Fixed Point Theorem in Fuzzy Metric Space Using (CLRg) Property

International Journal of Engineering Science Invention
ISSN (Online): 2319 – 6734, ISSN (Print): 2319 – 6726
www.ijesi.org ||Volume 5 Issue 4|| April 2016 || PP.35-39
www.ijesi.org 35 | Page
Fixed Point Theorem in Fuzzy Metric Space Using (CLRg)
Property
Smt. Preeti Malviya1
, Dr.Vandna Gupta2
&Dr. V.H. Badshah3
1
Govt. New Science College , Dewas (M.P.) ,455001, India
2
Govt. Kalidas Girl's College, Ujjain (M.P.) 456010, India
3
School of Studies in Mathematics,Vikram University, Ujjain(M.P), 456010,India
ABSTRACT: The object of this paper is to establish a common fixed point theorem for semi-compatible pair
of self maps by using CLRg Property in fuzzy metric space.
2010 Mathematics Subject Classification : 54H25, 47H10.
keywords: Common fixed point, fuzzy metric space , Semi compatible maps , Weakly compatible maps, CLRg
Property.
l. INTRODUCTION
Zadeh'
s [1 ] introduced the fuzzy set theory in 1965. Zadeh's [ 1 ] introduction of the notion of fuzzy set laid
the foundation of fuzzy mathematics . Sessa [2 ] has introduced the concept of weakly commuting and
Jungck [ 3] initiated the concept of compatibility. In , 1988 , Jungck and Rhoades [4 ] introduced the notion
of weakly compatible . The concept of fuzzy metric space introduced by kramosil and Mishlek [ 5 ] and
modified by George and Veramani [ 6 ]. In 2009, M. Abbas et. al. [7] introduced the notion of common
property E.A. B.Singh et. al. [8] introduced the notion of semi compatible maps in fuzzy metric space .
Recently in 2011 , Sintunavarat and Kuman [ 9] introduced the concept of common limit in the range
property . Chouhan et.al. [10 ] utilize the notion of common limit range property to prove fixed point
theorems for weakly compatible mapping in fuzzy metric space .
II. PRELIMINARIES
Definition 2.1 [ 11 ] Let X be any set . A Fuzzy set A in X is a function with domain X and Values in [ 0,1].
Definition 2.2[ 6] A Binary operation * : [0,1] ×[0,1] →[0,1] is called a continuous t-norms
if an topological monoid with unit 1 such that a*b≤ c*d whenever a≤c and b≤d , for all
a,b,c,d in [0,1].
Examples of t – norms are a*b = ab and a*b =min {a,b}.
Definition 2.3[ 6 ] The triplet ( X,M, *) is said to be a Fuzzy metric space if , X is an arbitrary set , * is a
continuous t- norm and M is a fuzzy set on X2
×(0,∞) satisfying the following conditions; for all x,y,z in X
and s,t > 0,
(i) M(x,y,0) = 0 , M(x,y,t)>0,
(ii) M(x,y,t) = 1 ,for all t > 0 if and only if x=y,
(iii) M(x,y,t) = M(y,x,t),
(iv) M(x,y,t) * M( y,z,s) ≤ M( x,z, t+s),
(v) M(x,y,t) : [ 0,∞ ) →[0,1] is left continuous.
Example 2.1 [ 6] Let (X,d) be a metric space . Define a*b = min {a,b} and
M(x,y,t) = t / t + d(x,y) for all x,y ∈ X and all t > 0 . Then (X,M, *) is a fuzzy
metric space . It is called the fuzzy metric space induced by the metric d.
Definition 2.4 [ 6 ] A sequence {xn} in a fuzzy metric space (X,M,*) is called a Cauchy
Sequence if , lim n→∞ M ( Xn+p,Xn,t) = 1 for every t.>0 and for each p>0.
A fuzzy metric space(X, M,*) is Complete if ,every Cauchy sequence in X converge
to X.
