Common Fixed Point Theorems in Metric Space by Altering Distance Function

(1)

http://www.scirp.org/journal/apm

ISSN Online: 2160-0384 ISSN Print: 2160-0368

Common Fixed Point Theorems in Metric Space

by Altering Distance Function

Vishnu Narayan Mishra

1,2*

, Balaji Raghunath Wadkar

3

, Ramakant Bhardwaj

4

, Idrees A. Khan

5

,

Basant Singh

3

1_{Applied Mathematics & Humanities Department, S.V. National Institute of Technology, Gujarat, India} 2_{Department of Mathematics, Indira Gandhi National Tribal University, Madhya Pradesh, India} 3_{Department of Mathematics, AISECT University, Bhopal, India}

4_{Department of Mathematics, TIT Group of Institutes, Anand Nagar, India} 5_{Department of Mathematics, Integral University, Lucknow, India}

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 com-mon fixed point theorem by considering four sub compatible maps under a contractive condition of integral type.

Keywords

Altering Distance Function, Sub Compatible

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 func-tion. 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

How to cite this paper: Mishra, V.N., Wadkar, B.R., Bhardwaj, R., Khan, I.A. and Singh, B. (2017) Common Fixed Point Theo-rems in Metric Space by Altering Distance Function. Advances in Pure Mathematics, 7, 335-344.

https://doi.org/10.4236/apm.2017.76020 Received: April 4, 2017

Accepted: June 6, 2017 Published: June 9, 2017

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under contraction rule of integral type in metric space by altering distance func-tion, 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

func-tions if ψ_{is continuous with monotone increasing in all variables and}

(

x x x1, 2, 3, ,xn

)

0

ψ

_ = _if x₁=x₂ =x₃==xn =0.

The collection of all altering distance is denoted by Ψn.

Now let us define a function m y

( )

by m y

( )

=

ψ

(

y y y y, , , ,,y

)

for

[

0,

)

y∈ ∞ , clearly m y

( )

=0 if and only if y=0.

Examples of ψ_are

ψ

(

f f1, 2,f3,,fn

)

=

µ

max

{

f f1, 2,f3,,fn

}

, for

0

µ

> _{, (1)}

(

)

1 2 3

1, 2, 3, , , 1, 2, , 1.

n

a a

n n n n n n

f f f f f f f f a a a

ψ _ = + + +_+ _ ≥ ₍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

{ }

en in E such that lim n lim n ,

n→∞p =n→∞q =t t∈E and which satisfies nlim→∞σ

(

pqe qpen, n

)

=0.

Example 1.3: Let E=

[ )

0,∞ we define p & q with metric σ as follows

( )

2

p e =e &

( )

₂ 6 if

[ ]

_[

4, 9

_]

(

27,

)

, if 9, 27

e e

q e

e e

 + ∈ ∞



=  _∈





(3)

Let us define the sequence

{ }

en in E as

1 3 n e

n

= + , for n=0,1, 2, then

(

)

2

lim n lim n 9 lim n lim n 6 ,

n→∞pe =n→∞e = =n→∞qe =n→∞ e + (4)

and

( )

(

)

(

)

2

lim _n lim _n 6 lim _n 6 81 when ,

n→∞pq e =n→∞p e + =n→∞ e + = n→ ∞ (5)

( )

2

( )

4 2 2

lim _n lim _n _n lim _n 81 when .

n→∞qpe =n→∞qe = e =n→∞ e = n→ ∞ (6)

Thus, we have lim

(

n, n

)

0

n→∞σ pqe qpe = . (7)

Hence maps p and q are sub-compatible.

On the other hand, we have pe=qe if and only if e=3,

( )

3

( )

9 81

pq = p = and qp

( )

3 =q

( )

9 = + =9 6 15.

