Volume 02, Issue 10 (October 2013)

A Related Fixed Point Theorem of Integral Type on Two Fuzzy 2-Metric Spaces

DOI : 10.17577/IJERTV2IS101122

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Pheiroijam Suranjoy Singh, 2013, A Related Fixed Point Theorem of Integral Type on Two Fuzzy 2-Metric Spaces, INTERNATIONAL JOURNAL OF ENGINEERING RESEARCH & TECHNOLOGY (IJERT) Volume 02, Issue 10 (October 2013),

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A Related Fixed Point Theorem of Integral Type on Two Fuzzy 2-Metric Spaces

Pheiroijam Suranjoy Singh Sagolband Takyel Kolom Leikai

Imphal West, Manipur – 795001, India.

Abstract

In this paper, a related fixed point theorem is obtained. It extends a result proved by R.K. Namdeo, N.K. Tiwari, B. Fisher and K. Tas [9]. The notion of fuzzy 2-metric spaces satisfying integral type inequalities is used.

Keywords : Fuzzy 2-metric space, fixed point, related fixed point, integral type inequality.

2000 AMS Subject Classification : 47H10, 54H25.

Intoduction

The concept of fuzzy sets was introduced by L. Zadeh [14] in 1965. Fuzzy metric space was introduced by Kramosil and Michalek [7] in 1975. Then, it was modified by George and Veeramani [4] in 1994. Fuzzy has been studied and developed by many mathematicians for many years. Introduction of fuzzy 2-metric space is one of such developments. Gahler [10, 11] investigated 2-metric spaces in a series of his papers. Fuzzy 2-metric space is studied in [6, 8, 12, 13] and many others. Related fixed point is studied in [1, 2, 3, 5, 9] and many more.

Some definitions are stated as follows:

Definition 1.1 : A binary operation : [0, 1] × [0, 1] [0, 1] is called a t – norm in ( [0, 1], ) if follo- wing conditions are satisfied:

For all a, b, c, d [0, 1],

  1. a 1 = a,

  2. a b = b a,

  3. a b c d whenever a c and b d,

  4. a (b c) = (a b) c.

Definition 1.2. The 3-tuple (X, , ) is called a fuzzy 2- metric space if X is an arbitrary set, is a continuous t – norm and is a fuzzy set in X 3 [0, ) satisfying the following conditions:

For all x, y, z, u X and t1, t2, t3 > 0, i. (x, y, z, 0) = 0,

  1. (x, y, z, t) = 1, t > 0 and when at least two of the three points are equal,

  2. (x, y, z, t) = (y, x, z, t) = (z, x, y, t) ( symmetry about three variables), iv. (x, y, z, t1+t2+t3) (x, y, u, t1) (x, u, z, t2) (u, y, z, t3)

  1. (x, y, z,) : [0, ) [0, 1] is left continuous,

  2. limt (x, y, z, t) = 1.

Definition 1.3 : Let (X, , ) be a fuzzy 2-metric space. A sequence {xn} X is said to:

  1. converge to x in X if and only if

    limt (xn, x, a, t) = 1 a X and t > 0.

  2. be a Cauchy sequence if and only if limt (xn+p, xn, a, t) = 1 a X, p > 0 and t > 0.

Definition 1.4 : A fuzzy 2-metric space (X, , )

sequence in X is convergent in X.

is aid to be complete if and only if every Cauchy

The following was proved in [9].

Theorem 1.1 : Let (X , d) and (Y, ) be complete metric spaces. Let T be a mapping of X into Y and S

be a mapping of Y into X satisfying the inequalities

d ( Sy, Sy /) d( STx, STx / ) c max{ d (Sy, Sy/ ) (Tx, Tx / ) , d (x / , Sy) ( y / , Tx) ,

d (x, x / ) d (Sy, Sy/ ) , d (Sy, STx) d (Sy/ , STx/ ) }

(Tx, Tx / ) (TSy, TSy/ )

c max{ d (Sy, Sy/ ) (Tx, Tx / ) , d (x / , Sy) ( y / , Tx) ,

( y, y / ) (Tx, Tx / ) , (Tx,TSy) (Tx / , TSy/ ) }

for all x , x / in X and y , y / in Y ,where 0 c < 1. If either S or T is continuous, then ST has a unique fixed point z in X and TS has a unique fixed point w in Y . Further, Tz = w and Sw = z.