Definition 2.5[6 ] A sequence {Xn } in a fuzzy metric space ( X,M,*) is said to be Convergent to x in X if
, limn→∞M( Xn,X, t) = 1 , for each t>0.
Definition 2.6 [12] Two self mappings P and Q of a fuzzy metric space (X,M,*) are said to be
Compatible , if limn→∞M(PQxn,QPxn,t) =1 whenever {xn} is a sequence such that
limn→∞Pxn = limn→∞ Qxn = z , for some z in X .
Definition 2.7 [ 13 ] Self maps A and S of a Fuzzy metric space (X,M,*) are said to be Weakly Compatible if
they commute at their coincidence points,
if, AP=SP for some pєX then ASp=SAp.

Fixed Point Theorem in Fuzzy Metric Space…
Lemma 2.1 [ 8 ] Let {yn} is a sequence in an FM- space . If there exists a positive number k<1
such that M(yn+2, yn+1 , kt) ≥ M( yn+1, yn,t) , t>0 , n 𝜖 N,
then {yn} is a Cauchy sequence in X .
Lemma 2.2 [ 8 ] If for two points x, y in X and a positive number k < 1
M(x,y,kt) ≥ M(x,y,t) , then x = y.
Lemma 2.3 [ 14] For all x,y ∈ X , M(x,y,.) is a non – decreasing function.
Definition 2.8 [ 8 ] A pair (A,S) of self maps of a fuzzy metric space (X,M,*) is said to be
Semi compatible if limn→∞ ASxn = Sx , whenever {xn} is a sequence such that
limn→∞ Axn = limn→∞ Sxn = x , for some x ∈ X .
It follows that (A,S) is semi compatible and Ay = Sy then ASy = SAy
Example 2.2 Let X = [ 0,1] and (X,M, t) be the induced fuzzy metric space with
M (x,y,t) = t / t +│x-y│. Define self maps P and Q on X as follows :
2 , if 𝜃 ≤ x ≤ 1 2 , if x=1
Px= x/2, if 1< x≤ 2 and Qx = x+ 3/5 . if 1 < x≤ 2
And xn = 2 – 1/ 2n
. Then we have P (1) = Q(1)= 2 and S(2) = A(2) = 1.
PQ(1) =QP(1) =1 and PQ(2) = QP(2) = 2 . Hence Pxn→1 and Qxn→1 and QPxn→1, as n→∞.
Now ,
limn→∞ M (PQxn, Qy, t ) = M(2,2,t) = 1
limn→∞ M(PQxn, QPxn,t) = M(2,1,t) = t / 1+t < 1.
Hence (P,Q) is semi compatible but not compatible.
Definition 2.9 [ 9 ] A pair of self mapping P and Q of a fuzzy metric space (X,M,*) is said
to satisfy the (CLRg) property if there exists a sequence {xn} in X such that
limn→∞ Pxn = limn→∞ Qxn = Qu , for some u ∈ X .
Definition 2.10 [ 9 ] Two pairs (A,S) and (B,T) of self mappings of a fuzzy metric
Space (X,M, *) are said to satisfy the (CLRST) property if there exist two sequence
{xn} and {yn} in X such that
limn→∞ Axn = limn→∞ Sxn = limn→∞ Byn = limn→∞ Tyn = Sz,
for some z 𝜖 S(X) and z 𝜖 T(X).
Definition 2.11 [ 9 ] Two pairs (A,S) and (B,T) of self mappings of a fuzzy metric
Space (X,M, *) are said to share CLRg of S property if there exist two sequence
{xn} and {yn} in X such that
limn→∞ Axn = limn→∞ Sxn = limn→∞ Byn = limn→∞ Tyn = Sz,
for some z 𝜖 X.
Proposition 2.1 [ 4 ] In a fuzzy metric space (X,M,*) limit of a sequence is unique.
