## Generalised Common Fixed Point Theorems of A-Compatible and S-Compatible Mappings

**Shahidur Rahman**^{1}, **Yumnam Rohen**^{1,}, **M. Popeshwar Singh**^{1}

^{1}National Institute of Technology Manipur, Imphal, India

### Abstract

In this paper we prove a common fixed point theorem of four self mappings satisfying a generalized inequality using the concept of A-compatible and S-compatible mappings. Our result generalizes many earlier related results in the literature.

**Keywords:** common fixed point, complete metric space, compatible mappings, compatible mappings of type (A), A-compatible mappings, S-compatible mappings

*American Journal of Applied Mathematics and Statistics*, 2013 1 (2),
pp 27-29.

DOI: 10.12691/ajams-1-2-2

Received February 26, 2013; Revised March 12, 2013; Accepted April 15, 2013

**Copyright:**© 2013 Science and Education Publishing. All Rights Reserved.

### Cite this article:

- Rahman, Shahidur, Yumnam Rohen, and M. Popeshwar Singh. "Generalised Common Fixed Point Theorems of A-Compatible and S-Compatible Mappings."
*American Journal of Applied Mathematics and Statistics*1.2 (2013): 27-29.

- Rahman, S. , Rohen, Y. , & Singh, M. P. (2013). Generalised Common Fixed Point Theorems of A-Compatible and S-Compatible Mappings.
*American Journal of Applied Mathematics and Statistics*,*1*(2), 27-29.

- Rahman, Shahidur, Yumnam Rohen, and M. Popeshwar Singh. "Generalised Common Fixed Point Theorems of A-Compatible and S-Compatible Mappings."
*American Journal of Applied Mathematics and Statistics*1, no. 2 (2013): 27-29.

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### 1. Introduction

The first important result in the theory of fixed point of compatible mappings was obtained by Gerald Jungck in 1986 ^{[2]} as a generalization of commuting mappings. In 1993 Jungck, Murthy and Cho ^{[3]} introduced the concept of compatible mappings of type (A) by generalizing the definition of weakly uniformly contraction maps. Pathak and Khan ^{[6]} introduced the concept of A-compatible and S-compatible by splitting the definition of compatible mappings of type (A). Fixed point results of compatible mappings are found in [1-8]^{[1]}.

Sharma and Sahu ^{[8]} proved the following theorem.

**THEOREM 1.****1** Let *A*, *S* and *T* be three continuous mappings of a complete metric space (*X*, *d*) into itself satisfying the following conditions:

(i) *A* commutes with *S* and *T* respectively

(ii) *S* (*X*) ⊆ *A*(*X*) and *T*(*X*) ⊆ *A*(*X*)

(iii) [*d*(*Sx*, *Tx*)]^{2} ≤ *a*_{1}*d*(*Ax*, *Sx*)*d*(*Ay*, *Ty*)+*a*_{2}*d*(*Ay*, *Sx*)*d*(*Ax*, *Ty*)+*a*_{3}*d*(*Ax*, *Sx*)*d*(*Ax*, *Ty*) +*a*_{4}*d*(*Ay*, *Ty*)*d*(*Ay*, *Sx*)*+a*_{5}*d*^{2}(*Ax*, *Ay*)

For all *x*, *y* ∈ *X*, where *a*_{i }≥ 0, *i *= 1, 2, 3, 4, 5 and *a*_{1}+*a*_{4}+* a*_{5}< 1, 2*a*_{1}+3*a*_{3}+2*a*_{5}<2.

Then *A*, *S* and *T* have a unique common fixed point in *X*.

Murthy ^{[6]} pointed out that the constraints taken by Sharma and Sahu in condition (iii) of theorem 1.1 is not true and suggested the corrected replacement as max {*a*_{1}+2*a*_{3}+* a*_{5}, *a*_{1}+2*a*_{4}+*a*_{5}, *a*_{2}+* a*_{5}} < 1 and proved a new fixed point theorem.

The aim of this paper is to prove a common fixed point theorem of S-compatible mappings in metric space by considering four self mappings. Further we give another common fixed point theorem of A-compatible mappings.

### 2. Preliminaries

Following are definitions of types of compatible mappings.

**Definition ****2****.1 **^{[2]}: Let *A* and *S* be mappings from a complete metric space *X* into itself. The mappings *A* and *S* are said to be compatible if *d*(*ASx*_{n},* SAx*_{n})* *= 0 whenever {*x*_{n}} is a sequence in *X* such that *Ax*_{n} = *Sx*_{n} = *t* for some *t* ∈ *X*.

**Definition ****2****.2 **^{[3]}: Let *A* and *S* be mappings from a complete metric space *X* into itself. The mappings *A* and *S* are said to be compatible of type (A) if *d*(*ASx*_{n},* SSx*_{n})* *= 0 and *d*(*SAx*_{n},* AAx*_{n})* *= 0 whenever {*x*_{n}} is a sequence in *X* such that for*Ax*_{n} = *Sx*_{n} = *t* for some *t* ∈ *X*.

