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 Authors : A. Sangeetha
 Paper ID : IJERTV2IS3118
 Volume & Issue : Volume 02, Issue 03 (March 2013)
 Published (First Online): 20032013
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Fixed Point Theorems For Continuous Mappings Based On Commutativity In A Complete 2 Metric Space
A. Sangeetha
Assistant Professor, Department of Mathematics, EBET Group of Institutions Nathakadaiyur, Tirupur D.T.
ABSTRACT
The notion of 2metric space was first introduced by S.Gahler in 1963. Gahler gave the definition of 2metric space. After that Ehret (1969), Iseki (1975), Diminie and White (1976), Singh (1979), Rhoades (1979), SharmaAshok K(1980), M.S.Khan and Fisher (1982) S V R Naidu (1986), Ganguly and Chandel (1987) and number of other Mathematicians have worked in this field. The aim of this paper is to study about the fixed point theorems for continuous self – maps satisfying the commuting property in a complete 2metric space.
INTRODUCTION
In mathematics, a metric space is a set where a notion of distance (called a metric) between elements of the set is defined. The metric space which most closely corresponds to our intuitive understanding of space is the 3dimensional Euclidean space. In fact, the notion of "metric" is a generalization of the Euclidean metric arising from the four long known properties of the Euclidean distance. The Euclidean metric defines the distance between two points as the length of the straight line segment connecting them. The geometric properties of the space depend on the metric chosen, and by using a different metric we can construct interesting nonEuclidean geometries such as those used in the theory of general relativity. The 2metric space was shown to have a unique nonlinear structure, quite different from a metric space. As in other spaces, the fixed point theory of operators has been developed in this space also. A 2metric is a real function of triple points which abstracts the properties of the area function for Euclidean triangles.
In this paper, we have some useful definitions and two fixed point theorems in complete 2metric space using the concept of continuous selfmapping and commutative mapping. Throughout this paper,

(X,d) is a complete 2metric space ,

is a nonnegative real function mapping from R+ to R+ such that is a non decreasing function, (t) < t for any t > 0 and lim n = 0 .
n

N is the set of all positive integers .
The paper begins with the necessary definitions that are used to prove the theorems.
Definition 1. 2metric space :
Let X be a set consisting of aleast three points. A 2metric on X is a mapping d from XxXxX to the set of nonnegative real numbers that satisfies the following conditions:

There exists three points x, y, z such that d(x, y, z) 0,

d(x, y, z) = 0 if atleast two of the three points are equal ,

d(x, y, z) = d(y, z, x) = d(x, z, y) for all x, y, z in X,

d(x, y, z) d(x, y, a) + d(x, a, z) + d(a, y, z) for all x, y, z, a in X The pair (X,d) is called a 2metric space .

Definition 2
A sequence {xn} in a 2metric space (X,d) is said to be convergent with lim x in X if, lim d(xn, x, a) = 0 for each aX.
n
Definition 3
A sequence {xn} in a 2metric space (X,d) is said to be Cauchy sequence if, lim d(xn, xm, a) = 0 for all aX , m, nN .
m, n
Definition 4
A 2metric space (X,d) is said to be complete if every Cauchy sequence is convergent in X.
Definition 5
A mapping T of X into itself is said to be continuous at a point xX, if whenever a sequence {xn} of X converges to xX, then the sequence {Txn} converges to Tx.
Definition 6
The function is called upper semicontinuous if,
lim sup (x) (a) for all aR+
xa
and is called lower semicontinuous if,
lim inf (x) (a) for all aR+ .
xa
Definition 7
Two mappings S, T : X X are said to be commuting if ,
(ST) (x) = (TS) (x) for each xX.
The first fixed point theorem is for for two continuous self maps S, T commuting with an another self map B .