Example 2.3 Let X = [ 0,∞) be the usual metric space . Define g , h : X→X by
gx = x+ 3 and gx = 4x , for all x 𝜖 X . We consider the sequence {xn} = { 1 + 1/n }.
Since , limn→∞ gxn = limn→∞ hxn = 4 = h(1) 𝜖 X .
Therefore g and h satisfy the (CLRg) property.
Lemma 2.4 Let A, B , S and T be four self mapping of a fuzzy metric space (X,M,*)
Satisfying following
1. The pair (A,S) ( or (B,T) ) satisfies the common limit in the range of S property
(or T property )
2. There exists a constant k 𝜖 (0,1) such that
( M( Ax,By, Kt )) 2
≥ min (( M( Sx,Ty,t ))2
, M(Sx,Ax,t ), M(Sx,By,2t),M(Ty,Ax,t)
M(Sx,By,2t), M(Ty,By,t) ), For all x,y ∈ X and t > 0
3. A(X) ⊆T(X) ( or B(X) ⊆S(X) ) .
Then the pairs (A,S) and (B,T) share the common limit in the range property.
Singh and Jain [ 8] proved the following results.
Theorem 2.1 Let A,B,S and T be self maps on a complete fuzzy metric space (X,M,*)
Satisfying
1. A(X) ⊂ T(X) , B(X) ⊂ T(X)
2. One of A and B is continuous.
3. (A,S) is semi compatible and (B,T) is weak compatible.
4. For all x,y ∈ X and t > 0
M(Ax,Bx,t) ≥ r ( M ( Sx,Ty,t) ),
Where r : [0,1]→[0,1] is a continuous function such that r(t) > t , for each 0< t<1 . Then
A,B,S and T have a unique common fixed point.

III. MAIN RESULT
In the following theorem we replace the continuity condition by using (CLRg) property.
Theorem 3.1 Let A, B, S and T be self mapping on a complete fuzzy metric space (X,M,*),
where * is a continuous t – norm definied by ab = min [a,b] satisfying
( i) A(X) ⊆ T (X) , B(X) ⊆ S(X).
(ii) (B,T) is semi compatible ,
(iii) Then for all x,y ∈ X and t > 0 .
M ( Ax, By, kt ) ≥ 𝜙 [ min ( M ( Sx , Ty ,t) , { M(Sx,Ax,t) . M(By,Ty,t) } ,
½ ( M ( Ax, Ty,t ) + M(By,Ax,t) ]
Where 𝜙 : [0,1] → [0,1] is a continuous function such that 𝜙 (1) = 1 , 𝜙 (0) = 0
and 𝜙 (b) = b , for 0 < b<1.
If the pair (A,S ) and ( B, T) share the common limit in the range of S property ,
then A, B, S and T have a unique common fixed point
Proof –Let x0 be any arbitrary point for which there exist two sequences {xn} and {yn}
in X such that
y2n+1 = Ax2n = Tx2n+1 and y2n+2 = Bx2n+1 = Sx2n+2, for n= 0,1,2,…
Now, M(y2n+1,y 2n+2,kt) = M(Ax2n,Bx 2n+1,kt)
≥ 𝜙 [ min ( M(( Sx2n,Tx 2n+1 , t ) ,{ M( Sx2n A,x 2n ,t . M( Bx2n+1,Tx 2n+! , t)},
½ (M ( Ax2n, T2n+1,t ) + M( Bx2n+1,Ax 2n,t ) )]
≥ 𝜙 [min (M(y2n,y 2n+1, t ), { M(y2n, y2n+1 ,t) . M( y2n+2, y2n+1 ,t) },
½ ( M( y2n+1 ,y 2n+1,t) + M( y2n+2,y 2n+1,t )) ]
M(y2n+1,y 2n+2,kt) > M( y2n, y2n+1,t )
Similarly, we can proved M(y2n+2, y2n+3,t) > M(y2n+1,y 2n+2,t)
In general , M(yn+1,y n ,t) > M (yn,y n+1,t)
Thus, from this we conclude that { M ( yn+1,y n,t ) is an increasing sequence of positive
real numbers in [ 0,1] and tends to limit l≤1 .
