapping such that T = I. If there exists ≤ Υ < such that

22)
for any x, y ∈ C then we can nd at least one point u ∈ C such that Tu = u.
Proof. Let x be an arbitrary but xed point in C and the sequence {xp} de ned by (2.15). On the other hand, by (2.16) we have: Thus, and we distinguish two cases: Case 1. max xp − Tx p+ , Txp − Tx p+ = Txp − Tx p+ . Then, by (2.22), keeping in mind (2.16) and following the same arguments as in Corollary 9 we get We have Consequently, since Υ < , we get (2.24) and letting κ = max Υ , aΥ , from (2.23) and (2.24) we have Moreover, choosing c = Υ , c = Υ and λ = Υ , for p su ciently large, we get (2.11) and then, by Theorem 6 it follows that u is a xed point of T.