Then p

( )

3 = =9 q

( )

3 but pq

( )

3 =81 15≠ =qp

( )

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

( )

( ) ( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

_{( )}

( )

( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

1 1

2

1 1

, , , ,, , , , , , , , ,

2 2

0 0

1 1

, , ,, , , , , , , , ,

2 2

0

d d

d ,

Uc Vd c d Uc c Vd d Vd c Uc d c d Uc c

c d Uc c Vd d Vd c Uc d c d Uc c

y y y y

y y

φ σ ψ σ σ σ σ σ σ σ

ψ σ σ σ σ σ σ σ

η η

η

 ₊ ₊ 

 

 

 ₊ ₊ 

 

 

≤

−

∫

(8)

where ψ ψ ∈ Ψ1, 2 5 with

φ ψ

1=

(

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 dy 0

(3)

and V.

Proof: Consider arbitrary point e0 of E, for n=1, 2, 3, we have 2n1 2n

e + =Ue

and e2n+2=Ve2n+1.

Let rn=

σ

(

e en, n+1

)

(9)

Substituting c=e2n and d =e2n+1 in Equation (8), then for all n=1, 2, 3,

we have

( )

( ) ( ) ( ( ))

_{( )}

( )

( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

( )

( ) ( ) ( ) ( )

1 2 2 1 1 2 1 2 2

1 2 2 1 2 2 2 1 2 1 2 1 2 2 2 1 2 2 1 2 2

2 2 2 1 2 2 2 1 2 1 2 1 2 2

, , 0 0 1 1 , , , , , , , , , , , 2 2 0 1 , , , , , , , 2 0 d d d d

n n n n

n n n n n n n n n n n n n n

n n n n n n n n

Ue Ve e e

e e Ue e Ve e Ve e Ue e e e Ue e

e e Ue e Ve e Ve e Ue

y y y y

y y

φ σ φ σ

ψ σ σ σ σ σ

η η η η + + + + + + + + + + + + +  ₊ ₊      + ≤ ≤ −

∫

( ) { } { ( ) ( )}

( )

( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

( )

( ) ( ) ( ) ( )

2 1 2 2 1 2 2

1 2 2 1 2 1 2 2 2 2 1 2 2 2 2 1 2 1 2 2 1 2 1 2

2 2 2 1 2 1 2 2 2 2 1 2 2 2 2 1 1 , , , , 2 1 1 , , , , , , , , , , , 2 2 0 1 , , , , , , , , 2 0 d d

n n n n n n

n n n n n n n n n n n n n n

n n n n n n n n n

e e e Ue e

e e e e e e e e e e e e e e

e e e e e e e e e e

y y

σ σ

ψ σ σ σ σ σ

η η + + + + + + + + + + + + + + + + +  ₊       ₊ ₊      + ≤ −

∫

( )

{ 2 1} { (2 2 1) (2 1 2 )}

1

, , ,

2

n+ σ enen+ σ en+ en

 ₊ 

 

 

∫

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

( )

( ) { } { }

_{( )}

( )

{ } { }

1 2 1 1 2 2 2 1 2 1 2 2 2

2 2 2 2 1 2 1 2 2 2

1 1

, , 0 ,

2 2

0 0

1 1

, , 0 ,

2 2

0

d d

d

n n n n n n n n

n n n n n n n

r r r r r r r r

r r r r r r r

y y y y

y y φ ψ ψ η η η + + + + +  ₊ ₊ ₊       ₊ ₊ ₊      ≤ −

∫

(10)

As r2n+1>r2n implies that r2n+1+r2n ≤2r2n+1, so we have

( )

1 2 1 1 2 ,2,2 1 2 1 2, 1 2 1

0 d 0 d 0 d

n n n n n n n

r r r r r r r

y y y y y y

φ ψ φ

η η η

+ + + +

≤ =

∫

(11)

Now by monotone increase of ψ1 in all variables and using the property that

(

)

2 r2n,r2n,r2n1,r2n1,r2n 0

ψ

+ + ≠ whenever r2n+1≠0, we get a contradiction i.e.

