APM  Vol.7 No.6 , June 2017
Common Fixed Point Theorems in Metric Space by Altering Distance Function
ABSTRACT
In the present paper, we prove two theorems. In first theorem, we prove fixed point result for self-maps in the metric space under contractive condition of integral type by altering distance. In second result, we prove a unique common fixed point theorem by considering four sub compatible maps under a contractive condition of integral type.

1. Introduction and preliminaries

In [1] , Khan introduced and proved fixed point results by the altering distance in metric space. Aliouche [2] proved common fixed point results in symmetric space for weakly compatible mappings under contractive condition of integral type. In [3] , Babu generalized and proved fixed point results using control function. Later Bouhadjera and Godet [4] generalized concept of pair sub compatible maps and proved fixed point results. Also Chaudhari [5] [6] , Chugh & Kumar [7] , Naidu [8] , Sastry et al. [9] generalized and proved some fixed point results. Recently in [10] [11] , Hosseni used contractive rule of integral type by altering distance and generalized common fixed point results. Many authors proved fixed point results with different techniques in different spaces (see [12] - [17] ). In [18] [19] [20] [21] , Wadkar et al. proved fixed point theorems using the concept of soft metric space. In the present paper, we prove two theorems on fixed point under contraction rule of integral type in metric space by altering distance function, first for self map and second for a pair of sub compatible maps. Our results are motivated by V. R. Hosseni, Neda Hosseni.

Definition 1.1: A function ψ : R + R + = [ 0 , 1 ) is an altering distance functions if ψ is continuous with monotone increasing in all variables and ψ ( x 1 , x 2 , x 3 , , x n ) = 0 if x 1 = x 2 = x 3 = = x n = 0.

The collection of all altering distance is denoted by Ψ n .

Now let us define a function m ( y ) by m ( y ) = ψ ( y , y , y , y , , y ) for y [ 0 , ) , clearly m ( y ) = 0 if and only if y = 0 .

Examples of ψ are ψ ( f 1 , f 2 , f 3 , , f n ) = μ max { f 1 , f 2 , f 3 , , f n } , for μ > 0 , (1)

ψ ( f 1 , f 2 , f 3 , , f n ) = f n a 1 + f n a 2 + f n a 3 + + f n a n , a 1 , a 2 , , a n 1. (2)

Definition 1.2: The maps p , q : E E of metric space ( E , σ ) are called as sub compatible if and only if the sequence { e n } in E such that lim n p n = lim n q n = t , t E and which satisfies lim n σ ( p q e n , q p e n ) = 0.

Example 1.3: Let E = [ 0 , ) we define p & q with metric σ as follows

p ( e ) = e 2 & q ( e ) = { e + 6 if e [ 4 , 9 ] ( 27 , ) e 2 , if e [ 9 , 27 ] (3)

Let us define the sequence { e n } in E as e n = 3 + 1 n , for n = 0 , 1 , 2 , then

lim n p e n = lim n e n 2 = 9 = lim n q e n = lim n ( e n + 6 ) , (4)

and lim n p q ( e n ) = lim n p ( e n + 6 ) = lim n ( e n + 6 ) 2 = 81 when n , (5)

lim n q p e n = lim n q e n 2 = ( e n 2 ) 2 = lim n ( e n ) 4 = 81 when n . (6)

Thus, we have lim n σ ( p q e n , q p e n ) = 0 . (7)

Hence maps p and q are sub-compatible.

On the other hand, we have p e = q e if and only if e = 3 , p q ( 3 ) = p ( 9 ) = 81 and q p ( 3 ) = q ( 9 ) = 9 + 6 = 15.

Then p ( 3 ) = 9 = q ( 3 ) but p q ( 3 ) = 81 15 = q p ( 3 ) , hence p and q are not OWC (Oscillatory weakly commuting).

