Some New Contractive Mapping Theorem in Partially Ordered Metric Spaces

Some New Contractive Mapping Theorem in Partially Ordered Metric Spaces

Rita Shukla

Department of Mathematics RSR-RCET, Kohka, Bhilai, Chhattisgarh, India

Abstract : In this paper ,we establish some coincidence , common fixed point theorems for monotone f-non-decreasing self-mappings satisfying certain rational type contraction in the context of a metric spaces with partial order . these results generalize and extend well known existing results in the literature .

Keyword: Compatible mappings Partially ordered metric spaces ,Weakly compatible mappings .

Introduction : In fixed point theory ,the classical Banach contraction principle plays a valid role to obtain an unique solution of the result ,lot of variety of generalizations of this Banach contraction principle[1] have been taken place in a metric fixed point theory by improving the underlying contraction condition .some contractive conditions in a partially ordered set which guarantee the existence of fixed points have been recently established in [5] and [6]

Preliminaries:

The following definitions are frequently used in results given in upcoming sections .

1. The triple (X.d,) is called a partially ordered metric space ,if (X , ) is a partially ordered set together with (X , d ) is a metric space .

2. The triple (X , d , ) is called a partially ordered complete metric space if (X,d ) is a complete metric space .

3. Let (X , ) be a partially ordered set . A self mapping : said to be strictly increasing if f(x) () for all x, with x < and is also said to be strictly decreasing if f(x) > f(y) for all x, with x < .

4. Let (X , d) is a metric space and A . Then a point x is called a common

   fixed point (coincidence) point of two self mappings f and T if fx = Tx=x (fx=Tx)

5. The two self mapping f and T defined over a subset A of a metric space (X, d) are called commuting iff fTx = Tfx for x .

6. Two self mappings f and T defined over A are compatible ,if for any sequence

   { }

7. Two self mapping f and T defined over A are said to be weakly compatible ,if they commute at their coincidence point

   i.e. if fx=Tx then fTx =Tfx

8. Let f and T be two self mappings defined over a partially ordered set (X,). A mapping T is called a monotone f non- deceasing if fx fy implies Tx

   , x,y

9. Let A be a non-empty subset of a partially ordered set ( X, ). If very two elements of A are comparable then it is called well ordered set.

   =0

10. A partially ordered metric space (X,d,) is called an ordered complete ,if for each convergent sequence {} ,one of the following condition holds.

    1. If {} is a non-decreasing sequence in X such that implies

       ,for all n N that is x=sup{}

    2. If {} is a non increasing sequence in X such that implies

,for all n N that is x=inf{}.

MAIN RESULTS :

We prove coincidence ,common fixed point theorem in the context of ordered metric space

Theorem (1) Let (X,d,) be a complete partially ordered metric space .suppose that the self mappings f and T on X are continuous T is a monotone f-nondecreasing T(x) =f(X) and satisfying the following condition

d(T+1, ) (, +1) +

(, 1)

(1-) d(T+1, ) (, 1)

d(Tx,Ty) .(,).(,) + . (, )

()

d(T+1

,

) (

1

)(

, 1)

– (1)

for all x, y in X with f(x) () are comparable ,where , [0,1) with 0

+ < 1.

continuing the same process up to (n-1) times,

we get

d(T , ) ( ) d(T , T )

If there exists a point 0

such that f(0)

+1

1 1 0

(0) and the mapping T and f are compatible then T and f have a coincidence point in X .

Let k = (

1

) [0,1] then from triangular

Proof : Let 0 such that f(0) (0) since from hypotheses , we have () () then we can choose a point 1 such that f1=T0 but T1 () then again there exists another point 2 such that f2= T1 by continuing the same way, we can

construct a sequence {} in X such that f+1= T for all n.

Again ,by hypotheses we have f(0)

(0) = f(1) and T is a monotone f- nondecreasing mapping then ,we get T(0)

T(1) . Similarly ,we obtain T(1)T(2) since f(1)f(2) and then by continuing the same procedure ,we obtain that

T(0) T(1)T(2) T()

T(+1)

The equality T(+1) = T() is impossible

inequality for m .we have

d(T , ) d(T , 1) + d(T1 , 2) + d(T1 , 2)+——-

-+

d(T+1, ) (1 + 2+—————

—

–+) d(T1, 0)

1 0

( ) d(T , T )

1

as m, n , d(T , ) 0 which shows that the sequence {T} is a Cauchy sequence in X . so by the completeness of X , there exist appoint the continuity of T ,we have

lim ()= T( lim )=T

because f(+2) f(+1) for all n N. Thus

d(T() , T(+1))> 0 for all n 0.therefore from contraction condition (1) ,we have

But f+1 = T , then f+1 as n

and from the compatibility for T and f

d(T

,T

) (+1,,+1).(,). +

lim ((), ()) = 0

+1

(

+1

,)

[(+1

, )]

d(T, )=d(T,Tf ) + d(T(f),f(T))+d(f(T),f)

+1

d(T , ) (,+1).(1 ,,)

(,1)

on taking limit as

in both sides of

+ (, 1)

above equation and using the fact that T and f are continuous then we get d(T, ) =0 thus

T = .Hence is coincidence point of T and f in X.

Theorem 2 : Let (X, d ,) be a complete partially ordered metric space .suppose that f and T are self-mapping on X ,T is a monotone f-non-decreasing , T(X) () and satisfying

d(Tx, Ty) .(,).(,) +

(,)

. (, ) .for all x, y in X with f(x)f(y) are compatible and for some , [0,1] with 0 + < 1.If there exist a point 0 such that f(0) (0) and {} is a non- decreasing sequence in X such that

then for all n .If f(X) is a complete

Sup f() f(v) and sup T() () for all n .

f() f(u) f(1) f() f(v)

Case I st Suppose if there exists some 0

1 such that f(0 ) = f(0 ) then, we have f(0 ) =f(u) =f(0 )= f(1) = T(u) .

