International Journal of Scientific & Engineering Research Volume 2, Issue 10, Oct-2011 1
ISSN 2229-5518
Common Fixed Point Theorems for Multivalued
Compatible Maps in IFMS
Anil Rajput , Namrata Tripathi, Sheel Kant Gour, Seema Chouhan, Rajamani S.
Abstract-The aim of this paper is to obtain the notion of multivalued weakly compatible (mwc) maps and prove common fixed point theorems for single and multi valued maps by using a contractive condition of integral type in intuitionistic fuzzy metric spaces.
Index Terms- Fixed Points , intuitionistic fuzzy metric space, multivalued weakly compatible maps, compatible maps.
—————————— ——————————
A fundamental result in fixed point theory is intuitionistic fuzzy metric spaces which is stated in theorem Through out
t)*H(By, Ty, t)*H(By, Sx,(2-α)t)],
[H(Ax, Ty, αt)*H(By, Ty, t)*H(Sx, By, t)]}
the paper X will represent the intuitionistic fuzzy metric space (X, M, N, *, ) and CB( X) , the set of all non-empty closed and bounded sub-sets of X . For A, B CB( X )and
for every t>0,
denote H(A, B, t)=sup{M(a, b, t);a A, b B} and H(A, B,
N (Sx,Ty, kt)
0
where
(t )dt
n( x, y,t )
(t )dt
0
(1.2)
t)=inf{N(a, b, t);a A, b B}
and δM(A, B, t)=Inf{ M(a, b, t);a A, b B},
δN(A, B, t)=sup{N(a, b, t);a A, b B}
If A consists of a single point a, we write
δM(A, B, t)= δM(a, B, t) and δN(A, B, t)= δN(a, B, t). If B
also consists of a single point b, we write
δM(A, B, t)= M(A, B, t) and δN(A, B, t)= N(A, B, t)
It follows immediately from definition that
δM(A, B, t)= δM(B, A, t)≥0 and
δN(A, B, t)= δN(B, A, t)≥0 δM(A, B, t)=1 A=B={a}
δN(A, B, t)=0 A=B={a} for all A,B CB(X)¸
be multivalued weakly compatible (mwc) if there exists some point x X such that
Ax Bx and ABx BAx.
Clearly weakly compatible maps are multivalued weakly
compatible (mwc).
Now, we prove our main result.
Theorem 1. Let (X,M, N, *, ) be a complete intuitionistic
fuzzy metric space with continuous
t-norm * and continuous t-corm defined by t*t=t and (1 - t) (1 - t) ≤ (1 - t) for all t [0, 1] such that, A :X →X and
B: X→ CB (X) be single and multi valued mappings
reMsp(eSctxiv,Teyly,kstu)ch that thme( x,my,at )ps (A,S) and (B,T) are (mwc)
n(x, y, t)=max{[H(Ax, Sx, t)+H(By, Ty,t)], [N(Ax, By, t)◊H(By, Ty, t)◊H(By, Sx,(2-α)t)],
[H(Ax, Ty, αt)◊H(By, Ty, t)◊H(Sx, By, t)]}
is a function which is sum able, Lebseque integrable, non-
(t )dt 0
negative and such that 0
for each ε > 0 .for every x, y X and t > 0,
α (0,2). Then A, B, S and T have unique common fixed point in X.
elements , u v in X such that Au Su ¸ ASu SAu and Bv
Tv , BTv TBv .First we prove that Au =Bv. As Au Su so AAu ASu SAu , Bv Tv ¸ so BBv BTv TBv
and hence
M(A2u,B2v,t)≥δM (SAu,TBv, t), N(A2u,B2v,t)≤δN
(SAu,TBv, t) and if Au≠ Bv then
δM (SAu,TBv, t)< 1 , δN (SAu,TBv, t)<1 .
