A Generalization of Common Fixed Point in 2-Metric Space

 

Krishnadhan Sarkar1*, Kalishankar Tiwary2

1Department of Mathematics, Raniganj Girls’ College, Raniganj-713358, West Bengal, India,

2Department of Mathematics, Raiganj University, Raiganj-733134, West Bengal, India,

*Corresponding Author E-mail: sarkarkrishnadhan@gmail.com

 

ABSTRACT:

In this paper, we use the concept of semi-compatibility and weak compatibility to prove common fixed point theorems of four self maps in 2-metric spaces. Which is also the generalization result of V.H. Badshah et al. [1] and D. Sharma [11].

 

MATHEMATICS SUBJECT CLASSIFICATION:  47H10, 54H25

KEYWORDS: 2-metric space, semi compatible maps, weak compatible maps, compatible maps, fixed point.

 

 


INTRODUCTION:

The notion of a 2-metric and its basic properties were initially introduce by Gahler [4]. Then after a decade Iseki [5] started investigation of fixed points of mappings in a setting of 2-metric space. Then many researchers like Rhodes [10], De sarkar and K. Tiwary [3], G. Jungck {[6],[7],[8]} N.S. Simoniya [12] worked on several generalization version of fixed point theorems for different types of contractive type mappings in 2-metric spaces. It is interesting to note that every convergent sequences in a 2-metric space need not be a Cauchy sequence[9]. A 2-metric space d is said to be continous when it is continous in two of its arguments[10]. Cho et al. [2] introduced the notion of semi compatible maps in d-topological space. B.Singh and S.Jain {[13],[14]} introduced the concept of semi compatibility in fuzzy metric space, D metric space, 2-metric space and proved fixed point results using implicit relations in these spaces.

 

In this paper we generalize the result of V.H. Badshah et al. [1] by using weak compatibility and semi compatibility mappings.

 

PRELIMANARIES DEFINATIONS:

Let X be a non-empty set and let d: X×X×Xà[0,∞) be such that

(i)               To each pair of point x, y in X with x≠y there exists a point z in X such that d(x,y,z)≠0.

(ii)             d(x,y,z)=0 when at least two of the three points are equal.

(iii)           For any x,y,z in X , d(x,y,z)=d(x,z,y)=d(y,z,x).

(iv)            For any x,y,z,w in X , d(x,y,z)≤ d(x,y,w)+d(x,w,z)+d(w,y,z),

Then d is called a 2-metric [4] and (X,d) is called a 2-metric space[4].       

 

In this note X will denote a complete 2-metric space unless or otherwise stated instead of (X,d).

               A sequence {xn} in X is called a Cauchy sequence[10] when d(xn,xm,a)→0  as n,m →∞

               A sequence {xn} in X is said to be converge[10] to an element x in X when d(xn,x,a)→0 as n→∞

               A 2-metric space (X,d) is said to be complete if every Cauchy sequence in X converges to a point of X.

Definition 1: Two self maps A and B of a 2-metric space (X,d) are said to be compatible if                                             d( ABxn,BAxn,a) =0 for all a in X , where {xn} is a sequence in X such that if                                                        Axn= Bxn=x for some x in X.

 

Definition 2: Let A and B be mappings from a metric space (X,d) intoitself.A and B are said to be weakly compatible if they commute at their coincidence point    i.e, Ax=Bx for some x in  X implies ABx=BAx

 

Definition 3: Two self maps A and B of a 2-metric space (X,d) are said to be semi-compatible if                                 d( ABxn,Bxn,a) =0 for all a in X , where {xn} is a sequence in X such that if                                                          Axn= Bxn=x for some x in X.

 

MAIN RESULTS:

Theorem: Let, P,Q,S and T be mappings from a complete 2-metric space (X,d) to itself satisfying the conditions:

(A) S(X)  Q(X), T(X)  P(X)

(B) d(Sx,Ty,a) ≤ α [{ d(Px,Sx,a)3 + d(Qy,Ty,a)3} / { d(Px,Sx,a)2 + d(Qy,Ty,a)2}] + β d(Px,Qy,a) + γ d(Sx,Px,a) +                    δ d(Sx,Qy,a) + μ d(Ty,Px,a) for all x,y in X, where α,β,γ,δ,μ ≥ 0 and 2α+β+γ+δ+2μ < 1

(C) One of S and Pare continuous function

(D) The pair (S,P) semi compatible and (T,Q) weak compatible on X.

    Then, P,Q,S and T have a unique common fixed point in X.