2n1

r + not greater than r2n. Hence we have r2n+1≤r2n, for

0,1, 2, 3,

n=  (12)

Substituting c=e2n−1,d=e2n in Equation (8) we have

( )

1 2 1 2 1 2, 1 2, ,2 ,2 1 2 2 1 2, 1 2, ,2,2 1

0 d 0 d 0 d

n n n n n n n n n n n

r r r r r r r r r r r

y y y y y y

φ ψ ψ

η _≤ − − − η ₋ − − − η

∫

(13)

By using (12) we consider

2n 2 2n1

r + ≤r + (14)

From (10) and (12) we obtain

1

n n

r+ ≤r (15)

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

( )

1 1 1 2

0 d 0 d 0 d

n n n

r r r

y y y y y y

φ ψ ψ

η η η

+ ≤ −

∫

then

( )

2 1 1 1 1

0 d 0 d 0 d

n n n

r r r

y y y y y y

φ φ φ

η η η

+ +

≤ −

(4)

Taking summation in above equation we obtain

( )

( ) ( )

_{( )}

2 1 1 0

0 0

0

d d

n

r r

y y y y

φ φ

η

+ ∞

≤ < ∞

∑∫

∫

,

which implies

φ

2

( )

r →0 as n→ ∞. (16)

Now from (13) sequence

{ }

rn is convergent and as n→ ∞, rn →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

(

en+1,en

)

→0. (17)

We now show that the sequence

{ }

en is a Cauchy sequence in E. Keeping in

mind Equation (15) it is require to show that

{ }

e2s s1

{ }

en ∞

= ⊂ is a Cauchy

se-quence. If

{ }

e2s s1

∞

= is not a Cauchy sequence of natural number

( )

{

2m k

}

, 2

{

n k

( )

}

such that n k

( )

>m k

( )

, σ

(

e₂_{m k}_{( )},e₂_{n k}_{( )}

)

≥ ∈

( ) ( )

(

e2m k ,e2n k 1

)

σ ₋ < ∈ (18)

Hence using (16)

( ) ( )

(

)

( ) ( )

(

)

(

( ) ( )

)

( ) ( )

(

)

2 2

2 2 1 2 2 1

2 2 1

,

, ,

, .

m k n k

m k n k n k n k

n k n k

e e

e e e e

e e

σ

σ σ

σ

− −

−

∈ <

≤ +

< ∈ +

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

ar-rive at

( ) ( )

(

2 2

)

lim _{m k}, _{n k}

n→∞σ e e = ∈. (19)

For all k=1, 2, 3,

( ) ( )

(

e2n k 1,e2m k

)

(

e2n k( )1,e2n k( )

)

(

e2n k( ),e2m k( )

)

σ ₊ ≤σ ₊ +σ ₍₂₀₎

Also for k=1, 2, 3,

( ) ( )

(

e2n k ,e2m k

)

(

e2n k( ),e2n k( )1

)

(

e2n k( )1,e2 ( )m k

)

σ ≤σ ₊ +σ ₊ _{. (21)}

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

( ) ( )

(

2 1 2

)

lim _{n k} , _{m k} k→∞σ e + e ≤ ∈

and ∈ ≤_klim_→∞σ

(

e2n k( )+1,e2m k( )

)

Therefore, lim_k_→∞σ

(

e2n k( )+1,e2m k( )

)

= ∈, for k=1, 2, 3, (22) ( ) ( )

(

e2n k ,e2m k 1

)

(

e2n k( ),e2m k( )

)

(

e2m k( ),e2m k( )1

)

,

σ ₋ ≤σ +σ ₋

( ) ( )

(

e2n k ,e2m k

)

(

e2n k( ),e2m k( ) 1

)

(

e2m k( )1,e2m k( )

)

.