2. Main Result

Theorem 2.1: Let us consider the mappings U , V : E E of complete metric space ( E , σ ) be such that for all c , d E

0 ϕ 1 ( σ ( U c , V d ) ) η ( y ) d y 0 ψ 1 ( σ ( c , d ) , σ ( U c , , c ) , σ ( V d , d ) , 1 2 { σ ( V d , c ) + σ ( U c , d ) } , 1 2 { σ ( c , d ) + σ ( U c , c ) } ) η ( y ) d y 0 ψ 2 ( σ ( c , d ) , σ ( U c , , c ) , σ ( V d , d ) , 1 2 { σ ( V d , c ) + σ ( U c , d ) } , 1 2 { σ ( c , d ) + σ ( U c , c ) } ) η ( y ) d y , (8)

where ψ 1 , ψ 2 Ψ 5 with ϕ 1 = ψ ( e , e , e , e , e ) , e [ 0 , ) and Lebesgue-integr- able mapping η : R + R + , which is positive, sum able, and for each > 0 , 0 η ( y ) d y > 0 , then there exist a unique common fixed point in E for U and V.

Proof: Consider arbitrary point e 0 of E, for n = 1 , 2 , 3 , we have

e 2 n + 1 = U e 2 n

and e 2 n + 2 = V e 2 n + 1 .

Let r n = σ ( e n , e n + 1 ) (9)

Substituting c = e 2 n and d = e 2 n + 1 in Equation (8), then for all n = 1 , 2 , 3 , we have

0 ϕ 1 ( σ ( U e 2 n , V e 2 n + 1 ) ) η ( y ) d y 0 ϕ 1 ( σ ( e 2 n + 1 , e 2 n + 2 ) ) η ( y ) d y 0 ψ 1 ( σ ( e 2 n , e 2 n + 1 ) , σ ( U e 2 n , e 2 n ) , σ ( V e 2 n + 1 , e 2 n + 1 ) , 1 2 { σ ( V e 2 n + 1 , e 2 n ) + σ ( U e 2 n , e 2 n + 1 ) } , 1 2 { σ ( e 2 n , e 2 n + 1 ) + σ ( U e 2 n , e 2 n ) } ) η ( y ) d y 0 ψ 2 ( σ ( e 2 n , e 2 n + 1 ) , σ ( U e 2 n , e 2 n ) , σ ( V e 2 n + 1 , e 2 n + 1 ) , 1 2 { σ ( V e 2 n + 1 , e 2 n ) + σ ( U e 2 n , e 2 n + 1 ) } , 1 2 { σ ( e 2 n , e 2 n + 1 ) + σ ( U e 2 n , e 2 n ) } ) η ( y ) d y 0 ψ 1 ( σ ( e 2 n , e 2 n + 1 ) , σ ( e 2 n + 1 , e 2 n ) , σ ( e 2 n + 2 , e 2 n + 1 ) , 1 2 { σ ( e 2 n + 2 , e 2 n ) + σ ( e 2 n + 1 , e 2 n + 1 ) } , 1 2 { σ ( e 2 n , e 2 n + 1 ) + σ ( e 2 n + 1 , e 2 n ) } ) η ( y ) d y 0 ψ 2 ( σ ( e 2 n , e 2 n + 1 ) , σ ( e 2 n + 1 , e 2 n ) , σ ( e 2 n + 2 , e 2 n + 1 ) , 1 2 { σ ( e 2 n + 2 , e 2 n ) + σ ( e 2 n + 1 , e 2 n + 1 ) } , 1 2 { σ ( e 2 n , e 2 n + 1 ) + σ ( e 2 n + 1 , e 2 n ) } ) η ( y ) d y

Using Equation (9) for all n = 1 , 2 , 3 , we get

0 ϕ 1 ( r 2 n + 1 ) η ( y ) d y 0 ψ 1 ( r 2 n , r 2 n , r 2 n + 1 1 2 { r 2 n + 1 + r 2 n + 0 } , 1 2 { r 2 n + r 2 n } ) η ( y ) d y 0 ψ 2 ( r 2 n , r 2 n , r 2 n + 1 1 2 { r 2 n + 1 + r 2 n + 0 } , 1 2 { r 2 n + r 2 n } ) η ( y ) d y (10)