Theorem 1
Let S, T be two continuous selfmappings on a complete 2metric space (X,d) into itself satisfying the following conditions:
(1.1) is lower semi continuous , (1.2) lim n (t) = 0 ,
n
(1.3) S and T are surjective ,
(1.4) S and T commutes with an another the self map B and (1.5) d(Bx, By, a) [max {d(Sx, Ty, a), d(Bx, Sx, a), d(By, Ty, a),
(d(Bx, Ty, a) d(By, Sx, a))}] for all x, y, a X.
Then S, T and B have a common fixed point.
Proof:
Since the mappings S and T commutes with the map B we have, (SB) (x) = (BS) (x)
(TB) (x) = (BT) (x) for each x X.
Now, we find an arbitrary point xo in X such that , Bxo = Sx1 and
Bx1 = Tx2 for x1, x2X.
Since S and T are surjective, there exists yo, y1X such that,
yo = Sx1 and y1 = Tx2
Therefore,
yo = Sx1 = Bxo and y1 = Tx2 = Bx1
In general,
yn = Sxn+1 = Bxn n N U {0} and
yn+1 = Txn+2 = Bxn+1 n N ( I )
This theorem can be proved by proving,
(T.1.a) {yn} and {yn+1} converges to any point say z in X as n , (T.1.b) Sz = z ,
(T.1.c) Tz = z and (T.1.d) Bz = z .
Proof of (T.1.a) :
Taking x = x1, y = x2 in condition (1.5),
d(Bx1, Bx2, a) [max {d(Sx1, Tx2, a), d(Bx1, Sx1, a), d(Bx2, Tx2, a),
(d(Bx1, Tx2, a) d(Bx2, Sx1, a))}]
Therefore, d(y1, y2, a) [max {d(yo, y1, a), d(y1, yo, a), d(y2, y1, a),
(d(y1, y1, a) d(y2, yo, a))}] [ by ( I ) ]
We know that d(y1, y1,a) = 0 [by condition (ii) of Definition 1].
That is, d(y1, y2, a) [max {d(yo, y1, a), d(y1, yo, a), d(y2, y1, a), 0}]
= [max {d(y1, yo, a), d(y2, y1, a)}] Suppose that, d(y2, y1, a) > d(y1, yo, a)
Then, d(y1, y2, a) [d(y1, y2, a)]. This is a contradiction.
Therefore, d(y2, y1, a) [d(yo, y1, a)] ( II )
Again taking x = x3 , y = x2 in condition (1.5),
d(Bx3, Bx2, a) [max {d(Sx3, Tx2, a), d(Bx3, Sx3, a),d(Bx2, Tx2, a),
(d (Bx3, Tx2, a) d (Bx2, Sx3, a))}]
This implies, d(y3, y2, a) [max {d(y2, y1, a), d(y3, y2, a), d(y2, y1, a),
(d(y3, y1, a) d(y2, y2, a))}] [ by ( I ) ]
= [max {d(y2, y1, a), d(y3, y2, a), d(y2, y1, a), 0}]
= [max {d(y2, y1, a), d(y3, y2, a)}]
Therefore, d(y3, y2, a) [d(y2, y1, a)] ( III )
Similarly for x = x3, y = x4 the condition (1.5) becomes,
d(Bx3, Bx4, a) [max {d(Sx3, Tx4, a), d(Bx3, Sx3, a),d(Bx4, Tx4, a),
(d (Bx3, Tx4, a) d (Bx4, Sx3, a))}]
This implies, d(y3, y4, a) [max {d(y2, y3, a), d(y3, y2, a), d(y4, y3, a),
(d(y3, y3, a) d(y4, y2, a))}] [ by ( I ) ]
= [max {d(y3, y2, a), d(y4, y3, a), 0}]
Therefore d (y4, y3, a) [d(y2, y3, a)] (IV)
By ( IV), d(y4, y3, a) [d(y2, y3, a)]
[ d(y1, y2, a)] [using ( III )]
= 2 d(y1, y2, a)
2 [ d(y1, y2, a)] [using ( II )]
= 3 d(y1, yo, a)
Continuing the process,
d(yn, yn+1, a) n d(y1, yo, a)
Taking limit as n ,
lim d(yn, yn+1, a) = 0 for m>n. [ since lim n = 0 ]
n n
Therefore, {yn} is a Cauchy sequence in X.
