Some Fixed Point Theorems under Generalized Expansion Principle with Control Function

ISSN: 2347-1557

Available Online: http://ijmaa.in/

ppli

tio ca

•I ns

SS N:

47 23

55 -1

•7

Inter

International Journal of Mathematics And its Applications

Some Fixed Point Theorems under Generalized Expansion Principle with Control Function

Research Article

Ritu Sahu¹, P L Sanodia^2∗ and Arvind Gupta³

1 Department of Mathematics, People’s College of Research & Technology, Bhopal, India.

2 Department of Mathematics, Institute for Excellence in Higher Education, Bhopal, India.

3 Department of Mathematics, Govt. Motilal Vigyan Mahavidhyalaya, Bhopal, India.

Abstract: We prove common fixed point theorems for semi and weak compatible mapping satisfying a generalized expansion principle by using a control function. Our theorems generalize recent results existing in the literature.

MSC: 47H10; 54H25.

Keywords: Generalized expansion principle, Weak compatible, Semi compatible, Commute map, Control function.

c

JS Publication.

1. Introduction

Generalizing Banach contraction principle in various ways has become a recent research interest and has been studied by many authors. For example, One may refer [2,3,7,10,12] and [14]. [1] has proved a generalization for weakly contractive mapping in Hilbert space which was proved by [10] in the setup of complete metric space.

On the other hand, [7] and [9] proved fixed point theorem for a self mapping by altering distances between the point and using a control function, whereas [12] extended the concept for weakly commuting pairs of self mapping and proved common fixed point theorem in a complete metric space by using the control function.

More recently, [3] have obtained a fixed point result by generalizing the concept of control function and the weakly contractive mapping. [4] proved a common fixed point theorem for commuting mapping generalizing the Banach’s contraction principle.

[13] introduced, “Weakly commuting mapping” which was generalized by [5] as, “Compatible mapping” [8] coined the notion of, “R-weakly commuting mapping”, whereas [6] defined a term called, “weakly compatible mapping”.

In this paper we prove some fixed point theorems using generalized expansion principal with control function and generalize the work of [11].

2. Definition and Preliminaries

Definition 2.1. Two self mappings T and F of a metric space (X, D) are said to be weak compatible, if T F x = F T x whenever F x = T x for all x ∈ X.

∗ E-mail: [email protected]

Definition 2.2 ([7, 9]). A control function φ is defined as φ : R⁺ → R⁺ which is continuous at zero, monotonically increasing and φ(t) = 0 if and only if t = 0.

Definition 2.3 ([2]). A self mapping T of metric space (X, D) is said to be weakly contractive with respect to a self mapping f : X → X, for each x, y ∈ X, d(T x, T y) ≤ d(f x, f y) − φ(d(f x, f y)), where φ : [0, ∞) → [0, ∞) is continuous and nondecreasing function such that φ is positive on (0, ∞), φ(0) = 0 and lim

t→∞φ(t) = ∞.

If F = I, the identity mapping, then the Definition2.3reduces to the definition of weakly contractive mapping given by [1]

and [10]. Combining the generalization of Banach contraction principle given by [7] and the generalization given by [3] and [10] obtained the following result.

Theorem 2.4 ([3]). Let (X, D) be a complete metric space and T : X → X be a self map mapping satisfying ϕ(d(T x, T y)) ≤ ϕ(d(f x, f y)) − φ(d(f x, f y)), where φ, ϕ : [0, ∞) → [0, ∞) are both continuous and monotone decreasing functions with ϕ(t) = 0 = φ(t) if and only if t = 0. Then T has a unique fixed point.

Here we see a following lemma which helps us to prove main result.

Lemma 2.5. Let (X, D) be a complete metric space and T : X → X or F : X → X be continuous self map satisfying ϕd(T x, T y) ≥ ϕd(F x, F y))+φd(T x, T y), where φ, ϕ : [0, ∞) → [0, ∞) are both continuous and monotone increasing functions with ϕ(t) = 0 = φ(t) ⇔ t = 0. If (F, T ) is semi compatible then T, F have unique common fixed point.

3. Main Results

Theorem 3.1. Let T andF be self mapping of metric space (X, D) with

(a) T (X) ⊂ F (X)

(b) ϕ[d(T x, T y)] ≥ ϕ[d(T x, F x)) + d(T x, F y)]φ[d(T x, T y)]

(c) Either T or F is continuous function.

(d) (T, F ) is semi compatible and weak compatible.

If ϕ and φ are monotonic increasing function such that φ, ϕ : [0, ∞) → [0, ∞) and ϕ(t) = 0 = φ(t) ⇔ t = 0 then z is unique common fixed point of F and T.