σ ≤σ ₋ +σ ₋

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

( ) ( )

(

2 2 1

)

lim _{n k} , _{m k}

(5)

Putting c=e2n k( ),d =e2m k( )−1 in (8), for all k=1, 2, 3,, we obtain

( )

( ) ( )

(

)

(

( ) ( )

)

_{( )}

( )

(

)

(

( )

)

(

( ) ( )

)

{

(

( ) ( )

)

(

( ) ( )

)

}

{

(

( ) ( )

)

(

( ) ( )

)

}

( )

(

)

(

( )

)

(

( ) ( )

)

1 2 1 2 1 2 2 1

1 2 2 1 2 1 2 2 2 1 2 2 2 1 2 1 2 2 1 2 1 2

2 2 2 1 2 1 2 2 2 1 2

, ,

0 0

1 1

, , , , , , , , , , ,

2 2

0

1

, , , , , ,

2 0

d d

n k m k n k m k

n m k n k n m k m k m k n k n k m k n k m k n k n

n m k n k n m k m k

e e Ue Ve

e e e e e e e e e e e e e e k

e e e e e e e

y y y y

y y

φ σ φ σ

ψ σ σ σ σ

η η

η

+ −

− + − + − − +

− + −

 ₊ ₊ 

 

 

=

≤

−

∫

( ) ( )

(

)

(

( ) ( )

)

{

2 2 1 2 1

}

{

(

2( ) 2 ( )1

)

(

2( )1 2 ( )

)

}

1

, , , , ,

2 n

m k en k σ en k+ em k− σ en k em k− σen k+ e k

 ₊ ₊ 

 

 

∫

Now in above inequality if we take k→∞ and by using results of (15), (20)

& (21) we get

( )

1 1 2

1 1

,0,0, , ,0,0, ,

2 2

0 y dy 0 y dy 0 y d .y

φ ψ ψ

η  η  η

∈ _≤ _∈ ∈ ∈_ ₋ _∈ ∈ ∈_

∫

Then 1 1 2 1

( )

1 1

( ) , 0, 0, , , 0, 0, , .

2 2

φ ∈ ≤ψ _∈ ∈ ∈ −_ ψ _∈ ∈ ∈ =_ φ ∈

   

This is due to monotone increasing fact of ψ1 in its variable and by using

property of ψ2 that

ψ

2

(

y y y y y1, 2, 3, 4, 5

)

=0, if and only if

1 2 3 4 5 0

y = y = y =y =y = .

From the above inequality we get a contradiction. So that ∈ =0. This

estab-lishes convergent sequence in

(

E,

σ

)

.

Let en→z as n→ ∞. (24)

Substituting c=e2n, d=z in (8) for all n=1, 2, 3,

( )

( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

( )

( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

1 2 1

1 2 2 1 2 2 2 1 2 2 1 2

2 2 2 1 2 2 2 1 2 2 1 2

, 0

1 1

, , , , , , , , , , ,

2 2

0

1 1

, , , , , , , , , , ,

2 2

0

d

d d n

n n n n n n n n

e Vz

e z e e Vz z Vz e e z e z e e

y y

φ σ

η

+

+ + +

 ₊ ₊ 

 

 

 ₊ ₊ 

 

 

≤

−

∫

Taking limit n tends to infinity in the above inequality and using continuity of

1

ψ _andψ₂_{and Equations (15), (22) we get}

( )

( ) ( ) ( ) { ( )} ( ) ( ) ( ) { ( )}

_{( )}

( )

( ) { ( )} ( ) { ( )}

_{( )}

1

1 2

( , ) 0

1 1

, , , , , , , ,0 , , , , , , , ,0

2 2

0 0

1 1

0,0, , , , ,0 0,0, , , , ,0

2 2

0 0

d

d d

z Vz

z z z z Vz z Vz z z z z z Vz z Vz z

Vz z Vz z Vz z Vz z

y y

y y y y

φ σ

ψ σ σ σ σ ψ σ σ σ σ

ψ σ σ ψ σ σ

η

η η

   

   

   

   

   

   

≤ −

∫

If

(

Vz z,

)

≠0 then monotone increasing ψ₁ and ψ₂ are monotone

in-creasing and

ψ

2

(

y y y y y1, 2, 3, 4, 5

)

=0, if and only if y1=y2=y3=y4=y5=0,

we obtain

( )