Since (X,d) is complete {yn} converges to any point say zX as n . Consequently, the sequence {yn+1} also converges to zX as n .
That is,
lim yn = lim yn+1 = z.
n n
Proof of (T.1.b) :
Taking limit as n in (I) and using (T.1.a),
lim yn = z = lim Sxn+1 = lim Bxn
n n n
lim yn+1 = z = lim Txn+2 = lim Bxn+1
n n n
Now settingx = SXn, y = xn+1 in condition 1.5,
(V)
d(BSxn, Bxn+1, a) [max{d(S2xn, Txn+1, a), d(BSxn, S2xn, a), d(Bxn+1, Txn+1, a),
(d(BSxn, Txn+1, a) d(Bxn+1, S2xn, a))}]
Since S is continuous and commutes with B, then {SSxn} and {BSxn} converges to Sz as n .
Now taking limit as n to the inequality and also using (V) ,
d(Sz, z, a) [max{d(Sz, z, a), d(Sz, Sz, a), d(z, z, a), (d(Sz, z, a) d(z, Sz, a))}]
= [max{d(Sz, z, a), 0, 0, d2(Sz, z, a), }] [by Definition 1.]
= [ d2 (Sz, z, a) ]
d(Sz, z, a) [ d2 (Sz, z, a) ]. This is a contradiction. This implies, d(Sz, z, a) = 0 aX. Hence Sz = z.
Proof of (T.1.c) :
Taking x = xn, y = Txn+1 in condition (1.5),
d(Bxn, BTxn+1, a) [max{d(Sxn, TTxn+1, a), d(Bxn, Sxn, a),
d(BTxn+1, TTxn+1, a),
(d(Bxn, TTxn+1, a) d(BTxn+1, Sxn, a))}]
Since T is continuous and commutes with B, then {TTxn +1} and {BTxn +1} converges to Tz as n .
Now taking limit as n and also using (V) ,
d(z, Tz, a) [max{d(z, Tz, a), d(z, z, a), d(Tz, Tz, a), (d(z, Tz, a) d(Tz, z, a))}]
= [max { d(Tz, z, a), 0, 0, d2(Tz, z, a)}] [by Definition 1.]
= [d2 (Tz, z, a)]
d(z, Tz, a) [d2 (Tz, z, a)] which is a contradiction.
Therefore, d(Tz, z, a) = 0 aX. Hence Tz = z.
Proof of (T.1.d) :
Consider,
d(Bz, Bxn+1, a) [max {d(Sz, Txn+1, a), d(Bz, Sxn+1, a), d(Bxn+1, Txn+1, a),
(d(Bz, Txn+1, a) d(Bxn+1, Sz, a))}] [by condition (1.5)]
Taking limit as n ,
d(Bz, z, a) [max{d(Sz, z, a), d(Bz, z, a), d(z, z, a),
(d(Bz, z, a) d(z, Sz, a) )}] [by (V)]
= [max{d(z, z, a), d(Bz, z, a), d(z, z, a), (d(Bz, z, a) d(z, z, a))}]
[by (T.1.b)]
= [max {0, d(Bz, z, a), 0, 0}] [by Definition 1.]
= [d(Bz, z, a)]
That is, d(Bz, z, a) [d(Bz, z, a)] . This is a contradiction. Therefore, d(Bz, z, a) = 0 aX. Hence Bz = z .
Therefore, z is the common fixed point of S, T and B.
The second fixed point theorem is for two continuous selfmaps S and T commuting with other two selfmaps A and B respectively.