and satisfy the inequality for all x, y
X where φ
:[0,1]→[0,1]
0
(t )dt
(t )dt
0
(1.1)
Using (1.2) for x = Au, y= Bv
where
m(x, y, t)=min{[H(Ax, Sx, t)+H(By, Ty,t)], [M(Ax, By,
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ISSN 2229-5518
m(Au, Bv, t)=min{[H(AAu, SAu, t)+H(B Bv, TBv,t)], [M(AAu, B Bv, t)*H(B Bv, T Bv, t)*H(B Bv, SAu,(2- α)t)],[H(AAu, TBv, αt)*H(BBv, TBv, t)*H(SAu, BBv, t)]}
≥min{[M(AAu, SAu,t)+M(B Bv, TBv,t)],
[M(AAu,BBv,t)*M(B Bv, T Bv, t)*
N (SAu,TBv, kt )
0
, a contradiction.
Hence Au= Bv .
(t )dt
n( Au , Bv,t )
(t )dt
0
N (SAu,TBv,t )
(t )dt
0
M(BBv,SAu,(2-α)t)],[M(AAu, TBv, αt)*M(BBv, T Bv,
t)*M(SAu, BBv, t)]} (1.3)
n(Au, Bv, t)=max{[H(AAu, SAu, t)+
H(BBv, TBv,t)],[N(AAu, BBv, t)◊H(BBv,TBv, t)◊H(B Bv, SAu,(2-α)t)], [H(AAu, TBv, αt)◊H(BBv, T Bv, t)◊H(SAu, BBv, t)]}
≤ max{[N(AAu, SAu, t)+N(B Bv, T Bv,t)], [N(AAu, B Bv, t)◊N(B Bv, TBv, t)
◊N(B Bv, SAu,(2-α)t)], [N(AAu, T Bv, αt)◊N(B Bv, T Bv,
t)◊N(SAu, BBv, t)]} (1.4)
Since, * and ◊ is continuous , letting α→1 in (1.3) and (1.4),
we get
m(Au, Bv, t) ≥ min{[M(A2u, SAu, t)+
M(B 2v, TBv,t)], [M(A2u, B 2v, t)* M(B 2v, T Bv, t)*M(B 2v, SAu, t)], [M(A2u, T Bv, t)*M(B 2v, T Bv, t)
*M(SAu, B2v, t)]}
≥ min{[1+1], [δM(SAu, TB v, t)*1*
δM(TB v, SAu, t)],[ δM(SAu, TBv, t)*1
* δM(SAu, TBv, t)]} = δM(SAu, T Bv, t) (1.5)
n(Au, Bv, t) ≤max{[N(A2u, SAu, t)+
N(B 2v, TBv,t)], [N(A2u, B 2v, t) ◊N(B 2v, T Bv, t) ◊N(B 2v,
Also M(A2u, Bu, t)≥δ M (SAu, Tu, t), N(A2u, Bu, t)≤δ N (SAu, Tu, t), M(A2u, Tu, t)≥δ M (SAu, Tu, t), N(A2u, Tu, t)≤δ N (SAu, Tu, t),
Now, we claim that Au= u . It not, then
δ M (SAu, Tu, t)<1, δ N (SAu, Tu, t)<1
Considering (1.1) and (1.2) for Au =x, u= y , α=1 m(Au, u, t)=min{[H(AAu, SAu, t)+H(Bu, Tu,t)], [M(AAu, Bu, t)*H(Bu, Tu, t)*H(Bu, S Au, t)], [H(AAu, Tu, t)*H(Bu, Tu, t)*H(S Au, Bu, t)]}
≥min{[M(A 2u, SAu, t)+M(Bu, Tu, t)],
[M(A 2u, Bu, t)*M(Bu, Tu, t)*M(Bu, S Au, t)] [M(A 2u, Tu, t)*M(Bu, Tu, t)*M(S Au, Bu, t)]}
≥min{[1+1], [δM(SA u, Tu, t)*1*δ M(Tu, S Au, t)], [δ M(SA u, Tu, t)*1*δ M(S Au, Tu, t)]}
m(Au, u, t)≥ δM(SA u, Tu, t) (1.7) n(Au, u, t)=max{[H(A Au, SAu, t)+H(Bu, Tu,t)], [N(A Au, Bu, t)◊H(Bu, Tu, t)◊H(Bu, S Au, t)],
[H(A Au, Tu, t)◊H(Bu, Tu, t)◊H(S Au, Bu, t)]}
≤max{[N(A 2u, S Au, t)+N(Bu, Tu,t)],
From (1.1) and (1.7), (1.2) and (1.8) we have
SAu, t)], [N(A2u, T Bv, t) ◊
N(B 2v, T Bv, t) ◊N(SAu, B2v, t)]}
≤max {[0+0], [δN(SAu, TB v, t) ◊0 ◊δN(TB v, SAu, t)], [ δN(SAu, T Bv, t) ◊0 ◊ δN(SAu, T Bv, t)]} = δN(SAu, T
M (SAu,Tu, kt)
0
N (SAu,Tu, kt )
(t )dt
m( Au,u,t )
(t )dt
0
n( Au,u,t )
M (SAu,Tu,t)
(t )dt
0
N (SAu,Tu,t)
Bv, t) (1.6)
(t )dt
0
(t )dt
0
(t )dt
0
From (1.1) and (1.5) , (1.2) and (1.6) we have
which is again a contradiction and hence A=u .
