WEAKLY T F TYPE CONTRACTIVE MAPPINGS

In this paper, the concept of weakly TF -contractive conditions are considered for the Banach, Kannan and Chatterjea fixed point theorems. It is shown that these mappings have a unique fixed point in a complete metric space.


Introduction and Preliminaries
In 1922, Banach proved his famous theorem which ensures the existence and uniqueness of the fixed point.
A mapping T : X → X, where (X, d) is a metric space, is said to be a contraction if there exists k ∈ [0, 1) such that for all x, y ∈ X, d (T x, T y) ≤ kd (x, y) . (1.1) If the metric space (X, d) is complete then the mapping satisfying (1.1) has a unique fixed point.question is that whether one can find a contractive condition which will imply the existence of the fixed point but will not imply continuity of the mapppıng.Kannan [2] established the following result in which the question has been answered in the affirmative.Theorem 1. [2] If a mapping T : X → X where (X, d) is a complete metric space, satisfies the inequality where a ∈ 0, 1 2 and x, y ∈ X, then T has a unique fixed point.
A similar contractive condition has been introduced by Chatterjea [3] as following: such that b ∈ 0, 1 2 and x, y ∈ X, then T has a unique fixed point.
In 2010, Moradi and Beiranvand gave the following result [8]: Theorem 3. ( T F -Contraction Mapping Theorem ) Let (X, d) be a complete metric space and T, f : X → X be self-mappings such that T is one-to-one and graph closed (or subsequentially convergent and continuous ).Let f satisfying the inequality where α ∈ [0, 1) and F : [0, ∞) → [0, ∞) is nondecreasing continuous from the right and F −1 (0) = {0} .Then f has a unique fixed point in the complete metric space (X, d) .
In this study our purpose is to introduce weakly T F contractive conditions for Banach fixed point theorem, Kannan fixed point theorem and Chatterjea fixed point theorem.Definition 1. [4] Let (X, d) be a metric space.A mapping T : X → X is said to be sequentially convergent if we have, for every sequence {y n }, if {T y n } converges then {y n } is also convergent.T is said to be subsequentially convergent if we have, for every sequence {y n }, if {T y n } converges then {y n } has a convergent subsequence.

Main Results
For the simplicity, we will use the following symbols.
3) Also, we denote by SSC (X) the set of all mappings T : X → X such that T is one-to-one , continuous and subsequentially convergent, by SC (X) the set of all mappings T : X → X such that T is one-to-one , continuous and sequentially convergent.Theorem 4. ( Weakly T F Contractive Mapping Theorem ) Let (X, d) be a complete metric space and f : X → X be a mapping.Let T ∈ SSC and f satisfy the inequality where F ∈ ̥, ψ ∈ Ψ.Then f has a unique fixed point in X.Also, if T is sequentially convergent then, for every x 0 ∈ X the sequence of iterates {f n x 0 } converges to the fixed point.
Proof.Let x 0 be an arbitrary point in X.We define the sequence {x n } by It is clear {d (T x n , T x n+1 )} is a monotone decreasing sequence, and consequently there exists an r ≥ 0 such that Letting n → ∞ in (2.2), we obtain that F (r) ≤ F (r) − ψ (r).This case implies that r = 0. Now,we will prove that {T x n } is a Cauchy sequence.Suppose that {T x n } is not a Cahucy sequence.Then there exists an ǫ > 0 and there exist subsequences (2.4) Further, corresponding to m (k), we can choose n (k) in such a way that it is the smallest integer with n (k) > m (k) and Also, using (2.4) we have Again, we have letting k → ∞ in (2.8) and (2.9), we have This implies that lim Substituting (2.4) in (2.1) use the (2.7), (2.11) in both (2.12) and (2.13)