1 ( , 1 ( ,

0 d 0 d .

z Vz z Vz

y y y y

φ σ φ σ

η ≤ η

∫

This contradiction, hence we obtain

(

Vz z,

)

=0. (25)

In similar way we prove that z=Uz. Hence z=Uz=Vz. (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

(6)

( )

( ) ( )

( )

( ) ( ) ( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

( )

( ) ( ) ( ) ( ) ( ) { ( ) ( )} { ( ) ( )} 1 1 2 , 0 1 1 , , , , , , , , , , , , , , , 2 2 0 1 1 , , , , , , , , , , , , , , , 2 2 0 d d d , ps qt

Us Vt Us qt ps Vt Us ps Vt qt qt Us ps Vt Us Vt ps Us

y y

φ σ

ψ σ σ σ σ σ σ σ σ σ

η η η  ₊ ₊       ₊ ₊      ≤ −

∫

(27)

for all s t, ∈E, where ψ ψ ∈ Ψ1, 2 7,

φ ψ

1=

(

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 dy>0.

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

Proof: Consider arbitrary point e0∈E, we construct the sequence

{ }

en and

{ }

wn in E such that

2n 2n1 2n

pe =Ve + =w and qe2n+1=Ue2n+2=wn+1, n=0,1, 2,

Let rn=

σ

(

w wn, n+1

)

, Substitution s=e2n and t=e2n+1 in (27) we have

( )

( ) ( ) ( ( ))

_{( )}

( )

( ) ( ) ( ) ( ) ( ) ( ( ) ( ) { } { ( ) ( )})

( )

1 2 2 1 1 2 1 2 2

1 2 2 1 2 2 1 2 1 2 1 2 2 2 1 2 1

2 1 2 2 2 1 2 2 1 2 2

2 2 2 1 2 2

, , 0 0 , , , , , , , , , , 1 1 , , , , , 2 2 0 , , , 0 d d d d

n n n n

n n n n n n n n n n

n n n n n n n n

n n n

pe qe e e

Ue Ve Ue qe pe Ve Ue pe Ve qe

qe Ue pe Ve Ue Ve pe Ue

Ue Ve Ue qe

y y y y

y y

φ σ φ σ

ψ σ σ σ σ σ

σ σ σ σ

ψ σ σ

η η η η + + + + + + + + + + + + + + + = ≤ −

∫

( ) ( ) ( ) ( ) ( ( ) ( )

{ 2 1 2 2 21 1} {2 (1 22 12 1) (2 22 2 )}2) 1 2 1

, , , , , , ,

1 1

, , , , ,

2 2

n n n n n n n

n n n n n n n n

pe Ve Ue pe Ve qe

qe Ue pe Ve Ue Ve pe Ue

σ σ σ

σ σ σ σ

+ + + + + + + + + +

∫

( )

( ) ( )

( )

( ) ( ) ( ) ( ) ( ) ( ( ) ( ) { } ( ) ( ))

( )

( ) ( ) ( ) ( ) ( ) ( 2 2 1

1 2 1 2 2 1 2 1 2 1 2 2 1 2 2 2 1

2 1 2 2 2 2 1 2 2 2 1

2 2 1 2 2 1 2 1 2 1 2 2 1 2 2 2 1 2 , 0 , , , , , , , , , , 1 1 , , , , , 2 2 0 , , , , , , , , , , 1 2 0 d d d n n

n n n n n n n n n n

n n n n n n n n

n n n n n n n n n n

n

w w

w w w w w w w w w w

w w w w w w w w

w w w w w w w w w w

w

y y

φ σ

ψ σ σ σ σ σ

σ σ σ σ

ψ σ σ σ σ σ

σ η η η + − − + + − + + − − − − + + − + + + + ≤ −

∫

( ) ( )

{ 1 2 2 2 } ( 2 1 2 ) ( 2 2 1))

1

, , , , ,

2

n n n n n n n

w +σ w w σ w − w +σ w w −

∫

( )

( ) { } { }

_{( )}

( )