**Definition ****2****.3 **^{[5]}: Let *A* and *S* be mappings from a complete metric space *X* into itself. The mappings *A* and *S* are said to be A-compatible if *d*(*ASx*_{n},* SSx*_{n})* *= 0 whenever {*x*_{n}} is a sequence in *X* such that for*Ax*_{n} = *Sx*_{n} = *t* for some *t* ∈ *X*.

**Definition ****2****.4 **^{[5]}: Let *A* and *S* be mappings from a complete metric space *X* into itself. The mappings *A* and *S* are said to be S-compatible if *d*(*SAx*_{n},* AAx*_{n})* *= 0 whenever {*x*_{n}} is a sequence in *X* such that for*Ax*_{n} = *Sx*_{n} = *t* for some *t* ∈ *X*.

**Proposition ****2****.5**** **^{[6]}: Let *A *and *S* be mappings from a complete metric space (*X*,* d*) into itself. If a pair (*A*,* S*)* *is A-compatible on *X* and *St* = *At* for *t *∈* X*, then *ASt* = *SSt*.

**Proposition ****2****.6**** **^{[6]}: Let *A *and *S* be mappings from a complete metric space (*X*,* d*) into itself. If a pair (*A*,* S*)* *is S-compatible on *X* and *St* = *At* for *t *∈* X*, then *S**At *= *AAt*.

**Proposition ****2****.7**** **^{[6]}: Let *A *and *S* be mappings from a complete metric space (*X, d*) into itself. If a pair (*A*,* S*)* *is A-compatible on *X* and *Ax*_{n} =*Sx*_{n} = *t* for *t *∈* X*, then *SSx*_{n} → *At* if *A* is continuous at *t*.

**Proposition ****2****.8**** **^{[6]}: Let *A *and *S* be mappings from a complete metric space (*X, d*) into itself. If a pair (*A*,* S*)* *is S-compatible on *X* and *Ax*_{n} =*Sx*_{n} = *t* for *t *∈* X*, then *AAx*_{n} → *St* if *S* is continuous at *t*.

Now we prove the following theorem.

**LEMMA 2.****9** Let *A*, *B*, *S* and *T* be mapping from a metric space (*X*, *d*) into itself satisfying the following conditions:

(1) *A*(*X*) ⊆ *T*(*X*) and *B*(*X*) ⊆ *S*(*X*)

(2) [*d*(*Ax*, *Bx*)]^{2} ≤ *a*_{1}*d*(*Ax*, *Sx*)*d*(*By*, *Ty*)+*a*_{2}*d*(*By*, *Sx*)*d*(*Ax*, *Ty*)+*a*_{3}*d*(*Ax*, *Sx*)*d*(*Ax*, *Ty*)+*a*_{4}*d*(*By*, *Ty*)*d*(*By*, *Sx*)* +a*_{5}*d*^{2}(*Sx*, *Ty*)

where *a*_{1}+* a*_{2}* *+2*a*_{3}* *+*a*_{4}+* a*_{5}< 1 and *a*_{1},* a*_{2},* a*_{3},* a*_{4},* a*_{5 }≥ 0

(3) Let *x*_{0} ∈ *X* then by (1) there exists *x*_{1}∈ *X* such that *Tx*_{1} = *Ax*_{0} and for *x*_{1} there exists *x*_{2}∈ *X* such that *Sx*_{2} = *Bx*_{1} and so on. Continuing this process we can define a sequence {*y*_{n}} in *X* such that

then the sequence {*y*_{n}} is Cauchy sequence in *X*.

**Proof.** By condition (2) and (3), we have

Where

Since *a*_{1}+* a*_{2}* *+2*a*_{3}* *+*a*_{4}+* a*_{5}< 1 and *a*_{1},* a*_{2},* a*_{3},* a*_{4},* a*_{5 }≥ 0.

In order to satisfy the inequation, one value of λ will be positive and the other will be negative. We also note that the sum and product of the two values of λ is less than 1 and -1 respectively. Neglecting the negative value, we have where 0<p<1.

Hence {*y*_{n}} is Cauchy sequence.

### 3. Main Results

We prove the following theorem.

**THEOREM 3.1:** Let *A*, *B*, *S* and *T* be self maps of a complete metric space (*X*, *d*) satisfying the following conditions:

(1) A (X) ⊆ T(X) and B(X) ⊆ S(X)

(2) [d(Ax, Bx)]2 ≤ a1d(Ax, Sx)d(By, Ty)+a2d(By, Sx)d(Ax, Ty)+a3d(Ax, Sx)d(Ax, Ty)+a4d(By, Ty)d(By, Sx) +a5d2(Sx, Ty)

where a1+ a2 +2a3 +a4+ a5< 1 and a1, a2, a3, a4, a5 ≥ 0

(3) Let x0 ∈ X then by (1) there exists *x*_{1}∈ *X* such that *Tx*_{1} = *Ax*_{0} and for *x*_{1} there exists *x*_{2}∈ *X* such that *Sx*_{2} = *Bx*_{1} and so on. Continuing this process we can define a sequence {*y*_{n}} in *X* such that

then the sequence {*y*_{n}} is Cauchy sequence in *X*.