Proof. Let x0 ∈ X is an arbitrary point. Since T (X) ⊂ F (X). Then x1 ∈ X such that T x1= F x0. Inductively we can define a sequence T xn+1= F xn.

Using (b) with x = xn, y = xn+1

φ[d(T xn, T xn+1)] ≥ φ[d(T xn, F xn)) + d(T xn, F xn+1)] + ϕ[d(T xn, T xn+1)

≥ φ[d(T xn, T xn+1)) + d(T xn, T xn+2)] + ϕ[d(T xn, T xn+1)]

By triangle inequality we have [d(T xn+1, T xn+2)] ≤ [d(T xn+1, T xn)) + d(T xn, T xn+2)]. Then

φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)] + ϕd(T xn, T xn+1)] (1) φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)].

Since φ is an increasing function then we have d(T xn, T xn+1) ≥ d(T xn+1, T xn+2). Therefore the sequence d(T xn, T xn+1) will be decreasing. Let r ≥ 0 such that

n→∞lim d(T xn, T xn+1) = r. (2)

Hence on taking lim n → ∞ we have by (1) φ(r) ≥ φ(r) + ϕ(r). It is only possible whenr = 0. Then by (2)

n→∞lim d(T xn, T xn+1) = 0 (3)

Now we shall show that {T xn} is Cauchy sequence. Let we assume contrary. Then there exist  > 0 such that for m, n → ∞ and for mi< ni< mi+1,

d(T xmi, T xni) ≥  and (4)

d(T xm_i, T xn_i−1) < 

Then it follows that

 ≤ [d(T xm_i, T xn_i)] ≤ d(T xm_i, T xn_i−1) + d(T xn_i−1, T xn_i) < ε + d(T xn_i−1, T xn_i)d(T xm_i, T xn_i) <  + d(T xn_i−1, T xn_i)

i→∞lim d(T xm_i, T xn_i) <  + lim

i→∞d(T xn_i−1, T xn_i) By (3)

i→∞lim d(T xm_i, T xn_i) <  (5)

By (4) and (5)

lim

i→∞d(T xm_i, T xn_i) =  (6)

Now by using (b) with x = xm_i, y = xn_i

φ[d(T xm_i, T xn_i)] ≥ φ[d(T xm_i, F xm_i) + d(T xm_i, F xn_i)] + ϕd(T xm_i, T xn_i) φ[d(T xm_i, T xn_i)] ≥ φ[d(T xm_i, T xm_i+1) + d(T xm_i, T xn_i+1)] + ϕd(T xm_i, T xn_i)

i→∞lim and by (3) & (6) we have φ(ε) ≥ φ(ε) + ϕ(ε). It is only possible when ε = 0. Which is contradiction to our assumption that ε > 0. Therefore for all m, n → ∞ we have d(T xn_i, T xm_i) < ε. Therefore {T xn} is a Cauchy sequence. Since (X, D) is complete metric space, then it will be converge at some point z ∈ X, or lim

n→∞T xn = z. So all of its subsequence also converge to z or lim

n→∞T xn+1= z, lim

n→∞F xn= z.

Case (1): When T is continuous map Since lim

n→∞T xn= z, therefore lim

n→∞T T xn= T z. Also lim

n→∞F xn= z, therefore lim

n→∞T F xn= T z. Since pair (T, F ) is semi compatible map then since lim

n→∞F xn= lim

n→∞T xn= z, therefore lim

n→∞F T xn= T z. Now using (b) with x = T xn, y = xn

φ[d(T T xn, T xn)] ≥ φ[d(T T xn, F T xn)) + d(T T xn, F xn)] + ϕ[d(T T xn, T xn)

Now limiting lim

n→∞we have

φ[d(T z, z)] ≥ φ[d(T z, T z) + d(T z, z)] + ϕd(T z, z)]

φ[d(T z, z)] ≥ φ[d(T z, z)] + ϕd(T z, z) ϕ[d(T z, z)] ≤ 0

It is only possible when, T z = z. Since T (X) ⊂ F (X). Then let u ∈ X such that T z = F u = z. Now by using (b) with x = u, y = xn

φ[d(T u, T xn)] ≥ φ[d(T u, F u) + d(T u, F xn)] + ϕd(T u, T xn)]

Taking lim n → ∞

φ[d(T u, z)] ≥ φ[d(T u, z) + d(T u, z)] + ϕd(T u, z) φ[d(T u, z)] ≥ φ[2d(T u, z)] + ϕd(T u, z)

Since φ and ϕ are increasing function therefore obtained inequality is only possible when d(T u, z) = 0 ⇒ T u = z or T z = F u = z. Since (F, T ) is weak compatible then F T u = T F u = z or T z = F z = z. Therefore z is common fixed point of F and T.