 

Proof:  Let x0 be any point in X and as  S(X)  Q(X), T(X)  P(X) then there exists x1,x2 in X  such that Sx0=Qx1 , Tx1=Px2

 

Inductively, we can construct sequences {xn} and {yn} in X such that yn=Sxn=Qxn+1 and yn+1=Txn+1=Pxn+2, n=0,1,2,….                                                                                                              ………(1)

Now, we will prove that {yn} is a cauchy sequence.

Now, d( y1,y2,a)= d(Sx1,Tx2,a) ≤ α [{ d(Px1,Sx1,a)3 + d(Qx2,Tx2,a)3} / { d(Px1,Sx1,a)2 + d(Qx2,Tx2,a)2}] + β d(Px1,Qx2,a) + γ d(Sx1,Px1,a) + δ d(Sx1,Qx2,a) + μ d(Tx2,Px1,a)

= α[ d(y0,y1,a) + d(y1,y2,a)] +β  d(y0,y1,a) + γ d(y1,y0,a) + δ d(y1,y1,a) + μ d(y2,y0,a)

≤ α[ d(y0,y1,a) + d(y1,y2,a)] +β  d(y0,y1,a) + γ d(y1,y0,a) + δ d(y1,y1,a) + μ d(y2,y0,y1) + μ d(y2,y1,a) + μ d(y1,y0,a)  ..(2)                                                                                                                                                     

Now, put a=y0 in (2) and get, d( y1,y2,y0) ≤ α d( y1,y2,y0) +2 μ d( y1,y2,y0)

 

Or, (1-α-2μ) d( y1,y2,y0) ≤ 0

Or, d( y1,y2,y0) ≤ 0  , i.e., d( y1,y2,y0) =0

 

Using d( y1,y2,y0) =0 , from (2) we get,

d( y1,y2,a) ≤  d(y0,y1,a) = K d(y0,y1,a)  where K= < 1 as 2α+β+γ+δ+2μ < 1             ……..(3)

now, d( y2,y3,a)= d(Sx2,Tx3,a) ≤ α [{ d(Px2,Sx2,a)3 + d(Qx3,Tx3,a)3} / { d(Px2,Sx2,a)2 + d(Qx3,Tx3,a)2}] + β d(Px2,Qx3,a) + γ d(Sx2,Px2,a) + δ d(Sx2,Qx3,a) + μ d(Tx3,Px2,a)

= α[ d(y1,y2,a) + d(y2,y3,a)] +β  d(y1,y2,a) + γ d(y2,y1,a) + δ d(y2,y2,a) + μ d(y3,y1,a)

≤ α[ d(y1,y2,a) + d(y2,y3,a)] +β  d(y1,y2,a) + γ d(y2,y1,a) + δ d(y2,y2,a) + μ d(y3,y1,y2) + μ d(y3,y2,a) + μ d(y2,y1,a) ..(4)

Put, a= y1 in (4) and we get, d( y2,y3,y1) ≤ α d(y2,y3,y1)] + 2 μ d(y3,y1,y2)

Or, (1-α-2μ) d( y2,y3,y1) ≤ 0  or, d( y2,y3,y1) ≤ 0   i.e., d( y2,y3,y1) =0

So, from (4) we get, d(y2,y3,a) ≤  d(y1,y2,a) = K d(y1,y2,a) ≤ K2 d(y0,y1,a)  ( by 3)                                            where K= < 1 as 2α+β+γ+δ+2μ < 1                                                                                       …………….(5)

 

Similarly, if we proced we will get

d(yn,yn+1,a)  ≤ K d(yn-1,yn,a)  ≤……….≤ Kn d(y0,y1,a)  where K < 1                                    ……………..(6)

and, d(yn,yn+1,yn-1)  =0                                                                                                                              ………….(7)

 

Now, taking lim n→∞ we get from (6),   d(yn,yn+1,a)  =0 as K < 1

d(yn,yn+2,a)  ≤ d(yn,yn+2,yn+1)  + d(yn,yn+1,a)  + d(yn+1,yn+2,a)

                    =   d(yn,yn+2,yn+1)  + d(yn+r,yn+r+1,a)

Using (7) on the above inequality we get,  d(yn,yn+2,a)  ≤  d(yn+r,yn+r+1,a)

Similarlyproceding as above we will get,

d(yn,yn+p,a)  ≤  d(yn+r,yn+r+1,a)

=(Kn + Kn+1 +…+ Kn+p-1 ) d(yn+r,yn+r+1,a) = Kn(1-Kp)/ (1-K) d(yn+r,yn+r+1,a) ≤ Kn/(1-K) d(yn+r,yn+r+1,a)

 

Taking lim n→∞ on the above inequlity we get,  d(yn,yn+p,a)   →0

Which shows that {yn}  is a cauchy sequence and hence converges to some point u in X.