As r 2 n + 1 > r 2 n implies that r 2 n + 1 + r 2 n 2 r 2 n + 1 , so we have

0 ϕ 1 ( r 2 n + 1 ) η ( y ) d y 0 ψ 1 ( r 2 n , r 2 n , r 2 n + 1 r 2 n + 1 , r 2 n ) η ( y ) d y = 0 ϕ 1 ( r 2 n + 1 ) η ( y ) d y (11)

Now by monotone increase of ψ 1 in all variables and using the property that ψ 2 ( r 2 n , r 2 n , r 2 n + 1 , r 2 n + 1 , r 2 n ) 0 whenever r 2 n + 1 0 , we get a contradiction i.e. r 2 n + 1 not greater than r 2 n . Hence we have r 2 n + 1 r 2 n , for

n = 0 , 1 , 2 , 3 , (12)

Substituting c = e 2 n 1 , d = e 2 n in Equation (8) we have

0 ϕ 1 ( r 2 n ) η ( y ) d y 0 ψ 1 ( r 2 n 1 , r 2 n 1 , r 2 n , r 2 n , r 2 n 1 ) η ( y ) d y 0 ψ 2 ( r 2 n 1 , r 2 n 1 , r 2 n , r 2 n , r 2 n 1 ) η ( y ) d y (13)

By using (12) we consider

r 2 n + 2 r 2 n + 1 (14)

From (10) and (12) we obtain

r n + 1 r n (15)

From (8) & (11) for all n = 1 , 2 , 3 , , we have

0 ϕ 1 ( r n + 1 ) η ( y ) d y 0 ψ 1 ( r n ) η ( y ) d y 0 ψ 2 ( r n ) η ( y ) d y

then

0 ϕ 2 ( r n + 1 ) η ( y ) d y 0 ϕ 1 ( r n ) η ( y ) d y 0 ϕ 1 ( r n + 1 ) η ( y ) d y

Taking summation in above equation we obtain

0 0 ϕ 2 ( r n + 1 ) η ( y ) d y 0 ϕ 1 ( r 0 ) η ( y ) d y < ,

which implies ϕ 2 ( r ) 0 as n . (16)

Now from (13) sequence { r n } is convergent and as n , r n r . We know that ϕ is continuous and from Equation (14) we obtain ϕ 2 ( r ) = 0 which implies that r = 0 , i.e. as

n , r = σ ( e n + 1 , e n ) 0. (17)

We now show that the sequence { e n } is a Cauchy sequence in E. Keeping in mind Equation (15) it is require to show that { e 2 s } s = 1 { e n } is a Cauchy sequence. If { e 2 s } s = 1 is not a Cauchy sequence of natural number { 2 m ( k ) } , { 2 n ( k ) } such that n ( k ) > m ( k ) , σ ( e 2 m ( k ) , e 2 n ( k ) )

σ ( e 2 m ( k ) , e 2 n ( k ) 1 ) < (18)

Hence using (16)

< σ ( e 2 m ( k ) , e 2 n ( k ) ) σ ( e 2 m ( k ) , e 2 n ( k ) 1 ) + σ ( e 2 n ( k ) , e 2 n ( k ) 1 ) < + σ ( e 2 n ( k ) , e 2 n ( k ) 1 ) .

Taking k in the inequality above & by result of Equation (15), we arrive at

lim n σ ( e 2 m ( k ) , e 2 n ( k ) ) = . (19)

For all k = 1 , 2 , 3 ,

σ ( e 2 n ( k ) + 1 , e 2 m ( k ) ) σ ( e 2 n ( k ) + 1 , e 2 n ( k ) ) + σ ( e 2 n ( k ) , e 2 m ( k ) ) (20)

Also for k = 1 , 2 , 3 ,

σ ( e 2 n ( k ) , e 2 m ( k ) ) σ ( e 2 n ( k ) , e 2 n ( k ) + 1 ) + σ ( e 2 n ( k ) + 1 , e 2 m ( k ) ) . (21)

Making k in (18) & (19) respectively by using (15) & (17) we have

lim k σ ( e 2 n ( k ) + 1 , e 2 m ( k ) )

and lim k σ ( e 2 n ( k ) + 1 , e 2 m ( k ) )

Therefore, lim k σ ( e 2 n ( k ) + 1 , e 2 m ( k ) ) = , for k = 1 , 2 , 3 , (22)

σ ( e 2 n ( k ) , e 2 m ( k ) 1 ) σ ( e 2 n ( k ) , e 2 m ( k ) ) + σ ( e 2 m ( k ) , e 2 m ( k ) 1 ) ,

σ ( e 2 n ( k ) , e 2 m ( k ) ) σ ( e 2 n ( k ) , e 2 m ( k ) 1 ) + σ ( e 2 m ( k ) 1 , e 2 m ( k ) ) .