(4) One of *A*, *B*, *S* or *T* is continuous.

(5) [*A*, *S*] and [*B*, *T*] are S-compatible mappings on *X*.

Then *A*, *B*, *S* and *T* have a unique common fixed point in *X*.

**Proof:** By lemma 2.9, {*y*_{n}} is Cauchy sequence. Since *X* is complete, there exists a point *z*∈ *X* such that lim *y*_{n} = *z* as *n* → ∞. Consequently subsequences *Ax*_{2n}, *Sx*_{2n}, *Bx*_{2n-1} and *Tx*_{2n+1} converges to *z*.

Let S be a continuous mapping. Since *A* and *S* are S-compatible mappings on *X*, then by proposition 2.8., we have *AAx*_{2n} → *Sz* and *SAx*_{2n} → *Sz* as *n* → ∞.

Now by condition (2) of lemma 2.9, we have

As *n*→∞, we have

which is a contradiction. Hence *Sz* = *z*,

Now

Letting *n*→∞, we have [*d*(*Az, z*)]^{2} ≤ *a*_{3}[*d*(*Az, z*)]^{2}. Hence *Az* = *z*.

Now since *Az* = *z*, by condition (1), *z* ∈ *T*(*X*). Also *T* is self map of *X* so there exists a point *u* ∈*X* such that *z* = *Az* = *Tu*. More over by condition (2), we obtain,

i.e., [*d*(*z*, *Bu*)]^{2} ≤ *a*_{4}[*d*(*z*, *Bu*)]^{2}.

Hence *Bu* = *z* i.e., *z* = *Tu* = *Bu*.

By condition (5), we have

Hence *d*(*Tz*, *Bz*) = 0 i.e., *Tz* = *Bz*.

Now,

i.e., [d(*z*, *Tz*)]^{2} ≤ *a*_{2}[*d*(*z*, *Tz*)]^{2} which is a contradiction. Hence *z* = *Tz* i.e, *z* = *Tz* = *Bz*.

Therefore *z* is common fixed point of *A*, *B*, *S* and *T*. Similarly we can prove that *z* is a common fixed point of *A*, *B*, *S* and *T* if any one of *A*, *B* or *T* is continuous.

Finally, in order to prove the uniqueness of z, suppose w be another common fixed point of A, B, S and T Then we have,

which gives [d(z, Tw)]^{2} ≤ a_{2} [d(z, Tw)]^{2}. Hence *z* = *w*.

This completes the proof.

**THEOREM 3.2:** Let *A*, *B*, *S* and *T* be self maps of a complete metric space (*X*, *d*) satisfying the following conditions:

(1) *A *(*X*) ⊆ *T*(*X*) and *B*(*X*)* *⊆* S*(*X*)*.*

(2) [*d*(*Ax*, *Bx*)]^{2} ≤ *a*_{1}*d*(*Ax*, *Sx*)*d*(*By*, *Ty*)+*a*_{2}*d*(*By*, *Sx*)*d*(*Ax*, *Ty*)+*a*_{3}*d*(*Ax*, *Sx*)*d*(*Ax*, *Ty*) +*a*_{4}*d*(*By*, *Ty*)*d*(*By*, *Sx*)* +a*_{5}*d*^{2}(*Sx*, *Ty*)

where *a*_{1}+* a*_{2}* *+2*a*_{3}* *+*a*_{4}+* a*_{5}< 1 and *a*_{1},* a*_{2},* a*_{3},* a*_{4},* a*_{5 }≥ 0.

(3) Let *x*_{0} ∈ *X* then by (1) there exists *x*_{1}∈ *X* such that *Tx*_{1} = *Ax*_{0} and for *x*_{1} there exists *x*_{2}∈ *X* such that *S**x*_{2} = *Bx*_{1} and so on. Continuing this process we can define a sequence {*y*_{n}} in *X* such that

then the sequence {*y*_{n}} is Cauchy sequence in *X*.

(4) One of *A*, *B*, *S* or *T* is continuous.

(5) [*A*, *S*] and [*B*, *T*] are A-compatible mappings on *X*.

Then A, B, S and T have a unique common fixed point in X.

**Proof:** Similar to theorem 3.1.

**Remark:**** **

(i) By taking *a*_{1}=* a*_{2}* =k*_{1 }and *a*_{3}=* a*_{4}* =k*_{2} and* a*_{5}=0 and (*A*, *S*) and (*B*, *T*) as compatible mappings theorem 3.1 reduces to theorem 1 of Bijendra and Chouhan ^{[1]}.

(ii) By taking *S* = *T* and (*A*, *S*) and (*A*, *T*) as commuting mappings or compatible mappings of type (*A*) theorem 3.1 reduce to results of Murthy ^{[6]} and Sharma and Sahu ^{[8]} under certain conditions.

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