If l < 1, then M(yn+1,yn ,t ) ≥ 𝜙 ( M ( yn, yn+1, t ),
Letting n→∞ , we get limn→∞ M(yn+1,yn ,t ) ≥ 𝜙 [ lim n→∞ M ( yn, yn+1, t ) ]
l ≥ 𝜙(l) =l ( Since 𝜙(b) > b )
a contradiction . Now for any positive integer q
M(yn,y n+q,t) ≥ M( yn, yn+1, yn+q ,t /2(q-1)+1 ) * M( yn+1,y n+2 ,yn+q ,t/2(q-1)+1)*…*
M(yn+q+1, yn+q , t/ 2(q-1)+1)
Taking limit, we get
limn→∞ M(yn,y n+q,t) ≥ limn→∞ M( yn, yn+1, yn+q ,t /2(q-1)+1 ) * llimn→∞M( yn+1,y n+2 ,yn+q ,t/2(q-
1)+1)*…* lim n→∞ M( yn+q+!, yn+q, t / 2 (q-1)+1 )
Limn→∞ M (yn,y n+q,t ) ≥ 1*1*1*….*1=1
Which means {yn} is a Cauchy sequence in X . Since X is complete , then yn→z in X.
That is { Ax2n} , {Tx2n+1}, {Bx2n+1} and {Sx2n} also converges to z in X.
Since, the pair (A,S) and (B,T) share the common limit in the range of S property , then there exist
two sequences {xn} and {yn} in X such that
limn→∞ Axn = limn→∞ Sxn = limn→∞ Byn = limn→∞ Tyn = Sz, , for some z 𝜖 X.
First we prove that Az= Sz
By (3.3) , putting x= z and y= yn , we get
M(Az,Byn,kt) ≥ ∅ [ min ( M ( Sz, Tyn,t ), { M( Sz,Az,t) .M(Byn, Tyn,t) },
½ ( M ( Az, Tyn, t ) + M( Byn, Az,t)) ]
Taking limit n→∞ , we get
M(Az,Sz,kt) ≥ ∅ [ min ( M( Sz, Sz, t) , { M (Sz,Az,t) . M(Sz,Sz,t) },
½( M (Az,Sz,t ) + M(Sz,Az,t) ) ]
≥ ∅ [ min ( 1, { M(Sz, Az,t) . 1 }, M(Sz,Az,t) ]
M(Az,Sz,kt) ≥ M ( Sz, Az,t )
Hence by Lemma 2.2 , we get Az= Sz …(1)
Since , A(X) ⊆ T(X) , therefore there exist u ∈ X , such that Az=Tu …(2)
Again , by inequality (iii), putting x=z and y= u , we get
M(Az,Su,kt) ≥ ∅ [ min ( M( Sz, Tu, t) , { M (Sz,Az,t) . M(Bu,Tu,t) },
½( M (Az,Tu,t ) + M(Bu,Az,t) ) ]
Using (1) and (2) , we get
M(Tu, Bu,kt) ≥ ∅ [ min ( M( Az, Tu, t ) , { M (Az,Az,t) . M(Bu, Tu, t) },
½ ( M ( Az,Az,t) + M( Bu, Tu,t) ) ]

≥ ∅ [ min M(Tu,Tu,t) , { 1. M(Bu, Tu,t) } , M (Bu,Tu,t)]
M(Tu,Bu,kt) ≥ M (Bu, Tu,t)
Hence , by Lemma 2.2 , we get Tu=Bu …(3)
Thus , from (1),(2) and (3) , we get Az= Sz= Tu = Bu …(4)
Now , we will prove that Az= z
By inequality(iii), putting x= z and y = x2n+1 , we get
M(Az,Bx2n+! ,kt) ≥ ∅ [ min ( M(Sz,Tx2n+1,t) , { M(Sz,Az,t) . M(Bx2n+1,Tx 2n+1,t) }
½( M(Az, Tx2n+1,t) + M(Bx2n+1,Az,t) )]
Taking limit n →∞ , using (1) , we get
M(Az,z,t) ≥ ∅ [ min ( M(Sz,z,t), { M(Az,Az,t) . M(z,z,t) }, ½ ( M(Az,z,t)+ M(z,Az,t))]
M(Az,z,t) ≥ ∅ [ min (M(Az,z,t), { 1,1} , M(Az,z,t) ]
M(Az,z,t) ≥ M(Az,z,t)
Hence , by Lemma 2.2 , we get Az=z
Thus , from (4), we get z= Tu=Bu
Now , Semi compatibility of (B,T) gives BTy2n+1 →Tz , i.e. Bz=Tz.