Case (2): When F is continuous map Since lim

n→∞T xn= z, therefore lim

n→∞F T xn= F z. Also lim

n→∞F xn= z, therefore lim

n→∞F F xn= F z. Since pair (T, F ) is semi compatible map then since lim

n→∞F xn = limn→∞T xn= z, therefore limn→∞T F xn = F z. Now using (b) with, x = F xn, y = xn

φ[d(T F xn, T xn)] ≥ φ[d(T F xn, F F xn)) + d(T F xn, F xn)] + ϕ[d(T F xn, T xn) Now limiting n → ∞ we have

φ[d(F z, z)] ≥ φ[d(F z, F z) + d(F z, z)] + ϕd(F z, z) φ[d(F z, z)] ≥ φ[d(F z, z)] + ϕd(F z, z)

ϕd(F z, z)] ≤ 0

It is only possible when F z = z. Again using (b) with x = z, y = xn

φ[d(T z, T xn)] ≥ φ[d(T z, F z) + d(T z, F xn)] + ϕd(T z, T xn)

Limiting n → ∞

φ[d(T z, z)] ≥ φ[d(T z, z) + d(T z, z)] + ϕd(T z, z) φ[d(T z, z)] ≥ φ[2d(T z, z)] + ϕd(T z, z)

Since φ and ϕ are increasing function therefore obtained inequality is only possible when d(T z, z) = 0 ⇒ T z = z or F z = T z = z.

Uniqueness: Let w be another fixed point of F and T, then F w = T w = w. By using (b) with x = z, y = w we have

φ[d(T z, T w)] ≥ φ[d(T z, F z) + d(T z, F w)] + ϕd(T z, T w) φ[d(z, w)] ≥ φ[d(z, z) + d(z, w)] + ϕd(z, w)

φ[d(z, w)] ≥ φ[d(z, w)] + ϕd(z, w) ϕ[d(z, w)] ≤ 0 ⇒ z = w

Hence z is a unique common fixed point of F and T. This completes the proof.

Corollary 3.2. Let T,F and S be self mapping of metric space (X, D) with (a) T (X) ⊂ F (X), S(X) ⊂ F (X),

(b) φ[d(T x, Sy)] ≥ φ[d(T x, F x) + d(T x, F y)] + ϕ[d(T x, Sy)]

(c) Either T or F is continuous function.

(d) (T, F ) is semi compatible and weak compatible.

(e) T S = ST , F S = SF .

If φ and ϕ are monotonic increasing function such that φ, ϕ : [0, ∞) → [0, ∞) and ϕ(t) = 0 = φ(t) ⇔ t = 0 then z is unique common fixed point of F and T.

Theorem 3.3. Let T and F be self mapping of metric space (X, D) with (a) T (X) ⊂ F (X)

(b) φ[d(T x, T y)] ≥ φ min[d(F x, F y), d(T y, F y) + d(F x, T y)] + ϕ[d(T x, T y]

(c) Either T or F is continuous function.

(d) (T, F ) is semi compatible and commute.

If φ and ϕ are monotonic increasing function such that φ, ϕ : [0, ∞) → [0, ∞) and ϕ(t) = 0 = φ(t) ⇔ t = 0, if T² is an identity map then z is unique common fixed point of F and T.

Proof. Let x0 ∈ X is an arbitrary point. Since T (X) ⊂ F (X). Then x1 ∈ X such that T x1= F x0. Inductively we can define a sequence T xn+1= F xn. Using (b) with x = xn, y = xn+1we have

φ[d(T xn, T xn+1)] ≥ φ min[d(F xn, F xn+1), d(T xn+1, F xn+1) + d(F xn, T xn+1)] + ϕ[d(T xn, T xn+1) φ[d(T xn, T xn+1)] ≥ φ min[d(T xn+1, T xn+2), d(T xn+1, T xn+2) + d(T xn+1, T xn+1)] + ϕ[d(T xn, T xn+1)]

φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)] + ϕd(T xn, T xn+2)] (7) φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)]

Since φ is an increasing function therefore d(T xn, T xn+1) ≥ d(T xn+1, T xn+2). Therefore the sequence d(T xn, T xn+1) will be decreasing. Let r ≥ 0 such that

n→∞lim d(T xn, T xn+1) = r. (8)

Hence on taking lim n → ∞ we have by (7) φ(r) ≥ φ(r) + ϕ(r). It is only possible when r = 0. Then by (8)

lim

n→∞d(T xn, T xn+1) = 0 (9)

