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This document presents a common fixed point theorem involving six self-mappings of a complete metric space that satisfy certain contractive conditions and compatibility properties. The theorem is proved over multiple steps: 1) a Cauchy sequence is constructed from the mappings and shown to converge, 2) the limit is shown to be a common fixed point using compatibility, and 3) uniqueness of the fixed point is proved using the contractive conditions. The theorem generalizes several existing common fixed point results in the literature involving compatible mappings.
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lim d(TSxn,SSxn) =0 and lim d(STxn, TTxn)=0 as n→∞ whenever
{xn} is a
sequence in X such that lim Sxn=lim Txn =t as n→∞ for some t in
X.
Definition 2.3. A function Φ: [0, ∞) → [0, ∞) is said to be a contractive modulus if Φ (0)
=0 and
Φ (t) <t for t > 0.
3.Main Result
Theorem 3.1. Let S, R, T, U, I and J are self mapping of a complete metric space (X,d)
into itself satisfying the conditions
(i) SR(X) ⊂ J(X) , TU(X) ⊂ I(X)
(ii) d(SRx,TUy)≤ α d(Ix,Jy) +β [d(Ix,SRx) + d(Jy,TUy)] +γ [d(Ix,TUy)+
d(Jy,SRx)]
for all x,y Є X and α,β and γ are non-negative reals such that
α+2β+2γ<1
(iii) One of S,R,T,U,I and J is continuous.
(iv) (SR,I) and (TU.J) are compatible of type (A).Then SR,TU,I,J have a unique
common
fixed point. Further if the pairs (S,R) , (S,I) , (R,I) , T,U) , (T,I) , (U.J)
are commuting
pairs then S,R,T,U,I and J have a unique common fixed point.
Proof: Let x0 Є X be arbitrary. Choose a point x1 in X such that SRx0 = Jx1.
This can be done since SR(X) ) ⊂ J(X).
Let x2 be a point in X such that TUx1 = Ix2. This can be done since TU(X) ⊂ I(X).
In general we can choose x2n , x2n+1, x2n+2 …, such that SRx2n =Jx2n+1 and TUx2n+1 = Ix2n+2.
So that we obtain a sequence SRx0, TUx1, SRx2, TUx3 …..
Using condition (ii) we have
d(SRx2n,TUx2n+1)≤ α d(I2n,Jx2n+1) +β [d(Ix2n, SRx2n) + d( Jx2n+1, TUx2n+1)] +γ [d(Ix2n,
TUx2n+1) +
d(Jx2n+1,SRx2n)]
= α d (TUx2n-1, SRx2n) + β [d(TUx2n-1,SRx2n)+d(SRx2n, TUx2n+1)] +
γ[d(TUx2n-1,TUx2n+1) + d(SRx2n, SRx2n)]
≤ α d(TUx2n-1, SRx2n) + β [d(TUx2n-1, SRx2n) +d(SRx2n, TUx2n+1)] +
γ [d(TUx2n-1 ,SRx2n) + d(SRx2n,TUx2n+1)]
= (α+β+γ) d (TUx2n-1, SRx2n) + (β+γ) (SRx2n, TUx2n+1)
Hence d(SRx2n, TUx2n+1) ≤ kd(SRx2n, TUx2n-1) where k=(α+β+γ)/ 1-(β+γ) < 1 ,
Similarly we can show d(SRx2n, TUx2n-1) ≤ k d(SRx2n-2, TUx2n-1)
Therefore d (SRx2n , TUx2n+1) ≤ k2 d(SRx2n-2 , TUx2n-1)
≤ k2n d(SRx0, TUx1)
Which implies that the sequence is a Cauchy sequence and since (X,d) is complete so the
sequence has a limit point z in X. Hence the subsequences {SRx2n} ={Jx2n-1}
and {TUx2n-1} ={Ix2n} also converges to the point z in X.
Suppose that the mapping I is continuous. Then I2x2n→ Iz and ISRx2n → Iz as n→ ∞.
Since the pair (SR,I ) is compatible of type (A). we get SRIx2n→ Iz as n→∞.