Theorem 2
Let S and T be two continuous selfmappings defined on a complete 2metric space (X,d) into itself which satisfies the following conditions :
(2.1) 
is upper semicontinuous, 

(2.2) 
S and T commutes with other two selfmaps A and B respectively 
and 
(2.3) 
d(Ax, By, a) [max{d(Sx, Ty, a), d(Sx, Ax, a), d(Sx, By, a), 
d(Ty, By, a), d(Ty, Ax, a), (d(Sx, Bx, a) d(Ty, Ay,a))}] for all x, y, a X then S, T, A and B have a common fixed point which are unique.
Proof:
Since the map S commutes with A and T commutes with B we have,
(SA) (x) = (AS) (x)
(TB) (x) = (BT) (x) for each xX.
Now, for an arbitrary point in x0 in X there exists a point x1X such that,
Sx1 = Axo and for x2X , Tx2 = Bx1.
We define two sequences {y2n} and {y2n+1} in X such that,
y2n+1 = Sx2n+1 = Ax2n nN {0} and
y2n = Tx2n = Bx2n1 nN ( I )
This theorem can be proved by proving,
(T.2.a) (X,d) is bounded ,
(T.2.b) The sequences {y2n} and {y2n+1}converges to any point say zX as n , (T.2.c) Sz = z ,
(T.2.d) Tz = z ,
(T.2.e) Az = z ,
(T.2.f) Bz = z and
(T.2.g) S, T, A and B have a unique fixed point z .
Proof of (T.2.a) :
Let 2r is a sequence of real number where 0 < < 1, r N.
First, let 2r be an increasing sequence. Suppose if {y2n} is not bounded then,
d(y2n+1, y2n+2, a) 2r ( II )
Now consider,
d(y2n+1, y2n+2, y2n) = d(Ax2n, Bx2n+1, y2n) [using (I)]
[max{d(Sx2n, Tx2n+1, y2n), d(Sx2n, Ax2n, y2n), d(Sx2n, Bx2n+1, y2n), d(Tx2n+1, Bx2n+1, y2n),d(Tx2n+1,Ax2n,y2n),
(d(Sx2n, Bx2n, y2n) d(Tx2n+1, Ax2n+1, y2n))}] [by condition (2.3)]
= [max{d(y2n, y2n+1, y2n), d(y2n, y2n+1, y2n), d(y2n, y2n+2, y2n), d(y2n+1, y2n+2, y2n), d(y2n+1, y2n+1, y2n),
(d(y2n, y2n+1, y2n) d(y2n+1, y2n+2, y2n))}] [by ( I )]
= [max {0, 0, 0, d(y2n+1, y2n+2, y2n), 0, 0}] [by Definition 1.]
= [d(y2n, y2n+1, y2n+2)]
< d(y2n, y2n+1, y2n+2) [since (t)<t]
This implies, d(y2n, y2n+1, y2n+2) = 0 .
Hence d(y2n, y2n+1, y2n+2) 2n . [ since 0 < < 1] This is a contradiction to the assumption ( II ).
Therefore, {y2n} is a bounded sequence in X.
Similarly { y2n} is a bounded sequence in X when 2r is a decreasing sequence. Hence (X,d) is bounded.
Proof of (T.2.b) :
Let p, m be two positive integers belonging to N. Let the bound of d(y2n, y2n+1, y2n+p) be 2n (2n).
That is,
d(y2n, y2n+1, y2n+p) 2n (2n) and
d(y2n, y2n+1, y2n+p+m) 2n (2n) ( III )
Consider,
d(y2n, y2n+p, y2n+p+m) d(y2n, y2n+1, y2n+p+m) + d(y2n, y2n+p, y2n+1) + d(y2n+1, y2n+p, y2n+p+m)
[by Definition 1.]2n (2n) + 2n (2n) + d(y2n+1, y2n+p, y2n+p+m) [by (III)]
= 2 2n (2n) + d(y2n+1, y2n+p, y2n+p+m)
2 2n (2n) + d(y2n+1, y2n+2, y2n+p+m) + d(y2n+1, y2n+p, y2n+2)
+ d(y2n+2, y2n+p, y2n+p+m) [by Definition 1.]