Now we shall show that {T xn} is Cauchy sequence. Let we assume contrary. Then there exist  > 0 such that for m, n → ∞ and for mi< ni< mi+1

d(T xm_i, T xn_i) ≥  (10)

d(T xm_i, T xn_i−1) < 

Then it follows that

 ≤ [d(T xmi, T xni)] ≤ d(T xmi, T xni−1) + d(T xni−1, T xni) < ε + d(T xni−1, T xni)d(T xmi, T xni) <  + d(T xni−1, T xni)

i→∞lim d(T xm_i, T xn_i) <  + lim

i→∞d(T xn_i−1, T xn_i). By (9)

i→∞lim d(T xm_i, T xn_i) <  (11)

By (10) and (11)

lim

i→∞d(T xm_i, T xn_i) =  (12)

by using (b) with x = xm_i, y = xn_i

φ[d(T xm_i, T xn_i)] ≥ φ min[d(F xm_i, F xn_i), {d(T xn_i, F xn_i) + d(F xm_i, T xn_i)}] + ϕd(T xm_i, T xn_i) φ[d(T xm_i, T xn_i)] ≥ φ min[d(T xm_i+1, T xn_i+1), {d(T xn_i, T xn_i+1)] + d(T xm_i+i, T xn_i)} + ϕd(T xm_i, T xn_i)

n→∞lim and By (9) & (12) we have

φ(ε) ≥ φ min ε, ε + ϕ(ε) φ(ε) ≥ φ(ε) + ϕ(ε).

It is only possible when ε = 0. Which is contradiction to our assumption that ε > 0. Therefore for all m, n → ∞ we have d(T xn_i, T xm_i) < ε. Therefore {T xn} is a Cauchy sequence. Since (X, D) is complete metric space, then it will be converge at some point z ∈ X, or lim

n→∞T xn= z So all of its subsequence also converge to z or lim

n→∞T xn+1= z, lim

n→∞F xn= z.

Case (1): When T is continuous map Since lim

n→∞T xn= z, therefore lim

n→∞T T xn= T z. Also lim

n→∞F xn= z, therefore lim

n→∞T F xn= T z. Since pair (T, F )is semi compatible map then since lim

n→∞F xn= lim

n→∞T xn= z, therefore lim

n→∞F T xn= T z. Now using (b) with x = T xn, y = xn

φ[d(T T xn, T xn)] ≥ φ min[d(F T xn, F xn), d(T xn, F xn) + d(F T xn, T xn)] + ϕ[d(T T xn, T xn)

limiting n → ∞ we have

φ[d(T z, z)] ≥ φ min[d(T z, z), {d(z, z) + d(T z, z)} + ϕd(T z, z)]

φ[d(T z, z)] ≥ φ min[d(T z, z), d(T z, z)] + ϕd(T z, z)

φ[d(T z, z)] ≥ φd(T z, z) + ϕd(T z, z) it is only possible when, d(T z, z) = 0 T z = z.

Again by using (b) with x = z, y = xn+1

φ[d(T z, T xn+1)] ≥ φ min[d(F z, F xn+1), {d(T xn+1, F xn+1) + d(F z, T xn+1)}] + ϕ[d(T z, T xn+1)

Liming n → ∞

φ[d(z, z)] ≥ φ min[d(F z, z), {d(z, z) + d(F z, z)}] + ϕd(z, z)0 ≥ φd(F z, z) ⇒ F z = z Therefore T z = F z = z. z is common fixed point of F and T.

Case (2) - When F is continuous map since lim

n→∞T xn= z, therefore lim

n→∞F T xn= F z. Also lim

n→∞F xn= z, therefore lim

n→∞F F xn= F z. Since pair (T, F ) is semi compatible map then since lim

n→∞F xn= limn→∞T xn= z, therefore lim

n→∞T F xn= F z. Now using (b) with, x = F xn, y = xn

φ[d(T F xn, T xn)] ≥ φ min[d(F F xn, F xn), {d(T xn, F xn) + d(F F xn, T xn)}] + ϕ[d(T F xn, T xn)

Now limiting n → ∞

φ[d(F z, z)] ≥ φ min[d(F z, z), {d(z, z) + d(F z, z)}] + ϕd(F z, z) φ[d(F z, z)] ≥ φ min[d(F z, z), d(F z, z)] + ϕd(F z, z)

φ[d(F z, z)] ≥ φ[d(F z, z)] + ϕd(F z, z)

It is only possible when, d(F z, z) = 0 ⇒ F z = z. By using (b) with x = T z, y = xn

φ[d(T²z, T xn)] ≥ φ min[d(F T z, F xn), {d(T xn, F xn) + d(F T z, T xn)}] + ϕd(T²z, T xn),

since T²= I and pair(F,T) is commute also limit n → ∞

φ[d(z, z)] ≥ φ min[d(T F z, z), {d(z, z) + d(T F z, z)}] + ϕd(z, z) 0 ≥ φ min[d(T z, z), d(T z, z)]

0 ≥ φd(T z, z)

Which is possible when d(T z, z) = 0 ⇒ T z = z. Therefore F z = T z = z. z is common fixed point of T and F. Uniqueness can be proved easily.