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Now by (ii)
d(SRIx2n, TUx2n+1) ≤ α d( I2x2n, Jx2n+1) + β [d( I2x2n, SRIx2n) +d( Jx2n+1,TUx2n+1)] +
γ [d(I2x2n,TUx2n+1) +d(Jx2n+1, SRIx2n)]
letting n→∞ , we get
d(Iz,z) ≤ α d(Iz,z) + β [d(Iz,z) +d(z,z)]+γ [d(Iz,z) + d(z,Iz) ]
=(α+2γ) d(Iz,z)
This gives d(Iz,z)=0 since 0≤ α+2γ<1, Hence Iz=z.
Further d(SRz,TUx2n+1) ≤ αd(Iz,Jx2n+1) +β[d(Iz,SRz) + d(Jx2n+1,TUx2n+1)] +
γ [d(Iz,TUx2n+1) + d(Jx2n+1 , SRz)]
Letting Jx2n+1, TUx2n+1 → z as n→∞ and Iz=z we get
d(SRz,z) ≤ α d(z,z) +β [d(z,SRz) + d(z,z)] +γ [d(z,z) + d(z,SRz)
= (β+γ) d(SRz,z)
Hence d(SRz,z) =0 i.e SRz=z , since 0≤ β+γ <1. Thus SRz=Iz=z
Since SR(X) ⊂ J(X) ,there is a point z1 in X such that z=SRz=-Jz1
Now by (ii)
d(z,TUz1) = d(SRz,TU z1)
≤ α d(Iz,J z1) + β [d(Iz,SRz) + d(J z1,TUz1)] +γ [d(Iz,TU z1) +d(Jz1, SRz)]
= α d(z,z) + β [ d(z,z) +d(z,TUz1)] + γ [d(z,TU z1) + d(z,z) ]
=(β+γ) d(z,TU z1)
Hence d(z,TUz1) =0 i.e TU z1 =z =Jz1 , since 0≤ β+γ <1, Take yn = z1 for n≥ 1
Then TUyn→ Tz1 =z and Jyn→ J z1=z as n→∞
Since the pair (TU,J) is compatible of type (A) , we get
Lim d( TUJyn , JJyn) =0 as n→∞ implies d(TUz,Jz)=0 since Jyn =z for all n≥ 1. Hence
TUz=Jz.
Now d(z,TUz) = d(SRz,TUz)
≤ α d(Iz,Jz) + β [d(Iz,SRz) + d( Jz,TUz)] + γ [d(Iz, TUz) + d( Jz,
SRz )
= α d(z,TUz) + β [ d(z,z) +d( TUz,TUz)]+ γ [d(z,TUz) + d(TUz,z) ]
=(α+2γ) d( z, TUz)
Since α+2γ<1, we get TUz=z, hence z=TUz=Jz therefore z is common fixed point of
SR,TU,I,J when the continuity of I is assumed .
Now suppose that SR is continuous then S2R x2n →SRz , SRIx2n → SRz as n→∞ .
By condition (ii) ,we have
d(S2Rx2n, TUx2n+1) ≤ α d( ISRx2n Jx2n+1) + β [d( ISRx2n , S2Rx2n) + d( Jx2n+1 , TUx2n+1) +
γ[d( ISRx2n,TUx2n+1) + d( Jx2n+1, S2Rx2n)]
letting n→∞ and using the compatibility of type (A) of the pair (SR,I), we get
d(SRz,z) ≤ αd(SRz,z) +β [d(SRz,SRz) +d(z,z)] +γ [d(SRz,z) +d(z,SRz)]
=(α+2γ) d(SRz,z)
Since α+2γ<1 we get SRz=z. But SR(X) ⊂ J(X) there is a point p in X such that
z=SRz=Jp ,Now by (ii)
d(S2Rx2n, TUp) ≤ α d(ISRx2n , Jp) +β [d(ISRx2n,S2Rx2n) +d(Jp,TUp)] +
γ[d( ISRx2n ,TUp)+d(Jp,S2Rx2n)
letting n→∞ we have
d(z,TUp) =d(SRz,TUp)
≤αd(z,z) +β[d(z,z) +d(z,TUp)] +γ[d(z,TUp)+ d(z,z)]
=(β+γ) d(z,TUp)
Since β+γ<1 , we get TUp=z.Thus z=Jp=TUp.