Now, theorem 1.1 is extended to two pairs of mappings in integral and fuzzy 2-metric space settings as follows.

Main result

Theorem 2.1 : Let (X, , a, t) and (Y, , a, t) be two complete fuzzy 2-metric spaces. Let A, B be mappings of X into Y and S, T be mappings of Y into X satisfying the inequalities

k(Sy, Ty/ , a, t ) (SAx, TBx/ , a, t)

1

(s) ds 1

min{(Sy, Ty/ , a, t)( Ax, Bx/ , a, t), (x/ , Sy, a, t)( y/ , Ax, a, t) ,

(x, x/ , a, t) (Sy, Ty/ , a, t), (Sy, SAx, a, t) (Ty/ , TBx/, a, t)}

(s) ds

(1)

k( Ax ,Bx/ , a,t)(BSy, ATy/ , a,t)

1

(s) ds

min{(Sy, Ty / , a,t)( Ax, Bx /, a,t), (x /, Sy, a,t)( y /, Ax, a,t),

( y, y /, a,t)( Ax, Bx /, a,t),( Ax, BSy, a,t)(Bx /, ATy /, a,t)}

1

(s) ds

(2)

for all x, x/ in X and y, y/ in Y, where k(0, 1). If A and S or B and T are continuous, then SA and TB have a unique common fixed point z in X and BS and AT have unique common fixed point w in Y. Further, Az = Bz = w and Sw = Tw = z.

Proof : Let x be any arbitrary point in X . We define sequences {xn} and {yn}in X and Y respectively as:

S y2n 1 =

x2n 1 ,

Bx2n 1 y2n ,

Ty2n x2n ,

Ax2n y2n 1 , for n = 1, 2, 3 ,

Applying inequality (1), we get

k(Sy , Ty

, a, t) (SAx , TBx , a, t)

1 2n 1 2n

2n 2n 1

(s) ds

= 1

k 2 (x

2n 1

, x , a, t) 2n

(s) ds

min{(Sy

, Ty

, a,t)( Ax

, Bx

, a,t), (x

, Sy

, a,t)( y

, Ax

, a,t),

2n 1 2n

2n 2n 1

2n 1

2n 1

2n 2n

(x , x

, a,t) (Sy

, Ty

, a,t),(Sy

, SAx

, a,t)(Ty

, TBx

, a,t)}

1

2n 2n1

2n 1 2n

2n 1 2n

2n 2n 1

(s) ds

min{(x

2n 1

, x , a, t)( y

2n

, y

2n 1

, a, t), (x

2n 2n 1

, x

2n 1

, a, t)( y

2n

, y

2n 1

, a, t),

(x

2n

1

, x

2n 1

, a, t) (x

2n 1

, x , a, t), (x

2n 2n 1

, x

2n 1

, a, t) (x

2n

, x , a, t)}

2n

(s) ds

from which it follows that

min{( y

, y , a, t), (x

, x , a, t)}

k(x , x

2n 1 2n

1

, a, t)

(s) ds

1

2n 1 2n

2n 1 2n

(s) ds

(3)

Applying inequality (2), we get

k( Ax , Bx

, a, t)(BSy

, ATy

, a, t)

2n 2n 1

1

2n 1 2n

(s) ds

= 1

k 2 ( y , y

2n 1 2n

, a, t)

(s) ds

min{(Sy

2n 1

, Ty

2n

, a, t)( Ax

2n

, Bx

2n 1

, a, t), (x

2n 1

, Sy

2n 1

, a, t)( y

2n

, Ax

2n

, a, t),

( y , y

2n 1 2n

1

, a, t)( Ax

2n

, Bx

2n 1

, a, t) , ( Ax

2n

, BSy

2n 1

, a, t)(Bx

2n 1

, ATy

2n

, a, t)}

(s) ds

max{(x

, x , a, t) ( y

, y , a, t), (x , x

, a, t)( y , y

, a, t),

2n 1 2n

2n 1 2n

2n 1

2n 1

2n 2n 1

( y

, y , a, t)( y

, y , a, t) ,( y

, y , a, t)( y , y

, a, t)}

1

2n 1 2n

2n 1 2n

2n 1 2n

2n 2n 1

(s) ds

from which it follows that

k( y , y , a, t)

min{( y , y , a, t), (x

, x , a, t)}

2n 1 2n

1

(s) ds

1

2n 1 2n

2n 1 2n

(s) ds

(4)