Taking k in the above two inequalities and using (15) & (17) we obtain

lim n σ ( e 2 n ( k ) , e 2 m ( k ) 1 ) = . (23)

Putting c = e 2 n ( k ) , d = e 2 m ( k ) 1 in (8), for all k = 1 , 2 , 3 , , we obtain

0 ϕ 1 σ ( e 2 n ( k ) + 1 , e 2 m ( k ) ) η ( y ) d y = 0 ϕ 1 σ ( U e 2 n ( k ) , V e 2 m ( k ) 1 ) η ( y ) d y 0 ψ 1 ( σ ( e 2 n , e 2 m ( k ) 1 ) , σ ( e 2 n ( k ) + 1 , e 2 n ) , σ ( e 2 m ( k ) , e 2 m ( k ) 1 ) , 1 2 { σ ( e 2 m ( k ) , e 2 n ( k ) ) + σ ( e 2 n ( k ) + 1 , e 2 m ( k ) 1 ) } , 1 2 { σ ( e 2 n ( k ) , e 2 m ( k ) 1 ) + σ ( e 2 n ( k ) + 1 , e 2 n ( k ) ) } ) η(y)dy 0 ψ 2 ( σ ( e 2 n , e 2 m ( k ) 1 ) , σ ( e 2 n ( k ) + 1 , e 2 n ) , σ ( e 2 m ( k ) , e 2 m ( k ) 1 ) , 1 2 { σ ( e 2 m ( k ) , e 2 n ( k ) ) + σ ( e 2 n ( k ) + 1 , e 2 m ( k ) 1 ) } , 1 2 { σ ( e 2 n ( k ) , e 2 m ( k ) 1 ) + σ ( e 2 n ( k ) + 1 , e 2 n ( k ) ) } ) η(y)dy

Now in above inequality if we take k and by using results of (15), (20) & (21) we get

0 ϕ 1 ( ) η ( y ) d y 0 ψ 1 ( , 0 , 0 , , 1 2 ) η ( y ) d y 0 ψ 2 ( , 0 , 0 , , 1 2 ) η ( y ) d y .

Then ϕ 1 ( ) ψ 1 ( , 0 , 0 , , 1 2 ) ψ 2 ( , 0 , 0 , , 1 2 ) = ϕ 1 ( ) .

This is due to monotone increasing fact of ψ 1 in its variable and by using property of ψ 2 that ψ 2 ( y 1 , y 2 , y 3 , y 4 , y 5 ) = 0 , if and only if y 1 = y 2 = y 3 = y 4 = y 5 = 0 .

From the above inequality we get a contradiction. So that = 0 . This establishes convergent sequence in ( E , σ ) .

Let e n z as n . (24)

Substituting c = e 2 n , d = z in (8) for all n = 1 , 2 , 3 ,

0 ϕ 1 ( σ ( e 2 n + 1 , V z ) ) η ( y ) d y 0 ψ 1 ( σ ( e 2 n , z ) , σ ( e 2 n + 1 , e 2 n ) , σ ( V z , z ) , 1 2 { σ ( V z , e 2 n ) + σ ( e 2 n + 1 , z ) } , 1 2 { σ ( e 2 n , z ) + σ ( e 2 n + 1 , e 2 n ) } ) η ( y ) d y 0 ψ 2 ( σ ( e 2 n , z ) , σ ( e 2 n + 1 , e 2 n ) , σ ( V z , z ) , 1 2 { σ ( V z , e 2 n ) + σ ( e 2 n + 1 , z ) } , 1 2 { σ ( e 2 n , z ) + σ ( e 2 n + 1 , e 2 n ) } ) η ( y ) d y