Let yn=p then TUyn → TUp=z and Jyn→ TUp =z
Since (TU,J) is compatible of type (A), we have
Lim d(TUJyn ,JJyn) =0 as n→∞
This gives TUJp=JTUp or TUz=Jz
Further
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d(SRx2n, TUz) ≤ αd(Ix2n,Jz) +β[d(Ix2n,SRz) +d(Jz,TUz)] +γ [d(Ix2n, TUz)+d(Jz,SRx2n)]
Letting n→∞, we get
d(z, TUz) ≤αd(z,TUz) +β [d(z,z) +d(TUz,TUz)] +γ[d(z,TUz) +d(TUz,z)]
=(α+2γ) d(z,TUz)
Since 0≤ α+2γ <1 we get z=TUz
Again we have TU(X) ⊂ I(X) there is a point q in X such that z=TUz=Iq
Now d(SRq,z) =d(SRq,TUz) ≤ α d(Iq,Jz) +β [d(Iq,SRq) +
d(Jz,TUz)]+γ[d(Iq,TUz)+d(Jz,SRq)]
=αd(z,z) +β [d(z,SRq) + d(z,z)]+γ [d(z,TUz) +d(z,SRq)]
=(β+γ) d(z,SRq)
Since 0≤ β+γ<1 we get SRq=z, take yn =q then SRyn→ SRq =z , Iyn→Iq=z
Since (SR,I) is compatible of type (A) , we get
Lim d( ISRyn , IIyn) =0 as n→∞
This implies that SRIq=ISRq or SRz=Iz.
Thus we have z=SRz=Iz=Jz=TUz Hence z is a common fixed point of SR,TU,I and J,
when S is continuous
The proof is similar that z is common fixed point of SR,TU,I and J when I is
continuous,R and U is continuous.
For uniqueness let z and w be two common fixed point os SR,TU,I and J , then by
condition (ii)
d(z,w)= d(SRz,TUw) ≤ αd(Iz,Jw) +β[d(Iz,SRz) +d(Jw,TUw)] +γ [d(Iz,TUw)
+d(Jw,SRz)]
=αd(z,w) +β [d(z,z) +d(w,w)]+γ [d(z,w) +d(w,z)]
= (α+2γ) d(z,w)
Since α+2γ<1 we have z=w.
Again let z be the unique common fixed point of both the pairs (SR,I) , (TU,J) then
Sz=S(SRz) = S(RSz) =SR (Sz)
Sz=S(Iz)=I(Sz)
Rz=R(SRz)=(RS)(RS) =(SR)(Rz)
Rz=R(Iz)=I(Rz)
Which shows that Sz and Rz is the common fixed point of (SR,I) yielding thereby
Sz=z=Rz=Iz=SRz
In view of uniqueness of the common fixed point of the pair (SR,I).
Similarly using the commutativity of (T,U), (T,J), (U,J) it can be shown that
Tz=z=Uz=Jz=TUz.Thus z is the unique common fixed point of S,R,T,U,I and J.
Hence the proof.
4. References
Aage C.T., Salunke J.N.(2009), “On Common fixed point theorem in complete metric
space,” IMF,Vol.3,pp.151-159.
Jungck.G.(1986), “Compatible mappings and common fixed points” Internat. J. Math.
Math. Sci., Vol. 9, pp.771-779.
Jungck.G., P.P.Murthy and Y.J.Cho,(1993), “Compatible mapping of type (A) and
common fixed points” Math.Japonica Vol.38,issue 2, pp. 381-390.
S.Sessa(1986), “On a weak commutativity condition in a fixed point consideration.”Publ.
Inst. Math. 32(46), pp. 149-153.