Theorem 3.4. Let T,F,S and A be self mapping of metric space (X, D) with (a) T (X) ⊂ F (X), S(X) ⊂ A(X),

(b) φ[d(T x, Sy)] ≥ φ[d(T x, F x)) + d(T x, Ay)] + ϕ[d(T x, Sy]

(c) Either T or F is continuous function.

(d) (T, F ) is semi compatible and weak compatible.

(e) T S = ST, F S = SF .

If φ and ϕ are monotonic increasing function such that φ, ϕ : [0, ∞) → [0, ∞) and ϕ(t) = 0 = φ(t) ⇔ t = 0 then z is unique common fixed point of F and T.

Proof. Let x0 ∈ X is an arbitrary point. Since T (X) ⊂ F (X), S(X) ⊂ A(X). Then there exist x1, x2 ∈ X such that T x1 = F x0 and sx2= Ax1. Inductively we can define a sequence T xn+1= F xn= ynand Sxn+2= Axn+1= yn+1. Using (b) with x = xn, y = xn+1

φ[d(T xn, Sxn+1)] ≥ φ[d(T xn, F xn)) + d(T xn, Axn+1)] + ϕ[d(T xn, Sxn+1) φ[d(T xn, T xn+1)] ≥ φ[d(T xn, T xn+1)) + d(T xn, T xn+2)] + ϕ[d(T xn, T xn+1)]

By triangle inequality we have

[d(T xn+1, T xn+2)] ≤ [d(T xn+1, T xn)) + d(T xn, T xn+2)]

Then

φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)] + ϕd(T xn, T xn+2)] (13) φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)]

Since φ is an increasing function then we have d(T xn, T xn+1) ≥ d(T xn+1, T xn+2). Therefore the sequence d(T xn, T xn+1) will be decreasing. Let r ≥ 0 such that

n→∞lim d(T xn, T xn+1) = r. (14)

Hence on taking lim

n→∞we have by (13) φ(r) ≥ φ(r) + ϕ(r). It is only possible when r = 0. Then by (14)

n→∞lim d(T xn, T xn+1) = 0 (15)

From Theorem 3.1 it can be easily shown that {T xn} is Cauchy sequence. Since (X, D) is complete metric space, then it will be converge at some point z ∈ X, or lim T xn = z So all of its subsequence also converge to z. Or lim

n→∞T xn+1 = z, lim

n→∞F xn= z, lim

n→∞Sxn+2= z and lim

n→∞Axn+1= z.

Case (1) - When T is continuous map Since lim

n→∞T xn= z, therefor lim

n→∞T T xn= T z. Also lim

n→∞F xn= z, therefore lim

n→∞T F xn= T z. Since pair (T, F ) is semi compatible map then since lim

n→∞F xn= lim

n→∞T xn= z, therefore lim

n→∞F T xn= T z. Now using (b) with x = T xn, y = xn

φ[d(T T xn, Sxn)] ≥ φ[d(T T xn, F T xn)) + d(T T xn, Axn)] + ϕ[d(T T xn, Sxn)

Now limiting lim n → ∞ we have

φ[d(T z, z)] ≥ φ[d(T z, T z) + d(T z, z)] + ϕd(T z, z)]

φ[d(T z, z)] ≥ φ[d(T z, z)] + ϕd(T z, z) ϕ[d(T z, z)] ≤ 0.