.14)
This implies that ǫ = 0.But this case is a contradiction.Thus {T x n } is a Cauchy sequence in the complete metric space X.Hence, there is v ∈ X such that lim Since T is subsequentially convergent, {x n } has a convergent subsequence.Thus there is an u ∈ X such that Since T is continous and x n(k) → u, we have (2.17) Since T x n(k) is a subsequence of {T x n }, so that T u = v.Now, we will show that u ∈ X is a fixed point of f.
Letting k → ∞ in (2.18) and using the (2.16), (2.17) we have this implies that F (d (T u, T f u)) = 0.As T is one-to-one, then f u = u.To prove the uniqueness of the fixed point, assume that u ′ ∈ X is an other fixed points of f.Thus we have f u ′ = u ′ and (2.20) Inequality (2.20) is a contradiction unless ψ (d (T u, T u ′ )) = 0.This implies that T u = T u ′ .Since T is one-to-one u = u ′ , that is, the fixed point is unique.Also, if T is sequentially convergent, by replacing {n} with {n (k)} we obtain that lim This implies that {x n } converges to the fixed point of f.Remark 1.In Theorem 4, if we take ψ (t) = kF (t) where t ∈ [0, ∞) and k ∈ (0, 1], we obtain the above result that given by Moradi and Beiranvand [8].
If we take T x = x, we obtain the following result given by Dutta and Choudhury [6].
Corollary 1.Let (X, d) be a complete metric space and let f : X → X be a self-mapping satisfying the inequality where ψ,φ : [0, ∞) → [0, ∞) are both continuous and monotone nondecreasing functions with ψ (t) = 0 = φ (t) if and only if t = 0. Then f has a unique fixed point.
Corollary 2. Let (X, d) be a complete metric spaces and let f : X → X be a mapping such that for each x, y ∈ X and c ∈ (0, 1), where ϕ : [0, ∞) → [0, ∞) is a Lebesgue-integrable mapping which summeble ( i.e., with finite integral ) on each compact subset of [0, ∞), nonnegative, and such that for each ǫ > 0, ǫ 0 ϕ (t) dt > 0; then f has a unique fixed point a ∈ X such that for each x ∈ X, lim n→∞ f n x = a.Theorem 5. ( Weakly T F Kannan Contractive Mapping Theorem ) Let (X, d) be a complete metric space and f : X → X be mapping.Let T ∈ SSC and f satisfies the inequality where F ∈ ̥, ψ ∈ Ψ.Then f has a unique fixed point.Also, if T is sequentially convergent then for x 0 ∈ X the sequence of iterates {f n x 0 } converges to this fixed point.
Proof.Let x 0 ∈ X and {x n } be a sequence in X defined by ). (2.24) Note that F is nondecreasing continuous, then we have As the sequence {d (T x n , T x n+1 )} is a monotone decreasing sequence of non-negative real numbers, thus By following the similar method in the proof of the Theorem 4, we obtain that {T x n } is a Cauchy sequence in the complete metric space (X, d) .Therefore there is an u ∈ X such that x n → u and T x n → v (as n → ∞).Also, from the (2.21), we have letting n → ∞ in (2.26), we have ψ (d (T x n , T f x n ) , d (T u, T f u)) ≤ 0. Thus we get u ∈ X is a fixed point of f.It is easy to see uniqness of the fixed point.
If we take F s = s, then we obtain the following result, given by Moradi and Davood [7].Corollary 3. Let (X, d) be a complete metric space and T, S : X → X be mappings such that T is continuous, one to one and subsequentially convergent.If µ ∈ 0, then, S has a unique fixed point.Also, if T is sequentially convergent then for every x 0 ∈ X the sequence of iterates {S n x 0 } converges to the fixed point.t.
Theorem 6. ( Weakly T F Chatterjea Contractive Mapping Theorem ) Let (X, d) be a complete metric space and f : X → X be a mapping.Let T ∈ SSC and f satisfies the inequality where F ∈ ̥, ψ ∈ Ψ.Then f has a unique fixed point.Also if T is sequentially convergent then for every x 0 ∈ X the sequence of iterates {f n x 0 } converges to this fixed point.
Proof.Let x 0 ∈ X and {x n } be a sequence in X defined by x n+1 = f x n , such that n = 1, 2, • • • .From the (2.27), we have From the (2.28), we have Letting n → ∞ respectively, in (2.32), (2.33), (2.34) we obtain Also, letting n → ∞ in the (2.28) we obtain that ψ (r, 0) = 0.This implies that r = 0. Following the similar process in the proof of the Theorem 4, we obtain that {T x n } is a Cauchy sequence in the complete metric space (X, d) and there is an u ∈ X such that x n → u and T x n → v (as n → ∞).
Also, from the 2.27, we have letting k → ∞ in 2.36, it follows that ψ (d (T u, T f u) , d (T u, T f u)) ≤ 0. This implies that T u = T f u.Note that T is one-to-one so f u = u.Also, it is easy to see uniqness of the fixed point.
If we take T x = x and F x = x then, we obtain the following result given by Choudhury [2].
It is clear that the inequality (1.1) implies the continuity of T. A natural Received: December 29, 2014 c 2015 Academic Publications, Ltd. url: www.acadpubl.eu§ Correspondence author

1 2
and x, y ∈ X, d (T Sx, T Sy) ≤ µ [d (T x, T Sy) + d (T y, T Sx)] a monotone decreasing sequence of non-negative real numbers.Hence there is an r ∈ R such that lim n→∞ d (T x n , T x n+1 ) = r.(2.31)

F
(d (T x n+1 , T f u)) = F (d (T f x n , T f u)) T x n , T f u) + d (T u, T f x n )] − ψ (d (T x n , T f u) , d (T u, T f x n ))(2.36)