Taking limit n tends to infinity in the above inequality and using continuity of ψ 1 and ψ 2 and Equations (15), (22) we get

0 ϕ 1 ( σ ( z , V z ) ) η ( y ) d y 0 ψ 1 ( σ ( z , z ) , σ ( z , z ) , σ ( V z , z ) , 1 2 { σ ( V z , z ) } , 0 ) η ( y ) d y 0 ψ 2 ( σ ( z , z ) , σ ( z , z ) , σ ( V z , z ) , 1 2 { σ ( V z , z ) } , 0 ) η ( y ) d y 0 ψ 1 ( 0 , 0 , σ ( V z , z ) , 1 2 { σ ( V z , z ) } , 0 ) η ( y ) d y 0 ψ 2 ( 0 , 0 , σ ( V z , z ) , 1 2 { σ ( V z , z ) } , 0 ) η ( y ) d y

If ( V z , z ) 0 then monotone increasing ψ 1 and ψ 2 are monotone increasing and ψ 2 ( y 1 , y 2 , y 3 , y 4 , y 5 ) = 0 , if and only if y 1 = y 2 = y 3 = y 4 = y 5 = 0 , we obtain

0 ϕ 1 ( σ ( z , V z ) η ( y ) d y 0 ϕ 1 ( σ ( z , V z ) η ( y ) d y .

This contradiction, hence we obtain ( V z , z ) = 0. (25)

In similar way we prove that z = U z . Hence z = U z = V z . (26)

Hence (25) & (26) shows that z is a common fixed point of U and V.

Theorem 2.2: Let ( E , σ ) be a complete metric space and p, q, U and V be four mappings from E to itself such that

0 ϕ 1 ( σ ( p s , q t ) ) η ( y ) d y 0 ψ 1 ( σ ( U s , V t ) , σ ( U s , q t ) , σ ( p s , V t ) , σ ( U s , p s ) , σ ( V t , q t ) , 1 2 { σ ( q t , U s ) + σ ( p s , V t ) } , 1 2 { σ ( U s , V t ) + σ ( p s , U s ) } ) η ( y ) d y 0 ψ 2 ( σ ( U s , V t ) , σ ( U s , q t ) , σ ( p s , V t ) , σ ( U s , p s ) , σ ( V t , q t ) , 1 2 { σ ( q t , U s ) + σ ( p s , V t ) } , 1 2 { σ ( U s , V t ) + σ ( p s , U s ) } ) η ( y ) d y , (27)

for all s , t E , where ψ 1 , ψ 2 Ψ 7 , ϕ 1 = ψ ( e , e , e , e , e , e , e ) , for e [ 0 , ) .

i: One of the four mappings p, q, U and V is continuous.

ii: (p, U) & (q, V) are sub compatible.

iii: The pairs p ( s ) V ( s ) and q ( s ) U ( s ) .

iv: Where η : R + R + is Lebesgue-integrable mappings, which is sum able, non negative and such that for each > 0 , 0 η ( y ) d y > 0 .

Then p, q, U and V have a unique common fixed point in E.

Proof: Consider arbitrary point e 0 E , we construct the sequence { e n } and { w n } in E such that

p e 2 n = V e 2 n + 1 = w 2 n and q e 2 n + 1 = U e 2 n + 2 = w n + 1 , n = 0 , 1 , 2 ,

Let r n = σ ( w n , w n + 1 ) , Substitution s = e 2 n and t = e 2 n + 1 in (27) we have