2 2n (2n) + 2n+1(2n) + 2n+1(2n) + d(y2n+2, y2n+p, y2n+p+m)
= 2 2n (2n) + 2 2n+1(2n) + d(y2n+2, y2n+p, y2n+p+m)
= 2 [2n (2n) + 2n+1 (2n)] + d(y2n+2, y2n+p, y2n+p+m)
[by (III)]Continuing the process,
2n +p 1
d(y2n, y2n+p, y2n+p+m) 2 i (2n) ( IV )
i =2n
Consider,
d(Ax2n+m, Bx2n+m+p, y2n+m) [ max { d(Sx2n+m, Tx2n+m+p, y2n+m),
d(Sx2n+m, Ax2n+m, y2n+m) ,d(Sx2n+m, Bx2n+m+p, y2n+m), d(Tx2n+m+p, Bx2n+m+p, y2n+m), d(Tx2n+m+p, Ax2n+m, y2n+m), (d(Sx2n+m, Bx2n+m, y2n+m) d(Tx2n+p+m, Ax2n+m+p, y2n+m ))}] [by condition (2.3)]
Therefore,
d(y2n+m+p , y2n+m+p+1, y2n+m ) [max{d(y2n+m, y2n+m+p, y2n+m), d(y2n+m, y2n+m+1, y2n+m),
d(y2n+m, y2n+m+p, y2n+m), d(y2n+m+p, y2n+m+p+1, y2n+m), d(y2n+m+p, y2n+m+1, y2n+m),
(d(y2n+m, y2n+1+m, y2n+m) d(y2n+m+p, y2n+m+p+1, y2n+m ))}] [by ( I )]
= [max {0, 0, 0, d(y2n+m+p, y2n+m+p+1, y2n+m),
d(y2n+m+p, y2n+m+1, y2n+m), 0 )}] [by Definition 1.]
= [max { d(y2n+m+p, y2n+m+p+1, y2n+m), d(y2n+m+p, y2n+m+1, y2n+m)}
Suppose that,
d(y2n+m+p , y2n+m+p+1, y2n+m ) > d(y2n+m+p, y2n+m+1, y2n+m)
Then,
d(y2n+m+p, y2n+m+p+1, y2n+m) [d(y2n+m+p , y2n+m+p+1, y2n+m )]
< d(y2n+m+p , y2n+m+p+1, y2n+m ) [since (t)<t]
This is a contradiction. Therefore,
d(y2n+m+p, y2n+m+p+1, y2n+m) [d(y2n+m+p, y2n+m+1, y2n+m)}]
2n+p+m1
[2 i (2n) ] [by ( IV )]
i =2n
2n+p+m1
= 2 i+1 (2n)
i =2n
= 2 [2n+1+ 2n+2 + +2n+p+m] Since p, m is finite and allowing limit as n ,
d(y2n+m, y2n+m+p+1, y2n+p+m) = 0 . [since lim n = 0]
n
Hence {y2n+m} is a Cauchy sequence for mN. Therefore, {y2n} is also a Cauchy sequence in X.
Since (X,d) is complete, {y2n} converges to point zX as n . The sequence {y2n+1} also converges to zX as n .
That is, lim yn = lim yn+1 = z.
n n
Proof of (T.2.c) :
Now taking limit as n in ( I ) and using (T.2.b),
lim y2n+1 = z = lim Sx2n+1 = lim Ax2n nN {0} and
n n n
lim y2n = z = lim Tx2n = lim Bx2n1 nN ( V )
n n n
Consider,
d(ASx2n, Bx2n+1, a) [max{ d(S2x2n, Tx2n+1, a), d(S2x2n, ASx2n, a), d(S2x2n, Bx2n+1, a), d(Tx2n+1, Bx2n+1, a), d(Tx2n+1, ASx2n, a),
(d(S2x2n, BSx2n, a) d(Tx2n+1, Ax2n+1, a))}] [by condition (2.3)]
Since S is continuous and S commutes with A the sequence {SSx2n} and {ASx2n} converges to the point Sz as n .