Consequently the sub sequences {Sxn} , {Pxn+1}, { Txn+1}, {Qxn+2} of sequence {yn} are also converges to u.

i.e.,  Sxn = Pxn+1  =xn+1   =xn+2 =u.

 

Let, P is continuous and (S,P) is semi compatible pair, then we have PSxn→Pu, PPxn→Pu and SPxn→PSxn→Pu

Now, d(SPxn,Txn+1,a) ≤ α [{ d(PPxn,SPxn,a)3 + d(Qxn+1,Txn+1,a)3} / {d(PPxn,SPxn,a)2 + d(Qxn+1,Txn+1,a)2}] +                            β d(PPxn,Qxn+1,a) + γ d(SPxn,PPxn,a) + δ d(SPxn,Qxn+1,a) + μ d(Txn+1,PPxn,a)

 

Taking  on both sides of the above inequality we get,

d(Pu,u,a) ≤ α { d(Pu,Pu,a) + d(u,u,a)} + β d(Pu,u,a) + γ d(Pu,Pu,a) + δ d(Pu,u,a) + μ d(u,Pu,a)

Or, (1-β-δ-μ) d(Pu,u,a) ≤ 0,  So, Pu=u                                                                                                     …………….(8)

Now, from (B) we get, d(Su,Txn+1,a) ≤ α [{ d(Pu,Su,a)3 + d(Qxn+1,Txn+1,a)3} / {d(Pu,Su,a)2 + d(Qxn+1,Txn+1,a)2}] +                  β d(Pu,Qxn+1,a) + γ d(Su,Pu,a) + δ d(Su,Qxn+1,a) + μ d(Txn+1,Pu,a)

 

Taking lim n→∞ on both sides of the above inequality we get,

d(Su,u,a) ≤ α { d(u,Su,a) + d(u,u,a)} + β d(u,u,a) + γ d(Su,u,a) + δ d(Su,u,a) + μ d(u,u,a)

Or, (1-α-γ-δ) d(Su,u,a) ≤ 0 So, Su=u                                                                                                       …………..(9)

So, from (8) and (9) we get, Pu=Su=u                                                                                                 ……………(10)

As, S(X)  Q(X) , then there exists v in X such that Su=Qv i.e., Pu=Su=Qv(=u)                               ………..(11)

 

From (B) we get,  d(Su,Tv,a) ≤ α [{ d(Pu,Su,a)3 + d(Qv,Tv,a)3} / {d(Pu,Su,a)2 + d(Qv,Tv,a)2}] +                                             β d(Pu,Qv,a) + γ d(Su,Pu,a) + δ d(Su,Qv,a) + μ d(Tv,Pu,a)

Or, d(u,Tv,a) ≤ α { d(u,u,a) + d(u,Tv,a)} + β d(u,u,a) + γ d(u,u,a) + δ d(u,u,a) + μ d(Tv,u,a)

OR, (1-α-μ) d(u,Tv,a) ≤ 0, Or, d(u,Tv,a)≤ 0, i.e., Tv=u                                                                          …………(12)

Since, (T,Q) are weakly compatible, then we have

TQv=QTv   i.e., Tu=Qu   [ by (11) and (12) ]                                                                                       …………..(13)                                                  

 

Now, from (B) we get, d(Sxn,Tu,a) ≤ α [{ d(Pxn,Sxn,a)3 + d(Qu,Tu,a)3} / {d(Pxn,Sxn,a)2 + d(Qu,Tu,a)2}] +                             β d(Sxn,Qu,a) + γ d(Sxn,Pxn,a) + δ d(Sxn,Qu,a) + μ d(Tu,Pxn,a)

Taking lim n→∞ on both sides of the above inequality we get,

d(u,Tu,a) ≤ α { d(u,u,a) + d(Tu,Tu,a)} + β d(u,Tu,a) + γ d(u,u,a) + δ d(u,Tu,a) + μ d(Tu,u,a)

Or, (1-β-δ-μ) d(u,Tu,a) ≤ 0

 

So, Tu=u which implies that Tu=Qu=u                                                                                               …………..(14)

 

From, (10) and (14) we can say that u is a common fixed point of S,P,Q and T

Again, if we take S is continous and (S,P) is semi compatible pair, we have SPxn→Su, SSxn→Su and PSxn→SPxn→Su

 