0 ϕ 1 ( σ ( p e 2 n , q e 2 n + 1 ) ) η ( y ) d y = 0 ϕ 1 ( σ ( e 2 n + 1 , e 2 n + 2 ) ) η ( y ) d y 0 ψ 1 ( σ ( U e 2 n , V e 2 n + 1 ) , σ ( U e 2 n , q e 2 n + 1 ) , σ ( p e 2 n + 1 , V e 2 n + 1 ) , σ ( U e 2 n , p e 2 n ) , σ ( V e 2 n + 1 , q e 2 n + 1 ) , 1 2 { σ ( q e 2 n + 1 , U e 2 n ) + σ ( p e 2 n , V e 2 n + 1 ) } , 1 2 { σ ( U e 2 n , V e 2 n + 1 ) + σ ( p e 2 n , U e 2 n ) } ) η ( y ) d y 0 ψ 2 ( σ ( U e 2 n , V e 2 n + 1 ) , σ ( U e 2 n , q e 2 n + 1 ) , σ ( p e 2 n + 1 , V e 2 n + 1 ) , σ ( U e 2 n , p e 2 n ) , σ ( V e 2 n + 1 , q e 2 n + 1 ) , 1 2 { σ ( q e 2 n + 1 , U e 2 n ) + σ ( p e 2 n , V e 2 n + 1 ) } , 1 2 { σ ( U e 2 n , V e 2 n + 1 ) + σ ( p e 2 n , U e 2 n ) } ) η ( y ) d y

0 ϕ ( σ ( w 2 n , w 2 n + 1 ) ) η ( y ) d y 0 ψ 1 ( σ ( w 2 n 1 , w 2 n ) , σ ( w 2 n 1 , w 2 n + 1 ) , σ ( w 2 n + 1 , w 2 n ) , σ ( w 2 n 1 , w 2 n ) , σ ( w 2 n , w 2 n + 1 ) , 1 2 { σ ( w 2 n + 1 , w 2 n ) + σ ( w 2 n , w 2 n ) } , 1 2 σ ( w 2 n 1 , w 2 n ) + σ ( w 2 n , w 2 n 1 ) ) η ( y ) d y 0 ψ 2 ( σ ( w 2 n 1 , w 2 n ) , σ ( w 2 n 1 , w 2 n + 1 ) , σ ( w 2 n + 1 , w 2 n ) , σ ( w 2 n 1 , w 2 n ) , σ ( w 2 n , w 2 n + 1 ) , 1 2 { σ ( w 2 n + 1 , w 2 n ) + σ ( w 2 n , w 2 n ) } , 1 2 σ ( w 2 n 1 , w 2 n ) + σ ( w 2 n , w 2 n 1 ) ) η ( y ) d y

0 ϕ 1 ( r 2 n ) η ( y ) d y 0 ψ 1 ( r 2 n 1 , r 2 n 1 + r 2 n , r 2 n , r 2 n 1 , r 2 n , 1 2 { r 2 n } , 1 2 { r 2 n 1 + r 2 n 1 } ) η ( y ) d y 0 ψ 2 ( r 2 n 1 , r 2 n 1 + r 2 n , r 2 n , r 2 n 1 , r 2 n , 1 2 { r 2 n } , 1 2 { r 2 n 1 + r 2 n 1 } ) η ( y ) d y

If r 2 n + 1 r 2 n then r 2 n + 1 + r 2 n 2 r 2 n and

0 ϕ 1 ( r 2 n ) η ( y ) d y 0 ψ 1 ( r 2 n 1 , 2 r 2 n , r 2 n , r 2 n 1 , r 2 n , 1 2 { r 2 n } , r 2 n 1 ) η ( y ) d y 0 ψ 2 ( r 2 n 1 , 2 r 2 n , r 2 n , r 2 n 1 , r 2 n , 1 2 { r 2 n } , r 2 n 1 ) η ( y ) d y < 0 ϕ 1 ( r 2 n ) η ( y ) d y . (28)

Thus we arrive at a contradiction. Hence r 2 n r 2 n 1 , similarly by substituting s = r 2 n + 2 , t = r 2 n + 1 in (27) we can prove that, r 2 n + 1 r 2 n , for n = 0 , 1 , 2 , . Thus r n + 1 r n , for n = 0 , 1 , 2 , . Hence the sequence { r n } is sequence of positive real numbers, which is decreasing and converges to r R .

Let m = lim n 1 2 d ( w n , w n + 2 ) . Taking n in (27) we have

0 ϕ 1 ( r 2 n ) η ( y ) d y 0 ψ 1 ( r , r , r , r , r , r , r ) η ( y ) d y 0 ψ 2 ( r , r , r , r , r , r , m ) η ( y ) d y 0 ϕ 1 ( r ) η ( y ) d y 0 ψ 2 ( r , r , r , r , r , r , m ) η ( y ) d y .

Thus ψ 2 ( r , r , r , r , r , r , m ) = 0 So that r = m = 0. Hence lim n d ( y n , y n + 1 ) = 0 (29)

In view of (29), to prove sequence { w n } is a Cauchy sequence it is sufficient to prove the subsequence { w 2 n } of sequence { w n } is a Cauchy sequence. If { w 2 n } is not a Cauchy sequence there exist > 0 & sequence of natural numbers { 2 m ( k ) } & { 2 n ( k ) } which are monotone increasing such that n ( k ) > m ( k ) .

σ ( w 2 m ( k ) , w 2 n ( k ) ) & σ ( w 2 m ( k ) , w 2 n ( k ) 2 ) < . (30)

Then from (29) we have

< σ ( w 2 m ( k ) , w 2 n ( k ) ) σ ( w 2 m ( k ) , w 2 n ( k ) 2 ) + σ ( w 2 n ( k ) 1 , w 2 n ( k ) 2 ) + σ ( w 2 n ( k ) 1 , w 2 n ( k ) ) < + σ ( w 2 n ( k ) 1 , w 2 n ( k ) 2 ) + σ ( w 2 n ( k ) 1 , w 2 n ( k ) ) . (31)

Taking k and using (29) we have

lim n σ ( w 2 m ( k ) , w 2 n ( k ) ) = . (32)

Taking k using (29) & (30) in

| σ ( w 2 m ( k ) , w 2 n ( k ) + 1 ) σ ( w 2 m ( k ) , w 2 n ( k ) ) | σ ( w 2 n ( k ) , w 2 n ( k ) + 1 ) . (33)

We get lim n σ ( w 2 n ( k ) + 1 , w 2 m ( k ) ) = . (34)

Letting k and from Equations (29) & (30) in

| σ ( w 2 m ( k ) 1 , w 2 n ( k ) ) σ ( w 2 m ( k ) , w 2 n ( k ) ) | σ ( w 2 m ( k ) , w 2 m ( k ) 1 ) .

We get lim k σ ( w 2 m ( k ) , w 2 m ( k ) 1 ) = . (35)

Putting s = x 2 m ( k ) , t = x 2 n ( k ) 1 in (27), for all k = 1 , 2 , 3 , we obtain

Taking k & using (29), (30), (32), (33) & (35) we get

0 ϕ 1 ( ) η ( y ) d y 0 ψ 1 ( , , , 0 , 0 , 1 2 [ + ] , 1 2 ) η ( y ) d y 0 ψ 2 ( , , , 0 , 0 , 1 2 [ + ] , 1 2 ) η ( y ) d y < 0 ψ 1 ( , , , , , , ) η ( y ) d y = 0 ϕ 1 ( ) η ( y ) d y .

This is contradiction. Hence { w 2 n } is a Cauchy sequence and is convergent. Since E is complete there exist z E such that as n we have w n z .

Case I: Assume that U is continuous then U p e 2 n U z , U 2 e 2 n U z . Since (p, U) is sub compatible, we have p U e 2 n U z .

Step I: Substituting s = U e 2 n , t = e 2 n + 1 in (27), we have

0 ϕ 1 σ ( p U e 2 n , q e 2 n + 1 ) η ( y ) d y 0 ψ 1 ( σ ( U 2 e 2 n , V e 2 n + 1 ) , σ ( U 2 e 2 n , q e 2 n + 1 ) , σ ( p U e 2 n , V e 2 n + 1 ) , σ ( U 2 e 2 n , p U e 2 n ) , σ ( V e 2 n + 1 , q e 2 n + 1 ) , 1 2 { σ ( q e 2 n + 1 , U 2 e 2 n ) + σ ( p U e 2 n , V e 2 n + 1 ) } , 1 2 { σ ( U 2 e 2 n , V e 2 n + 1 ) + σ ( p U e 2 n , U 2 e 2 n ) } ) η ( y ) d y 0 ψ 2 ( σ ( U 2 e 2 n , V e 2 n + 1 ) , σ ( U 2 e 2 n , q e 2 n + 1 ) , σ ( p U e 2 n , V e 2 n + 1 ) , σ ( U 2 e 2 n , p U e 2 n ) , σ ( V e 2 n + 1 , q e 2 n + 1 ) , 1 2 { σ ( q e 2 n + 1 , U 2 e 2 n ) + σ ( p U e 2 n , V e 2 n + 1 ) } , 1 2 { σ ( U 2 e 2 n , V e 2 n + 1 ) + σ ( p U e 2 n , U 2 e 2 n ) } ) η ( y ) d y ,