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11.0003www.iiste.org call for paper.common fixed point theorem for compatible mapping of type_a--21-24
- 1. Computer Engineering and Intelligent Systems www.iiste.org ISSN 2222-1719 (Paper) ISSN 2222-2863 (Online) Vol 2, No.8, 2011 Common Fixed Point Theorem for Compatible Mapping of Type (A) Vishal Gupta Department pf Mathematics, Maharishi Markandeshwar University, Mullana, Ambala, Haryana, India. vishal.gmn@gmail.com, vkgupta09@rediffmail.com Received: 2011-10-20 Accepted: 2011-10-29 Published:2011-11-04 Abstract The purpose of this paper is to prove a common fixed point theorem involving two pairs of compatible mappings of type (A) using six maps using a contractive condition. This article represents a useful generalization of several results announced in the literature. Key Words: Complete metric space, Compatible mapping of type (A), Commuting mapping, Cauchy Sequence, Fixed points. 1. Introduction The study of common fixed point of mappings satisfying contractive type conditions has been studied by many mathematicians.Seesa(1982) introduce the concept of weakly commuting mapping and proved some theorem of commutativity by useing the condition to weakly commutativity, Jungck(1988) gave more generalized commuting and weakly commuting maps called compatible maps and use it for compatibility of two mappings. After that Jungck Muthy and Cho(1993) made another generalization of weak commuting mapping by defining the concept of compatible map of type (A). We proposed to re-analysis the theorems of Aage C.T (2009) on common fixed point theorem compatibility of type (A) 2.Preliminaries Definition 2.1. Self maps S and T of metric space (X,d) are said to be weakly commuting pair iff d(STx,TSx)≤d(Sx,Tx) for all x in X. Definition 2.2. Self maps S and T of a metric space (X,d) are said to be compatible of type (A) if 21 | P a g e www.iiste.org
- 2. Computer Engineering and Intelligent Systems www.iiste.org ISSN 2222-1719 (Paper) ISSN 2222-2863 (Online) Vol 2, No.8, 2011 lim d(TSxn,SSxn) =0 and lim d(STxn, TTxn)=0 as n→∞ whenever {xn} is a sequence in X such that lim Sxn=lim Txn =t as n→∞ for some t in X. Definition 2.3. A function Φ: [0, ∞) → [0, ∞) is said to be a contractive modulus if Φ (0) =0 and Φ (t) <t for t > 0. 3.Main Result Theorem 3.1. Let S, R, T, U, I and J are self mapping of a complete metric space (X,d) into itself satisfying the conditions (i) SR(X) ⊂ J(X) , TU(X) ⊂ I(X) (ii) d(SRx,TUy)≤ α d(Ix,Jy) +β [d(Ix,SRx) + d(Jy,TUy)] +γ [d(Ix,TUy)+ d(Jy,SRx)] for all x,y Є X and α,β and γ are non-negative reals such that α+2β+2γ<1 (iii) One of S,R,T,U,I and J is continuous. (iv) (SR,I) and (TU.J) are compatible of type (A).Then SR,TU,I,J have a unique common fixed point. Further if the pairs (S,R) , (S,I) , (R,I) , T,U) , (T,I) , (U.J) are commuting pairs then S,R,T,U,I and J have a unique common fixed point. Proof: Let x0 Є X be arbitrary. Choose a point x1 in X such that SRx0 = Jx1. This can be done since SR(X) ) ⊂ J(X). Let x2 be a