Now, taking limit as n and also using ( V ),
d(Sz, z, a) [max{d(Sz, z, a), d(Sz, Sz, a), d(Sz, z, a), d(z, z, a), d(z, Sz, a), (d(Sz, Bz, a) d(z, z, a))}]
= [ max{d(Sz, z, a),0, d(Sz, z, a), 0, d(z, Sz, a), 0}] [by Definition 1.]
= [d(Sz, z, a)]
d(Sz, z, a) [d(Sz, z, a)] < d (Sz, z, a) [since (t)<t] This is a contradiction.
This implies d(Sz, z, a) = 0 aX. Therefore, Sz = z.
Proof of (T.2.d) :
Consider,
d(Ax2n, BTx2n+1, a) [max{ d(Sx2n, T2 x2n+1, a), d(Sx2n, Ax2n, a), d(Sx2n, BTx2n+1, a),
d(T2x2n+1, BTx2n+1,a), d(T2x2n+1, Ax2n, a),
(d(Sx2n, Bx2n, a), d(T2 x2n+1, ATx2n+1, a))}] [by condition(2.3)]
Since T is continuous and B commutes with T, then the sequences{TTxn+1} and
{BTxn+1} converges to Tz as n . Taking limit as n and also using ( V),
d(z, Tz, a) [max{d(z, Tz, a), d(z, z, a), d(z, Tz, a), d(Tz, Tz, a),
d(Tz, z, a), d((z, z, a) d(Tz, Az, a))}]
= [max{d(z, Tz, a), 0, d(z,Tz, a), 0, d(Tz, z, a), 0}] [by Definition 1.]
= [d(z, Tz, a)]
d(z, Tz, a) [d(z, Tz, a)] < d(z, Tz, a) [since (t)<t] This is a contradiction.
This implies d(z, Tz, a) = 0 aX. Therefore, Tz = z.
Proof of (T.2.e) :
Consider,
d(Az, Bx2n+1, a) [max {d(Sz, Tx2n+1, a), d(Sz, Az, a),
d(Sz, Bx2n+1, a), d(Tx2n+1, Bx2n+1, a), d(Tx2n+1, Az, a),
(d(Sz, Bz, a), d(Sx2n+1, Ax2n+1, a))}] [by condition (2.3)]
Taking limit as n ,
d(Az, z, a) [max{d(z, z, a), d(z, Az, a), d(z, z, a), d(z, z, a),
d(z, Az, a), (d(z, Bz, a) d(z, z, a))}] [by ( V ) and (T.2.c)]
= [max{0, d(z, Az, a), 0, 0, d(z, Az, a), 0}] [by Definition 1.]
= [d(z, Az, a)]
d(Az, z, a) [d(z, Az, a)] < d(z, Az, a) [since (t)<t] This is a contradiction.
This implies d(z, Az, a) = 0 aX. Therefore, Az = z.
Proof of (T.2.f) :
Consider,
d(Ax2n, Bz, a) [max{ d(Sx2n, Tz, a), d(Sx2n, Ax2n, a),
d(Sx2n, Bz, a), d(Tz, Bz, a), d(Tz, Ax2n, a),
(d(Sx2n, Bx2n, a) d(Tz, Az, a))}] [by condition (2.3)]
Taking limit as n ,
d(z, Bz, a) [max{d(z, z, a), d(z, z, a), d(z, Bz, a), d(z, Bz, a),
d(z, z, a), (d(z, z, a) d(z, Az, a))}] [by (V) and (T.2.d) ]
= [max{0, 0, d(z, Bz, a), d(z, Bz, a), 0, 0}] [by Definition 1.]
= [d(z, Bz, a)]
d(z, Bz, a) [d(z, Bz, a)] < d(z, Bz, a) [since (t)<t] This is a contradiction.
This implies, d(z, Bz, a) = 0 aX. Therefore, