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