A Generalized Common Fixed Point Theorem For Two Pairs Of Weakly Compatible Self-Maps

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A GENERALIZED COMMON FIXED POINT

THEOREM FOR TWO PAIRS OF WEAKLY

COMPATIBLE SELF-MAPS

Swatmaram

Assistant Professor, Department of Mathematics, C.B.I.T, Gandipet, Hyderabad, Adhra Pradesh, India.

Abstract: A common fixed point theorem of Rahman et al. [3] has been generalized under a weaker inequality through the notion of weak compatibility by dropping continuity and restricting the completeness of the space.

Keywords: Compatible and compatible maps of type A, A- and S-Compatible maps, weakly compatible self-maps and common fixed point.

2000 AMS mathematics subject classification: 54 H 25

I. INTRODUCTION

In this paper

 

X,d denotes a metric space, and A, B, S and T self-maps on X.

Definition 1:A and S are compatible [1] if



ASx ,SAx



0

d

lim n n

n





 … (1)

whenever

{

x

_n

}

_n1 is a sequence in X such that t

Sx lim Ax

lim n

n n n

 

  

 … (2)

for some

t



X

.

Definition 2:A and S are compatible of type A [3] if



ASx ,SSx



0

d

lim _{n n}

n





 and nlim d



ASxn,AAxn



0





 … (3)

whenever {x_n}_n_1X has the choice (2).

It is known that Definition 1 and Definition 2 are equivalent if both S and A are continuous, but are independent of each other in general. Individual conditions in (3) define the following two weaker forms of Definition 2, as given in [3].

Definition 3(a): A and S are A-compatible if lim d



ASx_n,SSx _n



0

n





 whenever (2) holds good for some

X }

x {

1 n

n _  .

Definition 3(b): A and S are S-compatible if lim d



SAx_n,AAx_n



0

n





 whenever (2) holds good for some

X

x

n n



 1

}

{

.

With these notions Rahman et al. [3] proved the following.

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     

X,BX SX T

) X (

A   ... (4)

and the inequality







dAx,By



2ad



Ax,Sx

 

dBy,T y



bd



By,Sx

 

dAx,Ty



cd



Ax,Sx

 

dAy,Ty





By,T y

 

dBy,Sx



fd



Sx,T y



ed  2



for all x,yX, ... (5)

where a, b, c, e and f are nonnegative number such that a + b + 2c + e + f <1.

Given X₀Xthere exist points x1, x2 , … in X such that

1 n 2 1 n 2 2 n

2 Tx y

Ax _  _  _ and Bx_2n1Sx_2ny_2n, n = 1, 2, 3, … ... (6)

and the sequence {y_n}_n1is a Cauchy sequence. Further suppose that X is complete,

(a) one of A, B, S and T is continuous, and

(b)

 

A,S and

 

B,T are either A- compatible or S-compatible. Then A, B, S and T will have a unique common fixed point.

In this paper we obtain a generalization of Theorem 1 by  relaxing the completeness of X,

 weakening the inequality (5),

 dropping the continuity condition (a), and  weakening the notion of A and S compatibilities.

II. MAIN RESULT

Definition 4 (cf. [2]): Self-maps A and S are weakly compatible if they commute at their coincidence point.

Remark 1: It is easily seen that Definition 1 and Definitions 3(a)-3(b) imply Definition 4. That is weak compatibility is a weaker than compatibility, A and S-compatibilities.

Theorem 2. Let A, B, S and T be self-maps on X satisfying the inclusions (4) and the inequality



Ax,By



qmaxd



Ax,Sx

 

dBy,T y



d2  ,



 



d





Ax,Sx

 

dAx,T y





2 1 T y , Ax d Sx , By

d 



By,T y

 

dBy,Sx



d



Sx,T y





d  2 for all x,yX, ... (7)

where 091. Suppose that

(c) T(X)S(X) is a complete subspace of X, and

(d)

 

A,S and

 

B,T are weakly compatible. Then A, B, S and T have a unique common fixed point.

Proof. Let

x

0



X

be arbitrary. Using (4) the sequence

 1

}

{

y

_{n n} can be inductively defined with the choice (6).

Consider t_n d



y_n1,y_n



for

n



1

,

2

,

3

,

...

We assume that

t

n



0

for all n.

Now taking xx_2n,yx_2n1 in (7) and then usuing (6),



_{2n 2n 1}





_{2n 2n}

 

_{2n 1 2n 1}



2 _{Ax ,Bx qmaxd Ax ,Sx d Bx ,T y}

d _  _{ }

,



2n 1 2n

 

2n 2n 1



d





Ax2n,Sx2n

 

d Ax2n,T x2n 1





2

1 , T x , Ax d Sx , Bx

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

 



2



_{2n 2n 1}





n 2 1 n 2 1 n 2 1 n

2 ,T x d Bx ,Sx d Sx ,T x

Bx

d _{  }  _



d

2



y

_2n1

,

y

_2n





q

max

d



y

_2n1

,

y

_2n

 

d

y

_2n

,

y

_2n1











 



2

2n1

,

2n

max

2n1

,

2n 2n

,

2n1

,

d

y

_

y



q

d y

_

y

d y

y

_

d y



_2n

,

y

_2n

 

d y

_2n1

,

y

_2n1



,

1



_{2 1}

,

₂

 

_{2 1}

,

_{2 1}



,

2

d





y

n

y

n

d y

n

y

n







 

 

2n 2n 1





2 n 2 n 2 1 n 2 n

2 ,y d y ,y ,d y ,y

y

d _{ }

       _

 _{ } 2

n 2 n 2 1 n 2 n 2 , 1 n 2 n 2 2 n

2 [t (t t )],t

2 1 t t max q

t . … (8)

If t_2n t_2n1 then t_2nt_2n1t₂2_n so that [t2_2n t_2n t_{2n 1}] t2_2n 2

1 _{  }

 or [t_2n t_{2n 1}] t2_2n 2

1 _{ }

 .