{ } { }

1 2 1 2 1 2 1 2 2 , 2 1 2 2 2 1 2 1

2 2 1 2 1 2 2 , 2 1 2 2 2 1 2 1 1 1 , , , , , 2 2 0 0 1 1 , , , , , 2 2 0 d d d

n n n n n n n n n n

n n n n n n n n n

r r r r r r r r r r

r r r r r r r r r

y y y y

y y φ ψ ψ η η η − − − − − − − − − −  ₊ ₊       ₊ ₊      ≤ −

∫

If r2n+1≤r2n then r2n+1+r2n≤2r2n and

( )

( ) { }

_{( )}

( )

{ }

( )

1 2 1 2 1 2 2 , 2 1 2 2 2 1

2 2 1 2 2 , 2 1 2 2 2 1

1 2

1 ,2 , , , ,

2

0 0

1 ,2 , , , ,

2 0 0 d d d d .

n n n n n n n n

r r r r r r r r

r r r r r r r

r

y y y y

y y y y φ ψ ψ φ η η η η − − − − − −             ≤ − <

∫

(28)

Thus we arrive at a contradiction. Hence r2n≤r2n−1, similarly by substituting

2n2, 2n1

s=r + t=r + in (27) we can prove that, r2n+1 ≤r2n, for n=0,1, 2,. Thus

1

n n

r+ ≤r , for n=0,1, 2,. Hence the sequence

{ }

rn is sequence of positive real

numbers, which is decreasing and converges to r∈R.

Let

(

2

)

1

lim ,

2 n n

n

m d w w₊

→∞

(7)

( )

( ) ( )

_{( )}

( )

1 2 1 2

1 2

, , , , , , , , , , , ,

0 0 0

, , , , , ,

0 0

d d d

d d .

n

r r r r r r r r r r r r r r m

r r r r r r r m

y y y y y y

y y y y

φ ψ ψ

φ ψ

η η η

η η ≤ − ≤ −

∫

(

)

(

)

2 1

Thus , , , , , , 0 So that 0.

Hence lim _n, _n 0

n

r r r r r r m r m

d y y

ψ

+ →∞

= = =

= (29)

In view of (29), to prove sequence

{ }

wn is a Cauchy sequence it is sufficient

to prove the subsequence

{ }

w2n of sequence

{ }

wn is a Cauchy sequence. If

{ }

w2n is not a Cauchy sequence there exist ∈ >0 & sequence of natural

num-bers

{

2m k

( )

}

&

{

2n k

( )

}

which are monotone increasing such that

( )

n k >m k .

( ) ( )

(

w2m k ,w2n k

)

&

(

w2m k( ),w2n k( )2

)

.

σ ≥ ∈ σ ₋ < ∈ (30)

Then from (29) we have

( ) ( )

(

)

( ) ( )

(

)

(

( ) ( )

)

(

( ) ( )

)

( ) ( )

(

)

(

( ) ( )

)

2 2

2 2 2 2 1 2 2 2 1 2

2 1 2 2 2 1 2

,

, , ,

, , .

m k n k

m k n k n k n k n k n k

n k n k n k n k

w w

w w w w w w

w w w w

σ

σ σ σ

σ σ

− − − −

− − −

∈ <

≤ + +

< ∈ + +

(31)

Taking k→ ∞ and using (29) we have

( ) ( )

(

2 2

)

lim _{m k} , _{n k} .

n→∞σ w w = ∈ (32)

Taking k→ ∞ using (29) & (30) in

( ) ( )

(

w2m k ,w2n k 1

)

(

w2m k( ),w2n k( )

)

(

w2n k( ),w2n k( )1

)

.

σ ₊ −σ ≤σ ₊ (33)

We get lim

(

2n k( )1, 2m k( )

)

.

n→∞σ w + w = ∈ (34)

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

( ) ( )

(

w2m k 1,w2n k

)

(

w2m k( ),w2n k( )

)

(

w2m k( ),w2m k( )1

)

.