point in X such that TUx1 = Ix2. This can be done since TU(X) ⊂ I(X). In general we can choose x2n , x2n+1, x2n+2 …, such that SRx2n =Jx2n+1 and TUx2n+1 = Ix2n+2. So that we obtain a sequence SRx0, TUx1, SRx2, TUx3 ….. Using condition (ii) we have d(SRx2n,TUx2n+1)≤ α d(I2n,Jx2n+1) +β [d(Ix2n, SRx2n) + d( Jx2n+1, TUx2n+1)] +γ [d(Ix2n, TUx2n+1) + d(Jx2n+1,SRx2n)] = α d (TUx2n-1, SRx2n) + β [d(TUx2n-1,SRx2n)+d(SRx2n, TUx2n+1)] + γ[d(TUx2n-1,TUx2n+1) + d(SRx2n, SRx2n)] ≤ α d(TUx2n-1, SRx2n) + β [d(TUx2n-1, SRx2n) +d(SRx2n, TUx2n+1)] + γ [d(TUx2n-1 ,SRx2n) + d(SRx2n,TUx2n+1)] = (α+β+γ) d (TUx2n-1, SRx2n) + (β+γ) (SRx2n, TUx2n+1) Hence d(SRx2n, TUx2n+1) ≤ kd(SRx2n, TUx2n-1) where k=(α+β+γ)/ 1-(β+γ) < 1 , Similarly we can show d(SRx2n, TUx2n-1) ≤ k d(SRx2n-2, TUx2n-1) Therefore d (SRx2n , TUx2n+1) ≤ k2 d(SRx2n-2 , TUx2n-1) ≤ k2n d(SRx0, TUx1) Which implies that the sequence is a Cauchy sequence and since (X,d) is complete so the sequence has a limit point z in X. Hence the subsequences {SRx2n} ={Jx2n-1} and {TUx2n-1} ={Ix2n} also converges to the point z in X. Suppose that the mapping I is continuous. Then I2x2n→ Iz and ISRx2n → Iz as n→ ∞. Since the pair (SR,I ) is compatible of type (A). we get SRIx2n→ Iz as n→∞. 22 | P a g e www.iiste.org
- 3. Computer Engineering and Intelligent Systems www.iiste.org ISSN 2222-1719 (Paper) ISSN 2222-2863 (Online) Vol 2, No.8, 2011 Now by (ii) d(SRIx2n, TUx2n+1) ≤ α d( I2x2n, Jx2n+1) + β [d( I2x2n, SRIx2n) +d( Jx2n+1,TUx2n+1)] + γ [d(I2x2n,TUx2n+1) +d(Jx2n+1, SRIx2n)] letting n→∞ , we get d(Iz,z) ≤ α d(Iz,z) + β [d(Iz,z) +d(z,z)]+γ [d(Iz,z) + d(z,Iz) ] =(α+2γ) d(Iz,z) This gives d(Iz,z)=0 since 0≤ α+2γ<1, Hence Iz=z. Further d(SRz,TUx2n+1) ≤ αd(Iz,Jx2n+1) +β[d(Iz,SRz) + d(Jx2n+1,TUx2n+1)] + γ [d(Iz,TUx2n+1) + d(Jx2n+1 , SRz)] Letting Jx2n+1, TUx2n+1 → z as n→∞ and Iz=z we get d(SRz,z) ≤ α d(z,z) +β [d(z,SRz) + d(z,z)] +γ [d(z,z) + d(z,SRz) = (β+γ) d(SRz,z) Hence d(SRz,z) =0 i.e SRz=z , since 0≤ β+γ <1. Thus SRz=Iz=z Since SR(X) ⊂ J(X) ,there is a point z1 in X such that z=SRz=-Jz1 Now by (ii) d(z,TUz1) = d(SRz,TU z1) ≤ α d(Iz,J z1) + β [d(Iz,SRz) + d(J z1,TUz1)] +γ [d(Iz,TU z1) +d(Jz1, SRz)] = α d(z,z) + β [ d(z,z) +d(z,TUz1)] + γ [d(z,TU z1) + d(z,z) ] =(β+γ) d(z,TU z1) Hence d(z,TUz1) =0 i.e TU z1 =z =Jz1 , since 0≤ β+γ <1, Take yn = z1 for n≥ 1 Then TUyn→ Tz1 =z and Jyn→ J z1=z as n→∞ Since the pair (TU,J) is compatible of type (A) , we get Lim d( TUJyn , JJyn) =0 as n→∞ implies d(TUz,Jz)=0 since Jyn =z for all n≥ 1. Hence TUz=Jz. Now d(z,TUz) = d(SRz,TUz) ≤ α d(Iz,Jz) + β [d(Iz,SRz) + d( Jz,TUz)] + γ [d(Iz, TUz) + d( Jz, SRz ) = α d(z,TUz) + β [ d(z,z) +d( TUz,TUz)]+ γ [d(z,TUz) + d(TUz,z) ] =(α+2γ) d( z, TUz) Since α+2γ<1, we get TUz=z, hence z=TUz=Jz therefore z is common fixed point of SR,TU,I,J when the continuity of I is assumed . Now suppose that SR is continuous then S2R x2n →SRz , SRIx2n → SRz as n→∞ . By condition (ii) ,we have d(S2Rx2n, TUx2n+1) ≤ α d( ISRx2n Jx2n+1) + β [d( ISRx2n , S2Rx2n) + d( Jx2n+1 , TUx2n+1) + γ[d( ISRx2n,TUx2n+1) + d( Jx2n+1, S2Rx2n)] letting