Using this in (8), we get





2

n 2 2 n 2 2 n 2 n 2 1 n 2 n 2 1 n 2 n 2 2 n

2 t t t ,t qt t

2 1 , t t max q t

0  

      _   _{ }

which is a contradiction since 0q1. Therefore

1 2

2n



t

n

t

for all n. … (9)

Then using (9) in (8), we see that

2 1 n 2 2 n 2 qt

t  _ for all n2. … (10)

With a similar argument, we can show that

2 2 n 2 2 1 n 2 qt

t _  _ for all n2. … (11)

Repeatedly applying (10) and (11), it follows that

2 1 n 2 2 3 n 2 3 2 2 n 2 2 1 n 2 n

2 qt q t q t ... q t

t  _  _  _    for all n2.

Applying the limit as n, (11) gives

lim

₂



0

  n

n

t

, that is





0 y

y

d _{2n, 2n1}  as

n





,

which implies that

 1

}

{

y

_{n n} is Cauchy sequence in T(X)S(X).

Then by (c), we see that

z Sx lim Bx lim y lim Tx lim Ax lim y

lim _2n

n 1 n 2 n n 2 n 1 n 2 n 2 n 2 n 1 n 2 n                     

 … (12)

For some zT(X)S(X) so that zT(X) and zS(X).

Now zT(X) implies that zTp for some pX. … (13)

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

Ax ,Bp



qmaxd



Ax ,Sx

 

dBp,T p



d2 _2n  _{2n 2n}

,

d



Bp,T p

 

d Bp,Sx_2n



d2



Sx_2n,T p







 



d





Ax ,Sx

 

dAx ,T p





2 1 T p , Ax d Sx , Bp

d _{2n 2n}  _{2n 2n 2n}

Applying the limit as

n





,

in this and using (12) and (13), we get



z,Bp



qmax{0,0,0,d (Bp,z),0} qd (Bp,z)

d2  2  2 so that Bpz.

Thus

Bp



Tp



z

.

This in view of weak compatibility of B and T gives

Bz = Tz. … (14)

Again zS(X) implies that zS for some X. … (15)

Then we show that ASz.

Substituting

x





and

y



x

_2n1in (7), and using (12) and (15), we get



_{2n 1}





 

_{2n 1 2n 1}



2 _{A ,Bx qmaxd A ,S dBx ,T x}

d  _    _{ }

,

d



Bx_2n1,S

 

dA,Tx_2n1





 





 

 



2



_{2n 1}





1 n 2 1 n 2 1 n 2 1

n

2 ,d Bx ,T x d Bx ,S d S ,T x

T x , A d S , A d 2 1

 

 

   

  

Applying limit as n,









 





   

 

 dA ,zdA ,z,0,0 2

1 , 0 , 0 max q z , A

d2    =





  z , A d 2 1

q 2  , which is a contradiction. Hence Az. Thus ASz. By weak compatibility of

 

A,S , we get ASSA or

Az



Sz

. … (16)

Writing

x



y



z

in the inequality (7), and using (14) and (16), we get









 



2

,

max

,

,

d

Az Bz



q

d Az Sz d Bz Tz

, d



Bz,Sz

 

dAz,Tz



,

d





Az,Sz

 

dAz,T z





,d



Bz,T z

 

dBz,Sz



d



Sz,T z





2

1 _ 2

qmax{0,d



Bz,Az

 

dAz,Bz



,0,0,d2



Az,Bz



}



q d

2



Az Bz

,





Az



Bz

.

Therefore from (14) and (16), it follows that

Tz

Sz

Bz

Az







. … (17)

Taking

x



z

,

y



x

_2n1in (7), and using (12) and (17), we get



Az,Bx



qmax



d



Az,Sz

 

dBx ,T x

 

,dBx ,Sz

 

dAz,T x



,

d2 _2n1  _{2n1 2n1 2n1}  _2n1



 



















  

 

 

1 2n 1 2n 1 2n 1, 2 2n 1 n

2 ,dBz ,Tx dBx Sz,d Sz,Tx

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Applying limit as

_n





_,



Az,z



qmax{0,d



z,Az

 

dAz,z



,0,0,d



Az,z



}

d2  2



qd Az z



,





Az



z

. In view of (17) then it follows that z is a common fixed point of self-maps A, B, S and T.

Uniqueness of the common fixed point follows easily from the inequality (7).

Remark 2. The completeness of the space X is restricted to that of its subspace, namely T(X)S(X) (cf. Condition (c) of Theorem 2). Condition (b) imples (d) in view of Remark 1. Further the inequality (5) implies (7) with the choice

f e c 2 b a

q     . It is important to note that the continuity of none of the four maps is needed to obtain a fixed point. Thus Theorem 2 is a generalization of Theorem 1.

REFERENCES

[1] Gerald Jungck, Compatible mappings and common fixed points, Int. Jour. Math. & Math. Sci, 9 (1986), 771-779.

[2] Gerald Jungck and B. E. Rhoades, Fixed point for set-valued functions with out continuity, Indian J. pure appl. Math. 29 (3) (1998), 227-238. [3] S. Rahman, Y. Rohen and M. P. Singh, Generalized common fixed point Theorems of A-compatibility and S- compatible mappings, Amer. J.