(3) and (4) can be written as

k(x , x , a, t)

min{( y , y , a, t), (x

, x , a, t)}

n 1 n

1

(s) ds

1

n 1 n

n 1 n

(s) ds

min{( y , y , a, t), (x

, x , a, t)}

k( y , y , a, t)

n 1 n

n 1 n

n 1 n

1

1

(s) ds

which can be again written as

k(x , x , a, t)

min{( y , y , a, t), (x

, x , a, t)}

n 1 n

1

(s) ds 1

n 1 n

n 1 n

(s) ds

(5)

k( y , y , a, t)

min{( y , y , a, t), (x

, x , a, t)}

n 1 n

1

(s) ds 1

n1 n

n 1 n

(s) ds

(6)

From (5) and (6) , by induction, we get

1 min{( y , y , a, t), (x , x , a, t)}

(x , x , a, t)

kn 1 2 1 2

n 1 n

1

( y , y

, a, t)

(s) ds 1

1 min{( y , y , a, t), (x , x , a, t)}

(s) ds

Let

t1

1

t . Now,

p

n 1 n

(s) ds kn 1 2

1

1 2 (s) ds

(xn , xn p , a, t) (xn , xn p , a, t t … p times)

1 (s)ds

1 (s)ds

1 (s)ds

1 (s)ds

1 (s)ds

= 1

1 1 (s)ds

1 (s)ds

1 (s)ds

1

1

1

1

(xn , xn 1, a, t )

(xn1, xn 2, a, t )

(xn p1, xn p , a, t )

1

1

1 min{( y , y , a, t), (x , x , a, t)}

1 min{( y , y , a, t), (x , x , a, t)}

kn 1 2

1

1 2 (s) ds kn p 1 1 2

1

1 2 (s) ds

which implies that

n n p

n n p

(x , x , a, t)

lim (s)ds 1

1

(xn , xn p , a, t) 1

{ xn } is a Cauchy sequence with a limit z in X.

Similarly, { yn } is a Cauchy sequence with a limit w in Y.

Now, on using the continuity of A and S respectively, we get

w = lim y 2n1 = lim Ax 2 n

= Az and z = lim x 2 n

= lim Sy 2 n

= Sw

so that we get

Az = w (7)

Sw = z (8)

From (7) and (8), we get

SAz = z (9)

Again applying inequality (1), we get

k(SAx

2n

1

, TBx

2n 1

, a, t)

(s) ds

min{( Ax , Bx , a, t), ( y

, Ax

, a, t), (x , x , a, t)}

1

2n 2n 1

2n 2n

2n 1 2n

(s) ds

(10)

On letting n , we have

k(Sw, TBz, a, t)

1

( Az , w, a, t)

(s) ds 1

(s) ds

By (7), we have

k(Sw, TBz, a, t)

1

(s) ds 0

k(Sw, TBz, a, t) 1

which implies that

Sw = TBz

and from (8) , we get

z = TBz (11)

From (9) and (11), we get

SAz = z = TBz (12)

Now, (10) gives

k(x

2n 1

1

, Ty

2n

, a, t)

(s) ds

min{( Ax

, Bx

, a, t), ( y

, Ax

, a, t), (x

, x , a, t)}

1

2n 2n 1

2n 2n

2n 1 2n

(s) ds

On letting n , we get

k(z , Tw, a, t)

1

(t) dt 0

k(z , Tw, a, t) 1

which implies that

z = Tw (13)

Again, applying inequality (2), we get

k(BSy

2n 1

1

, ATy

2n

, a, t)

(s) ds

min{(Sy

, Ty

, a, t), (x

, Sy

, a, t), ( y

, y , a, t), ( Ax

, Bx

, a, t)}

1

2n 1 2n

2n 1

2n 1

2n 1 2n

2n 2n 1

(s) ds

(14)

On letting n , we get

k(BSw, ATw, a, t)

1

(s) ds 0

k(BSw, ATw, a, t) 1

which implies that

BSw = ATw (15)