Now, putting x=z and y= z in inequality (iii), we get
M(Az, Bz,t ) ≥ ∅ [ min (M( Sz,Tz,t) , { M (Sz,Az,t) .M(Bz,Tz,t) } ,
1/2 (M( Az,Tz,t )+ M(Bz,Az,t)) ]
M(Az, Bz,t ) ≥ ∅ [ min (M( Az,Bz,t) , { M (Az,Az,t) .M(Tz,Tz,t) } ,
1/2 (M( Az,Bz,t )+ M(Bz,Az,t)) ]
M(Az,Bz,t) ≥ M(Az,Bz,t)
Hence , by Lemma 2.2 , we get Az=Bz.
And, hence Az= Bz = z.
Combining ,all result we get z= Az=Bz=Sz=Tz.
From, this we conclude that z is a common fixed point of A,B,S and T.
Uniquness
Let z1 be another common fixed point of A,B,S and T. Then
z1 = Az1 = Bz1 =Sz1 = Tz1 , and z =Az=Bz=Sz=Tz
Then, by inequality ( iii) , putting x=z and y =z1,we get
M(z,z1,kt) = M(Az,Bz1,t) ≥ ∅ [ min ( M(Sz, Tz1, t), {M(Sz,Az,t) . M(Bz1,Tz1,t)},
½ (M( Az, Tz1,t ) + M(Bz1,Az,t))]
≥ ∅ [ min ( M (z,z1,t) , { M(z,z,t) . M(z1,z1,t) },
½ (M(z,z1,t) + M(z1,z,t)) ]
≥ ∅ [ min (M(z,z1,t) , 1, M(z,z1,t)
M(z,z1,t) ≥ M(z,z1,t)
Hence, from Lemma 2.2 , we get z=z1
Thus z is the unique common fixed point of A, B, S and T.
Corollary 3.2 Let (X,M,*) be complete fuzzy metric space . suppose that the mapping
A, B,S and T are self maps of X satisfying( i-ii) conditions and there exist k∈ (0,1) such
that
M(Ax,By,kt) ≥ M(Sx,Ty,t), M(Ax,Sx,t), M(By,Ty,t), M(By,Sx,2t), M(Ax,Ty,t)
For every x,y ∈ X , t> 0 . Then A,B,S and T have a unique common fixed point in X.
Corollary3.3 Let (X,M,*) be complete fuzzy metric space . suppose that the mapping
A, B,S and T are self maps of X satisfying (i-ii) conditions and there exist k∈ (0,1) such
that
M(Ax,By,kt) ≥ M(Sx,Ty,t), M(Sx,Ax,t), M(Ax,Ty,t)
For every x,y ∈ X , t> 0 . Then A,B,S and T have a unique common fixed point in X.
IV. ACKNOWLEDGEMENTS
The authors are grateful to the referees for careful reading and corrections.
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