It is only possible when, T z = z. Since T (X) ⊂ F (X). Then let u ∈ X such that T z = F u = z. Now by using (b) with x = u, y = xn

φ[d(T u, Sxn)] ≥ φ[d(T u, F u) + d(T u, Axn)] + ϕd(T u, Sxn)]

Taking lim n → ∞

φ[d(T u, z)] ≥ φ[d(T u, z) + d(T u, z)] + ϕd(T u, z) φ[d(T u, z)] ≥ φ[2d(T u, z)] + ϕd(T u, z)

Since φ and ϕ are increasing function therefore obtained inequality is only possible when d(T u, z) = 0 ⇒ T u = z or T z = F u = z. Since (F, T ) is weak compatible then F T u = T F u ⇒ F z = T z or F z = T z = z. Now by using (b) with x = Sz and y = xnwe have

φ[d(T Sz, Sxn)] ≥ φ[d(T Sz, F Sz) + d(T Sz, Axn)] + ϕd(T Sz, Sxn)]

Since T S = ST and F S = SF and limiting n → ∞ we have

φ[d(ST z, z)] ≥ φ[d(ST z, SF z) + d(ST z, z)] + ϕd(ST z, z) φ[d(Sz, z)] ≥ φ[d(Sz, Sz) + d(Sz, z)] + ϕd(Sz, z) φ[d(Sz, z)] ≥ φ[d(Sz, z)] + ϕd(Sz, z)

ϕ[d(Sz, z)] ≤ 0 ⇒ Sz = z

By using (b) with x = xn, y = z

φ[d(T xn, Sz)] ≥ φ[d(T xn, F xn) + d(T xn, Az)] + ϕd(T xn, Sz)]

Now liming n → ∞

φ[d(z, z)] ≥ φ[d(z, z) + d(z, Az)] + ϕd(z, z)

φ[d(z, Az)] ≤ 0 ⇒ Az = z or Sz = T z = F z = Az = z.

Therefore z is common fixed point T, F, S and A.

Case (2) - When F is continuous map since lim

n→∞T xn= z, therefore lim

n→∞F T xn = F z. Also lim

n→∞F xn= z, therefore lim

n→∞F F xn= F z Since pair (T, F ) is semi compatible map then since lim

n→∞F xn= limn→∞T xn= z, therefore lim

n→∞T F xn= F z Now using (b) with, x = F xn, y = xn

φ[d(T F xn, Sxn)] ≥ φ[d(T F xn, F F xn)) + d(T F xn, Axn)] + ϕ[d(T F xn, Sxn)

Now limiting n → ∞ we have

φ[d(F z, z)] ≥ φ[d(F z, F z) + d(F z, z)] + ϕd(F z, z) φ[d(F z, z)] ≥ φ[d(F z, z)] + ϕd(F z, z)

ϕd(F z, z)] ≤ 0

It is only possible whend(F z, z) ⇒ F z = z. Again using (b) with x = z, y = xn

φ[d(T z, Sxn)] ≥ φ[d(T z, F z) + d(T z, Axn)] + ϕd(T z, Sxn)

Limiting n → ∞

φ[d(T z, z)] ≥ φ[d(T z, z) + d(T z, z)] + ϕd(T z, z) φ[d(T z, z)] ≥ φ[2d(T z, z)] + ϕd(T z, z)

Since φ and ϕ are increasing function therefore obtained inequality is only possible when d(T z, z) = 0 ⇒ T z = z Or F z = T z = z. Now by using (b) with x = Sz and y = xnwe have

φ[d(T Sz, Sxn)] ≥ φ[d(T Sz, F Sz) + d(T Sz, Axn)] + ϕd(T Sz, Sxn)]

Since T S = ST and F S = SF and limiting n → ∞ we have

φ[d(ST z, z)] ≥ φ[d(ST z, SF z) + d(ST z, z)] + ϕd(ST z, z) φ[d(Sz, z)] ≥ φ[d(Sz, Sz) + d(Sz, z)] + ϕd(Sz, z) φ[d(Sz, z)] ≥ φ[d(Sz, z)] + ϕd(Sz, z)

ϕ[d(Sz, z)] ≤ 0 ⇒ Sz = z

By using (b) with x = xn, y = z

φ[d(T xn, Sz)] ≥ φ[d(T xn, F xn) + d(T xn, Az)] + ϕd(T xn, Sz)]

Now liming n → ∞

φ[d(z, z)] ≥ φ[d(z, z) + d(z, Az)] + ϕd(z, z)

φ[d(z, Az)] ≤ 0 ⇒ Az = z or Sz = T z = F z = Az = z

Therefore z is common fixed point of T, F, S and A. Uniqueness can easily proved.

Theorem 3.5. Let T and F be self mapping of metric space (X, D) with (a) T (X) ⊂ F (X)

(b) φ[d(T x, T y)] ≥ φ[d(T y, F x)) + d(T y, F y)] + ϕ[d(T x, T y)]

(c) Either T or F is continuous function.

(d) (T, F ) is semi compatible and weak compatible.

If φ and ϕ are monotonic increasing function such that φ, ϕ : [0, ∞) → [0, ∞) and ϕ(t) = 0 = φ(t) ⇒ t = 0 then z is unique common fixed point of F and T.