Now, (14) gives

k( y

2n

1

, ATy

2n

, a, t)

(s) ds

min{(Sy

2n 1

1

, Ty

2n

, a, t) , (x

2n 1

, Sy

2n 1

, a, t) , ( y , y

2n 1 2n

, a, t) , ( Ax

2n

, Bx

2n 1

, a, t)}

(s) ds

On letting n , we get

k(w, ATw, a, t)

1

(s) ds 0

k(w, ATw, a, t) 1

which implies that

w = ATw (16)

From (15) and (16), we get

BSw = w = ATw (17)

From (8) and (17) , we get

Bz = w (18)

From (7) and (18), we get

Az = Bz = w (19)

From (8) and (13) , we get

Sw = Tw = z (20)

Similarly, on using the continuity of B and T, the above results hold.

To prove the uniqueness, let SA and TB have a second distinct common fixed point z / in X and BS

and AT have a second distinct common fixed point w / in Y.

Applying inequality (1), we have

k 2 (z , z/ , ,a, t)

1

(s) ds

min{(z , z/ , a, t)( Az , Bz/ , a, t), (z/ , z/ , a, t)(Bz/ , Az, a, t),

(z, z/ , a, t) (z , z/ , a, t), (z/ , z/ , a, t) (z, z, a, t)}

1

(s) ds

k(z , z/ , a, t)

1

(s) ds

min{( Az , Bz/ , a, t), (Bz/ , Az, a, t)}

1 (s) ds

k(z , z/ , a, t)

1

(s) ds

( Az , Bz/ , a, t)

1 (s) ds

(21)

Applying inequality (2), we get

k2 ( Az ,Bz/ , a, t)

1

(s) ds

min{(z, z/ , a, t)( Az, Bz/ , a, t), (z / , z / , a, t)(Bz / , Az, a, t),

( Az, Bz / , a, t)( Az, Bz/ , a, t), ( Az, Bz / , a, t)(Bz / , Az, a, t)}

1

(s) ds

k( Az ,Bz/ , a, t)

1

/

(z , z , a, t)

(z , z , a, t)

(s) ds

1

(s) ds

(22)

From (21) and (22), we get

k2(z, z/ , a, t)

1

(s) ds

(z, z/ , a, t)

1

(s) ds

(z, z/ , a,t)

1

(z, z/ , a,t)

1

(s) ds

(s) ds

1 (z, z/ , a, t)

k 2

1

1 ( z, z/ , a, t)

k 2

1

(s) ds

(s) ds 1

1 (z, z/ , a, t)

k n

(s) ds

(z, z/ , a,t)

1

(s) ds

lim 1

1 (z, z/ , a, t)

k n

(s) ds 1

(z, z/ , a,t) 1

which implies that

z = z / .

This proves the uniqueness of z. Similarly, the uniqueness of w can be proved.

The following corollary is a fuzzy 2-metric space version of theorem 1.1 in integral setting.

Corollary 2.2 : Let (X, , a, t) and (Y, , a, t) be two complete fuzzy 2-metric spaces. Let S be mappings of X into Y and T be mappings of Y into X satisfying the inequalities

k(Ty, Ty/ , a, t ) (TSx, TSx/ , a, t)

1

(s) ds 1

min{(Ty, Ty/ , a, t)(Sx, Sx/ , a, t), (x/ , Ty, a, t)( y/ , Sx, a, t) ,

(x, x/ , a, t) (Ty, Ty/ , a, t), (Ty, TSx, a, t) (Ty / , TSx/ , a, t)}

(s) ds

k(Sx,Sx/ , a, t)(STy, STy/ , a, t)

1

(s) ds 1

min{(Ty,Ty/ , a, t)(Sx, Sx/ , a, t), (x/ , Ty, a, t)( y/ , Sx, a, t),

( y, y/ , a, t)(Sx, Sx/ , a, t), (Sx, STy, a, t)(Sx/ , STy/ , a, t)}

(s) ds

for all x, x/ in X and y, y/ in Y ,where k (0, 1). If either S or T is continuous, then TS has a unique fixed point z in X and ST has a unique fixed point w in Y . Further, Sz = w and Tw = z.

Proof : By putting A = B = S and S = T = T in theorem 2.1, the result easily follows.

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