M (SAu,TBv, kt)
0
(t )dt
m( Au, Bv,t )
(t )dt
0
M (SAu,TBv,t)
(t )dt
0
Similarly, we can get Bv= v.Thus A, B, S and T have a common fixed point in X.For uniqueness let u ≠u ‘‚ be another fixed point of A, B, S and T, then (1.1) and (1.2) gives
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International Journal of Scientific & Engineering Research Volume 2, Issue 10, Oct-2011 3
ISSN 2229-5518
m(u, u', t)=min{[H(Au, Su, t)+H(Bu', Tu',t)], [M(Au, Bu', t)*H(Bu', Tu', t)*H(Bu', Su,(2-α)t)], [H(Au, Tu', αt)*H(Bu', Tu', t)*H(Su, Bu', t)]}
n(u, u', t)=max{[ H(Au, Su, t)+H(Bu', Tu',t)],
[M(Au, Bu', t)*H(Bu', Tu', t)*H(Bu', Su,(2-α)t)], [H(Au, Tu', αt)*H(Bu', Tu', t)*H(Su, Bu', t)]} Letting α→1
m(u, u', t)=min{[δ M(Au, Su, t)+δ M (Bu', Tu',t)], [M(Au,
Bu', t)* δ M (Bu', Tu', t)* δ M (Bu', Su, t)],
[ δ M (Au, Tu', t)* δ M (Bu', Tu', t)* δ M (Su, Bu', t)]}
m(u, u', t)=min{[1+1], [M(Su, Tu', t)* 1* δ M (Tu', Su, t)], [ δ
M (Su, Tu', t)* 1* δ M (Su, Tu', t)]}
m(u, u', t)= δ M (Su, Tu', t) (1.9)
n(u, u', t)=max{[δ N(Au, Su, t)+δ N (Bu', Tu', t)],
[N(Au, Bu', t)◊ δ N (Bu', Tu', t)◊δ N (Bu', Su, t)],
[ δ N (Au, Tu', t)◊ δ N (Bu', Tu', t)◊ δ N(Su, Bu', t)]}
n(u, u', t)=max{[0+0], [N(Su, Tu', t)◊ 0◊ δ N (Tu', Su, t)],[ δ
N(Su, Tu', t)◊0◊ δ N (Su, Tu', t)]}
n(u, u', t)= δ N (Su, Tu', t) (1.10)
Again from (1.1) and (1.9), (1.2)and (1.10) we obtain
353.Received: October,
5. 2010.
6.
M (Su,Tu ' , kt)
0
(t)dt
m(u,u',t )
(t )dt
0
M (Su,Tu ' ,t)
(t)dt
0
N (Su,Tu ' , kt )
0
(t)dt
n(u ,u ',t )
(t)dt
0
N (Su,Tu ' ,t)
(t)dt
0
Which yields Su =Tu . i.e., u= u'.
Thus, A, B, S and T have unique common fixed point.
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