Proof. Let x0 ∈ X is an arbitrary point. Since T (X) ⊂ F (X). Then x1 ∈ X such thatT x1 = F x0. Inductively we can define a sequence T xn+1= F xn. Using (b) with x = xn, y = xn+1

φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, F xn)) + d(T xn+1, F xn+1)] + ϕ[d(T xn, T xn+1)

≥ φ[d(T xn+1, T xn+1)) + d(T xn+1, T xn+2)] + ϕ[d(T xn, T xn+1)]

φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)] + ϕd(T xn, T xn+1)] (16) φ[d(T xn, T xn+1)] ≥ φ[d(T xn+1, T xn+2)]

Since φ is an increasing function then we have d(T xn, T xn+1) ≥ d(T xn+1, T xn+2). Therefore the sequence d(T xn, T xn+1) will be decreasing. Let r ≥ 0 such that

n→∞lim d(T xn, T xn+1) = r. (17)

Hence on taking lim

n→∞we have by (16), φ(r) ≥ φ(r) + ϕ(r). It is only possible when r = 0. Then by (17) lim

n→∞d(T xn, T xn+1) = 0 (18)

Now we shall show that {T xn} is Cauchy sequence. Let we assume contrary. Then there exist  > 0 such that for m, n → ∞ and for mi< ni< mi+1

d(T xm_i, T xn_i) ≥  (19)

and d(T xm_i, T xn_i−1) < . Then it follows that

 ≤ [d(T xm_i, T xn_i)] ≤ d(T xm_i, T xn_i−1) + d(T xn_i−1, T xn_i) < ε + d(T xn_i−1, T xn_i) d(T xmi, T xni) <  + d(T xni−1, T xni)

lim

i→∞d(T xm_i, T xn_i) <  + lim

i→∞d(T xn_i−1, T xn_i) By (18)

lim

i→∞d(T xm_i, T xn_i) <  (20)

By (19) and (20)

lim

n→∞d(T xm_i, T xn_i) =  (21)

Now by using (b) with x = xm_i, y = xn_i

φ[d(T xm_i, T xn_i)] ≥ φ[d(T xn_i, F xm_i) + φ[d(T xn_i, F xn_i)]] + ϕ[d(T xm_i, T xn_i)]

φ[d(T xm_i, T xn_i)] ≥ φ[d(T xn_i, T xm_i+1) + φ[d(T xn_i, T xn_i+1)]] + ϕd(T xm_i, T xn_i)

i→∞lim and By (18) and (21) we have φ(ε) ≥ φ(ε) + ϕ(ε).s. It is only possible when ε = 0. Which is contradiction to our assumption that ε > 0. Therefore for all m, n → ∞ we have d(T xn_i, T xm_i) < ε. Therefore {T xn} is a Cauchy sequence.

Since (X, D)is complete metric space, then it will be converge at some point z ∈ X, or lim

n→∞T xn= z So all of its subsequence also converge to z. Or lim

n→∞T xn+1= z, lim

n→∞F xn= z.

Case (1) When T is continuous map Since lim

n→∞T xn= z, therefore lim

n→∞T T xn= T z. Also lim

n→∞F xn= z, therefore lim

n→∞T F xn= T z. Since pair (T, F ) is semi compatible map. since lim

n→∞F xn= lim

n→∞T xn= z, therefore lim

n→∞F T xn= T z. Now using (b) with x = T xn, y = xn

φ[d(T T xn, T xn)] ≥ φ[d(T xn, F T xn) + d(T xn, F xn) + ϕ[d(T T xn, T xn)

Now limiting n → ∞ we have

φ[d(T z, z)] ≥ φ[d(z, T z) + d(z, z)] + ϕd(T z, z)]

φ[d(T z, z)] ≥ φd(z, T z) + ϕd(T z, z) ϕd(T z, z) ≤ 0

It is only possible when d(T z, z) = 0 ⇒ T z = z. Since T (X) ⊂ F (X). Then let u ∈ X such that T z = F u = z. Now by using (b) with x = xn, y = u

φ[d(T xn, T u)] ≥ φd(T u, F xn), d(T u, F u) + ϕ[d(T xn, u).

Taking lim

n→∞

φ[d(z, T u)] ≥ φ[d(T u, z) + d(T u, z)] + ϕd(z, T u) φ[d(z, T u) ≥ φ[2d(T u, z)] + ϕd(z, T u).