Hence u is a coincidence point of T and f in X

.

Case IInd suppose that f(0 ) f(0 ) for all n then we have

d(f+1,f+1) = d(T ,T)

subset of X Then T and f have a coincidence

. (,).(,) + (

, )

point in X .Further, if T and f are weakly

(,)

compatible ,then T and f have a common fixed point in X . Moreover, the set of common fixed points of T and f is well ordered if and only if T and f have one and only one common fixed point in X .

Proof : Suppose f(X) is a complete subset of X the sequence {T} i a Cauchy sequence and hence {f} is also a Cauchy sequence in (f(X),d) as f+1= T and T(X)

().since f(X) is complete then there exists

some fu () such that lim () =

lim () = f(u) .The sequences {} and

{f} are non-decreasing and from hypotheses, we have T() () and f()

() for all n . But T is a monotone f- nondecreasing that ,we get T() () for all n. Letting n we obtain that f(u) T(u).

Suppose that f(u) () then define a sequence {} by 0 = u and f+1=T for all n .

taking limit as on both sides of the above inequality ,we get

d(fu, fv) . ((),()).((),()) +

((),())

d(f(u),f(v))

0 + (, )

< (, ) since < 1

Thus we have f(u)=f(v)=f(1)= T(u)

Hence ,we conclude that u is a coincidence point of T and f in X .Now suppose that T and f are weakly compatible .Let w be a coincidence point then

T(w) =T(f(z)) = f(T(z)) = f(w)

Since w =T(z) = f(z) for some z

Now by contraction condition ,we have d(T(z),T(w)) . (,).(,)

(,)

An argument similar to that in the proof of theorem (1) yields that {f} is a non-

decreasing sequence and lim () =

+. (, )

((), ())

lim ()=f(v) for some v so from

hypotheses it is clear that

As < 1 , d(T(z),T(w)) =0

Therefore , T(z) =T(w) =f(w)=w .Hence w is a common fixed point of T and f in X . Now

suppose that the set of common fixed points of T and f is well ordered ,we have to show that common fixed point of T and f is unique .let u and v be two common fixed points of T and f such that u then

d(u,v) . (,).(,) + (, )

(,)

(, )

< (, ) since < 1

Which is a contradiction , Thus u = v .

Conversely , suppose T and f have only one common fixed point then the set of common fixed points of T and f being a singleton I well ordered .This completes the proof .

Theorem 3 : Let (X, d, ) be a complete partially ordered metric space . suppose that f and T are self- mappigs on X ,T is a monotone f-non-decreasing T(X) () and satisfying

d(Tx,Ty) d(fx,fy) +[(, ) +

(, )] + [(, ) + (, )]

for all x, y in X with f(x) () are compatible ,where , , [0,1] with if there exists point 0 + 2 + 2 < 1 ,0

such that f(0)T(0) and the mappings T and f are compatible ,then T and f have a coincidence point in X .

Proof : Let 0 such that f(0)T(0) since from hypothesis ,we have T(X) () then ,we can choose a point 1 such that f1 =T 0 but T1 () then again there

exists another point 2 such that f2=T1.

Similarly ,we obtain T(1) (2) , since f(1) (2) and then by continuing the same process we obtain that

T(0) (1) ) (2) —- ()

(+1)

The equality T(+1) = T() is impossible because f(+2)f(+1) for all n . Thus

d(T, T+1) > 0 for all n 0

therefore, from contraction condition ,we have

d(T+1,T) (+1 , f)

+[(+1 , T +1) + d(f

,T)]+[(+1,T)+ ( , T+1)]

d(T+1,T) ( , T1)

+[(, T +1) + d(T1 ,T)]

+[(,T)+ (1 , T+1)]

d(T+1,T) ( , T1)

+[(, T +1) + d(T1 ,T)]+[

(1 , T+1)]

d(T+1,T) ( , T1)

+[(, T +1) + d(T1 ,T)] +[

(1 , T) + d(T, T+1)]

d(T+1,T) ( , T1) +(, T +1) + d(T1 ,T)+(1 , T)

+ d(T, T+1)

(1- ) d(T+1,T)= ( + + )

d(T,T1)

By continuing ,the same way , we can

d(T

,T

) ( + +) . d(T

,T )

construct a sequence {} in X such that

+1

(1 )

1

f+1=T for all n.

continuing the same process up to (n-1) times

Again by hypothese ,we have f(0) T(0)= f1 and T is a monotone f-nondecreasing

we get d(T+1 d(T1,T0)

,T

) (( + +))

(1 )

mapping ,then we get T(0) (1).

Let k =

( + +) (1 )

[0,1) then from

triangular inequality for m ,we have

d(T,T) d( T, 1)+ d(T1,

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   (1+ 2 + + ) d(T1

   ,T0)

   1 0

   ( ) d(T ,T )

   1

   as m, n d(T ,T) 0 which shows that the sequence {} is a Cauchy sequence in X so

   by the completeness of X there exists a point such that as .

   Again by the

   continuity of T , we have lim () =

   T( lim ) =T

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    But f+1

    = T

    then f+1

    as n

    Appl. 2009 .Art.ID 493965, 11 P.P.

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    and from the compatibility for T and f we have

    lim ((), ()) = 0

    Further by triangular inequality , we have d(T, f ) = d(T, T(f )) + d[T(f)

    ,f(T)] +d[f(T) ,f]

    on taking limit as n in both sides of the above equation and using the fact that T and f are continuous then ,we get d(T, )= 0 thus T = f. Hence is a coincidence point of T and f in X.

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