σ ₋ −σ ≤σ ₋

We get lim_k_→∞σ

(

w2m k( ),w2m k( )−1

)

= ∈. (35)

Putting s=x2m k( ),t=x2n k( )−1 in (27), for all k=1, 2, 3, we obtain

( )

( ) ( )

(

)

( )

( ) ( )

(

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

(

)

(

( ) ( )

)

{

}

{

(

( ) ( )

)

(

( ) ( )

)

}

( )

( ) ( )

(

)

( )

1 2 2 1

1 2 2 1 2 2 1 2 2 1 2 2 2 1 2 1

2 1 2 2 2 1 2 2 1 2 2

2 2 2 1 2

, 0 , , , , , , , , , 1 1 , , , , , 2 2 0 , , , 0 d d d m k n k

m k m k m k n k m k n k m k m k n k n k

n k m k m k n k m k n k m k m k

m k m k m k

px qx

Ux Vx Ux qx px Vx Ux px Vx qx

qx Ux px Vx Ux Vx px Ux

Ux Vx Ux qx

y y

φ

ψ σ σ σ σ σ

σ σ σ σ

ψ σ σ

η η η − − − − − − − − − −  + + _ ≤ −

∫

( )

(

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

(

)

(

( ) ( )

)

{

}

{

(

( ) ( )

)

(

( ) ( )

)

}

( )

( ) ( )

(

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

2 1 2 2 1 2 2 2 1 2 1

2 1 2 2 2 1 2 2 1 2 2

1 2 1 2 2 2 1 2 1 2 2 2 2 1 2 2 2 2 1

2 , , , , , , 1 1 , , , , , 2 2 , , , , , , , , , , 1 2 0 d

n k m k n k m k m k n k n k

n k m k m k n k m k n k m k m k

m k n k m k n k m k n k m k m k m k n k n

px Vx Ux px Vx qx

qx Ux px Vx Ux Vx px Ux

w w w w w w w w w w

w

y y

σ σ σ

σ σ σ σ

ψ σ σ σ σ σ

σ η − − − − − − − − − − − − − − −  + + _ ≤

∫

( ) ( )

(

)

(

( ) ( )

)

{

}

{

(

( ) ( )

)

(

( ) ( )

)

}

( )

( ) ( )

(

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

)

(

( ) ( )

(

)

(

( ) ( )

)

{

}

{

(

( ) ( )

)

(

( ) ( )

)

}

1 2 1 2 2 2 2 1 2 2 2 2 1

2 2 1 2 2 2 1 2 1 2 2 2 2 1 2 2 2 2 1

2 1 2 1 2 2 2 2 1 2 2 2 2 1

1 , , , , , 2 , , , , , , , , , , 1 1 , , , , , 2 2 0 d

k m k m k n k m k n k m k n k

m k n k m k n k m k n k m k m k m k n k

n k m k m k n k m k n k m k n k

w w w w w w w

w w w w w w w w w w

w w w w w w w w

y y

σ σ σ

ψ σ σ σ σ σ

σ σ σ σ

(8)

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

( )

( ) [ ]

_{( )}

( )

[ ]

( )

( ) ( )

_{( )}

1 1 2

1 1

1 1 1 1

, , ,0,0, , , , ,0,0, ,

2 2 2 2

0 0 0

, , , , , ,

0 0

d d d

d d .

y y y y y y

y y y y

φ ψ ψ

ψ φ

η η η

η η     ∈ _∈ ∈ ∈ ∈+∈ ∈_ _∈ ∈ ∈ ∈+∈ ∈_ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ≤ − < =

∫

This is contradiction. Hence

{ }

w2n is a Cauchy sequence and is convergent.

Since E is complete there exist z∈E such that as n→ ∞ we have wn →z.

Case I: Assume that U is continuous then Upe2n →Uz,

2

2n .

U e →Uz Since

(p, U) is sub compatible, we have pUe2n→Uz.