0 ϕ 1 ( σ ( U z , z ) ) η ( y ) d y 0 ψ 1 ( σ ( U z , z ) , σ ( U z , z ) , σ ( U z , z ) , σ ( U z , U z ) , σ ( z , z ) , 1 2 { σ ( z , U z ) + σ ( U z , z ) } , 1 2 { σ ( U z , z ) + σ ( U z , U z ) } ) η ( y ) d y 0 ψ 2 ( σ ( U z , z ) , σ ( U z , z ) , σ ( U z , z ) , σ ( U z , U z ) , σ ( z , z ) , 1 2 { σ ( z , U z ) + σ ( U z , z ) } , 1 2 { σ ( U z , z ) + σ ( U z , U z ) } ) η ( y ) d y 0 ψ 1 ( σ ( U z , z ) , σ ( U z , z ) , σ ( U z , z ) , 0 , 0 , σ ( U z , z ) , 1 2 { σ ( U z , z ) } ) η ( y ) d y 0 ϕ 1 ( σ ( U z , z ) ) η ( y ) d y .

It is contradiction if U z z . Hence U z = z .

Step II: Substituting s = z , t = e 2 n + 1 in (27) and taking limit as n tends to infinity we get p z = z .

Step III: We know that z = p z p ( e ) V ( e ) then there exist u E such that z = V u . Substituting s = e 2 n , t = u in (27) we get q z = z . Hence q z = z = V z and q V u = V q u , which gives q z = V z .

Step IV: Substituting s = z , t = z in (27) we have q z = z so that q ( z ) = z = V z . Hence p, q, U & V have a common fixed point z in E.

Case II: Assume that U is continuous then p 2 e 2 n p z , p U e 2 n p z . Similarly we can prove that z is common fixed point of p, q, U & V. When q or V is continuous, then the uniqueness of common fixed point follows easily from (27).

Example: Let E = [ 0 , 1 ] with the usual metric σ ( s , t ) = 1 2 | s t | . Define p , q , U , V : E E such that p s = s 4 , q t = t 4 , U s = s , V t = t .

Let ψ 1 ( y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 ) = max ( y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 ) , φ ( y ) = 2 y ,

ψ 2 = 1 4 ψ 1 then ψ 1 ( y ) = y , y [ 0 , )

| s 4 t 4 | 2 1 4 max { σ ( s , t ) , σ ( s , t 4 ) , σ ( s 4 , t ) , σ ( s , s 4 ) , σ ( t , t 4 ) , 1 2 { σ ( t 4 , s ) + σ ( s 4 , t ) } , 1 2 { σ ( s , t ) + σ ( s 4 , s ) } }

For all s , t E , it follows that the condition (27).

Let { e n } be a sequence in E such that p e n z & U e n z for some z in E. Then z = 0, σ ( p U e n , U p e n ) 0 . Hence { p , U } is sub compatible. We have common fixed point in E.

3. Conclusion

In this paper, we proved the fixed point theorem for four sub compatible maps under a contractive condition of integral type. These results can be extended to any directions and can also be extended to fixed point theory of non-expansive multi-valued mappings.

Acknowledgements

The authors would like to give their sincere thanks to the editor and the anonymous referees for their valuable comments and useful suggestions in improving the article.

Cite this paper
Mishra, V. , Wadkar, B. , Bhardwaj, R. , Khan, I. and Singh, B. (2017) Common Fixed Point Theorems in Metric Space by Altering Distance Function. Advances in Pure Mathematics, 7, 335-344. doi: 10.4236/apm.2017.76020.
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