Since φ and ϕ are increasing function therefore obtained inequality is only possible when d(T u, z) = 0 ⇒ T u = z or F u = T u = z. Since (F, T ) is weak compatible then F T u = T F u ⇒ F z = T z or F z = T z = z. Therefore z is common fixed point of F and T.

Case (2) - When F is continuous map since lim

n→∞T xn= z, therefore lim

n→∞F T xn = F z. Also lim

n→∞F xn= z, therefore lim

n→∞F F xn= F z Since pair (T, F ) is semi compatible map then since lim

n→∞F xn= lim

n→∞T xn= z, therefore, lim

n→∞T F xn= F z. Now using (b) with, x = F xn, y = xn

φ[d(T F xn, T xn)] ≥ φ[d(T xn, F F xn)) + d(T xn, F xn)] + ϕ[d(T F xn, T xn)

Now limiting n → ∞ we have

φ[d(F z, z)] ≥ φ[d(z, F z) + d(z, z)] + ϕd(F z, z) φ[d(F z, z)] ≥ φ[d(z, F z)] + ϕd(F z, z)

ϕd(F z, z)] ≤ 0.

It is only possible when d(F z, z) = 0 ⇒ F z = z. Again using (b) with x = xn, y = z

φ[d(T xn, T z)] ≥ φ[d(T z, F xn) + d(T z, F z)] + ϕd(T xn, T z)

Taking lim

n → ∞

φ[d(z, T z)] ≥ φ[d(T z, z) + d(T z, z)] + ϕd(z, T z) φ[d(T z, z)] ≥ φ[2d(T z, z)] + ϕd(z, T z)

Since φ and ϕ are increasing function therefore obtained inequality is only possible when d(T z, z) = 0 ⇒ T z = z Or F z = T z = z.

Uniqueness: Let w be another fixed point of F and T. Then F w = T w = w. By using (b) with x = z, y = w we have

φ[d(T z, T w)] ≥ φ[d(T w, F z) + d(T w, F w)] + ϕd(T z, T w) φ[d(z, w)] ≥ φ[d(w, z) + d(w, w)] + ϕd(z, w)

φ[d(z, w)] ≥ φ[d(w, z)] + ϕd(z, w) ϕd(z, w) ≤ 0 ⇒ z = w.

Hence z is a unique common fixed point of F and T. This complete the proof.

References

[1] Ya.I.Alber and S.Guerre Delabriere, Principle of weakly contractive maps in Hilbert spaces, New Results in Operator theory and its applications in I.Gohberg and Y.Lyubich(Eds.) Operator Theory : Advances and Applications , 98(1997), 7-22.

[2] I.Beg and M.Abbas, Coincidence point and invariant approximation for mappings satisfying generalized weak contractive condition, Fixed point theory and Appl., (2006), 1-7.

[3] P.N.Dutta and B.S.Choudhury, A generalization of contraction principle in metric spaces, Fixed point theory and Appl., (2008), 1-8.

[4] G.Jungck, Commuting mappings and fixed points, Amer. Math. Monthly, 83(1976), 261-263.

[5] G.Jungck, Compatible mappings and common fixed points, Internat. J. Math. Math. Sci., 9(1986), 43-49.

[6] G.Jungck and B.E.Rhoades, Fixed points for set valued functions without continuity, Indian J. Pure. Appl. Math., 29(3)(1998), 227-238.

[7] M.S.Khan, M.Swaleh and S.Sessa, Fixed point theorems by altering distances between the points, Bull. Austral. Math.

Soc. 30(1984), 1-9.

[8] P.R.Pant, Common fixed points of noncommuting mappings, J. Math. Anal. Appl., 188(1994), 436-440.

[9] S.Park, A unified approach to fixed points of contractive maps, J. Korean Math. Soc., 16(2)(1980), 95-105.

[10] B.E.Rhoades, Some theorems on weakly contractive maps, Nonlinear Analysis: Theory. Methods & Applications, 47(4)(2001), 2683-2693.

[11] R.Sumitra, V.R.Uthariaraj and R.Hemavathy, Common Fixed Point and Invariant Approximation Theorems for Map- pings Satisfying Generalized Contraction Principle, Journal of Mathematics Research, 2(2)(2010).

[12] K.P.R.Sastry and G.V.R.Babu, Some fixed point theorems by altering distances between the points, Indian J. Pure appl.

Math., 30(6)(1999), 641-647.

[13] S.Sessa, On weak commutative condition of mappings in fixed point considerations, Publ. Inst. Math. N.S., 32(46)(1982), 149-153.

[14] T.Suzuki, A generalized Banach contraction principle that characterizes metric completeness, Proc. Amer. Math. Soc., 136(5)(2008), 1861-1869.