Step I: Substituting s=Ue2n,t=e2n+1 in (27), we have

( )

(

) (

)

( )

(

)

( )

(

)

( )

{

}

{

(

) (

)

}

( )

(

) (

)

1 2 2 1

2 2 2

1 2 2 1 2 2 1 2 2 1 2 2 2 1 2 1

2 2 2

2 1 2 2 2 1 2 2 1 2 2

2 2

2 2 2 1 2 2 1

, 0 , , , , , , , , , , 1 1 , , , , , 2 2 0 , , , , 0 d d d n n

n n n n n n n n n n

n n n n n n n n

n n n n

pUe qe

U e Ve U e qe pUe Ve U e pUe Ve qe

qe U e pUe Ve U e Ve pUe U e

U e Ve U e qe pU

y y

φ σ

ψ σ σ σ σ σ

σ σ σ σ

ψ σ σ σ

η η η + + + + + + + + + + +  + + _  ≤ −

∫

( )

(

)

( )

(

)

( )

{

}

{

(

) (

)

}

2

2 2 1 2 2 2 1 2 1

2 2 2

2 1 2 2 2 1 2 2 1 2 2

, , , , , ,

1 1

, , , , ,

2 2

,

n n n n n n

n n n n n n n n

e Ve U e pUe Ve qe

qe U e pUe Ve U e Ve pUe U e

σ σ

σ σ σ σ

+ + + + + + + + _

∫

( )

( ) ( )

( )

( ) ( ) ( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

( )

( ) ( ) ( ) ( ) ( ) { ( ) ( )} { ( ) ( )}

( )

( ) ( ) ( ) ( ) 1 1 2 1 , 0 1 1 , , , , , , , , , , , , , , , 2 2 0 1 1 , , , , , , , , , , , , , , , 2 2 0 1 , , , , , ,0,0, , ,

2 0 d d d d Uz z

Uz z Uz z Uz z Uz Uz z z z Uz Uz z Uz z Uz Uz

Uz z Uz z Uz z Uz z Uz

y y

φ σ

ψ σ σ σ σ σ σ σ σ σ

ψ σ σ σ σ σ

η η η η  ₊ ₊       ₊ ₊      ≤ − ≤

∫

( ) { }

( )

( ) ( ) 1 , ,

0 d .

z Uz z y y φ σ η       ≤

∫

It is contradiction if Uz≠z. Hence Uz=z.

Step II: Substituting s=z t, =e2n+1 in (27) and taking limit as n tends to

in-finity we get pz=z.

Step III: We know that z= pz∈p e

( )

⊆V e

( )

then there exist u∈E such

that z=Vu. Substituting s=e2n,t=u in (27) we get qz=z. Hence qz= =z Vz and qVu=Vqu, which gives qz=Vz.

Step IV: Substituting s=z t, =z in (27) we have qz=z so that

( )

q z = =z Vz. Hence p, q, U & V have a common fixed point z in E. Case II: Assume that U is continuous then 2

2n ,

p e →pz pUe2n→ pz.

Si-milarly 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

( )

, 1 2

s t s t

σ = − . Define

, , , :

p q U V E→E such that 4 s ps= ,

4 t

qt= , Us=s, Vt=t.

Let

ψ

1

(

y y y y y y y1, 2, 3, 4, 5, 6, 7

)

=max

(

y y y y y y y1, 2, 3, 4, 5, 6, 7

)

,

ϕ

( )

y =2y, 2 1

1 4

(9)

( )

2

1

max , , , , , , , , , ,

4 4 4 4 4 4 4

1 1

, , , , ,

2 4 4 2 4

s t t s s t

s t s t s t

t s s

s t s t s

σ σ σ σ σ

σ σ σ σ

        

− ≤ _ _ _ _ _ _ _ _ _

        

   ₊    ₊    _ _ _ _  _ _

   

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

Let

{ }

en be a sequence in E such that pen→z & Uen→z for some z in E. Then z = 0,

σ

(

pUe Upen, 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 ano-nymous referees for their valuable comments and useful suggestions in improv-ing the article.

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