Now, d(SSxn,Txn+1,a) ≤ α [{ d(PSxn,SSxn,a)3 + d(Qxn+1,Txn+1,a)3} / {d(PSxn,SSxn,a)2 + d(Qxn+1,Txn+1,a)2}] +                             β d(PSxn,Qxn+1,a) + γ d(SSxn,PSxn,a) + δ d(SSxn,Qxn+1,a) + μ d(Txn+1,PSxn,a)

 

Taking lim n→∞ on both sides of the above inequality we get,

d(Su,u,a) ≤ α { d(Su,Su,a) + d(u,u,a)} + β d(Su,u,a) + γ d(Su,Su,a) + δ d(Su,u,a) + μ d(u,Su,a)

Or, (1-β-δ-μ) d(Su,u,a)≤ 0, i.e., d(Su,u,a)≤0  So, Su=u                                                                       ……………(15)

As, (S,P) are semi compatible so, PSxn→SPxn→Su=u                  [by (15)]

 

Taking lim n→∞ on both sides we get, Pu→u i.e., Pu=u                                                                     ……….(16)

As, S(X)  Q(X) , then there exists w in X such that Su=Qw i.e., Pu=Su=Qw(=u)                         …………..(17)             

Now, from (B) we get,  d(Sxn,Tw,a) ≤ α [{ d(Pxn,Sxn,a)3 + d(Qw,Tw,a)3} / {d(Pxn,Sxn,a)2 + d(Qw,Tw,a)2}] +                        β d(Sxn,Qw,a) + γ d(Sxn,Pxn,a) + δ d(Sxn,Qw,a) + μ d(Tw,Pxn,a)

 

Taking lim n→∞ on both sides of the above inequality we get,

d(u,Tw,a) ≤ α { d(u,u,a) + d(u,Tw,a)} + β d(u,u,a) + γ d(u,u,a) + δ d(u,u,a) + μ d(Tw,u,a)

Or, (1-α-μ) d(u,Tw,a)≤ 0 i.e., d(u,Tw,a) ≤ 0  So, Tw=u

 

So, from (17) we get, Tw=Qw=u=Su=Pu                                                                                             ………….(18)

Since, (T,Q) are weak compatible , then we have TQw=QTw i.e., Tu=Qu                                       ……….(19)

Now,  d(Sxn,Tu,a) ≤ α [{ d(Pxn,Sxn,a)3 + d(Qu,Tu,a)3} / {d(Pxn,Sxn,a)2 + d(Qu,Tu,a)2}] +   β d(Sxn,Qu,a) + γ d(Sxn,Pxn,a) + δ d(Sxn,Qu,a) + μ d(Tu,Pxn,a)

 

Taking  on both sides of the above inequality we get,

d(u,Tu,a) ≤ α { d(u,u,a) + d(Tu,Tu,a)} + β d(u,Tu,a) + γ d(u,u,a) + δ d(u,Tu,a) + μ d(Tu,u,a)

Or, (1-β-δ-μ) d(u,Tu,a) ≤ 0

 

So, Tu=u which implies that Tu=Qu=u                                                                                            …………..(20)

 

So, from (18) and (20) we get that Su=Pu=u=Tu=Qu, i.e., u is a common fixed point of S,P,Q and T

 

To prove uniqueness of the fixed point, let, z be another fixed point of S,P,Q and T such that u≠z

Using (B) we get, d(u,z,a)= d(Su,Tz,a) ≤ α [{ d(Pu,Su,a)3 + d(Qz,Tz,a)3} / { d(Pu,Su,a)2 + d(Qz,Tz,a)2}] +        

 β d(Pu,Qz,a) + γ d(Su,Pu,a) + δ d(Su,Qz,a) + μ d(Tz,Pu,a)

Or, d(u,z,a) ≤ α { d(u,u,a) + d(z,z,a)} + β d(u,z,a) + γ d(u,u,a) + δ d(u,z,a) + μ d(z,u,a)

Or, (1-β-δ-μ) d(u,z,a) ≤ 0 i.e, d(u,z,a) ≤0 So, u=z

i.e., the fixed point u is unique.

So, u is a unique common fixed point of S,P,Q and T.

 

CONCLUSION:

The main object of this paper is to generalize some fixed point theorems in 2-metric spaces using weak compatibility and semi comtability mappings. If we put γ=δ=μ=0 in our main result then we get the result of Badshah et al.[1], which is also the generalization of the result of Sharma D. [11]

 

ACKNOWLADGEMENT:

The authors are thankful to  the editor and the referees for their valuable suggestions.

 

REFERENCES

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Received on 22.03.2018            Accepted on 09.09.2018           

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Int. J. Tech. 2018; 8(2): 100-103.

DOI:10.5958/2231-3915.2018.00013.5