n→∞ and using the compatibility of type (A) of the pair (SR,I), we get d(SRz,z) ≤ αd(SRz,z) +β [d(SRz,SRz) +d(z,z)] +γ [d(SRz,z) +d(z,SRz)] =(α+2γ) d(SRz,z) Since α+2γ<1 we get SRz=z. But SR(X) ⊂ J(X) there is a point p in X such that z=SRz=Jp ,Now by (ii) d(S2Rx2n, TUp) ≤ α d(ISRx2n , Jp) +β [d(ISRx2n,S2Rx2n) +d(Jp,TUp)] + γ[d( ISRx2n ,TUp)+d(Jp,S2Rx2n) letting n→∞ we have d(z,TUp) =d(SRz,TUp) ≤αd(z,z) +β[d(z,z) +d(z,TUp)] +γ[d(z,TUp)+ d(z,z)] =(β+γ) d(z,TUp) Since β+γ<1 , we get TUp=z.Thus z=Jp=TUp. Let yn=p then TUyn → TUp=z and Jyn→ TUp =z Since (TU,J) is compatible of type (A), we have Lim d(TUJyn ,JJyn) =0 as n→∞ This gives TUJp=JTUp or TUz=Jz Further 23 | P a g e www.iiste.org
- 4. Computer Engineering and Intelligent Systems www.iiste.org ISSN 2222-1719 (Paper) ISSN 2222-2863 (Online) Vol 2, No.8, 2011 d(SRx2n, TUz) ≤ αd(Ix2n,Jz) +β[d(Ix2n,SRz) +d(Jz,TUz)] +γ [d(Ix2n, TUz)+d(Jz,SRx2n)] Letting n→∞, we get d(z, TUz) ≤αd(z,TUz) +β [d(z,z) +d(TUz,TUz)] +γ[d(z,TUz) +d(TUz,z)] =(α+2γ) d(z,TUz) Since 0≤ α+2γ <1 we get z=TUz Again we have TU(X) ⊂ I(X) there is a point q in X such that z=TUz=Iq Now d(SRq,z) =d(SRq,TUz) ≤ α d(Iq,Jz) +β [d(Iq,SRq) + d(Jz,TUz)]+γ[d(Iq,TUz)+d(Jz,SRq)] =αd(z,z) +β [d(z,SRq) + d(z,z)]+γ [d(z,TUz) +d(z,SRq)] =(β+γ) d(z,SRq) Since 0≤ β+γ<1 we get SRq=z, take yn =q then SRyn→ SRq =z , Iyn→Iq=z Since (SR,I) is compatible of type (A) , we get Lim d( ISRyn , IIyn) =0 as n→∞ This implies that SRIq=ISRq or SRz=Iz. Thus we have z=SRz=Iz=Jz=TUz Hence z is a common fixed point of SR,TU,I and J, when S is continuous The proof is similar that z is common fixed point of SR,TU,I and J when I is continuous,R and U is continuous. For uniqueness let z and w be two common fixed point os SR,TU,I and J , then by condition (ii) d(z,w)= d(SRz,TUw) ≤ αd(Iz,Jw) +β[d(Iz,SRz) +d(Jw,TUw)] +γ [d(Iz,TUw) +d(Jw,SRz)] =αd(z,w) +β [d(z,z) +d(w,w)]+γ [d(z,w) +d(w,z)] = (α+2γ) d(z,w) Since α+2γ<1 we have z=w. Again let z be the unique common fixed point of both the pairs (SR,I) , (TU,J) then Sz=S(SRz) = S(RSz) =SR (Sz) Sz=S(Iz)=I(Sz) Rz=R(SRz)=(RS)(RS) =(SR)(Rz) Rz=R(Iz)=I(Rz) Which shows that Sz and Rz is the common fixed point of (SR,I) yielding thereby Sz=z=Rz=Iz=SRz In view of uniqueness of the common fixed point of the pair (SR,I). Similarly using the commutativity of (T,U), (T,J), (U,J) it can be shown that Tz=z=Uz=Jz=TUz.Thus z is the unique common fixed point of S,R,T,U,I and J. Hence the proof. 4. References Aage C.T., Salunke J.N.(2009), “On Common fixed point theorem in complete metric space,” IMF,Vol.3,pp.151-159. Jungck.G.(1986), “Compatible mappings and common fixed points” Internat. J. Math. Math. Sci., Vol. 9, pp.771-779. Jungck.G., P.P.Murthy and Y.J.Cho,(1993), “Compatible mapping of type (A) and common fixed points” Math.Japonica Vol.38,issue 2, pp. 381-390. S.Sessa(1986), “On a weak commutativity condition in a fixed point consideration.”Publ. Inst. Math. 32(46), pp. 149-153. 24 | P a g e www.iiste.org
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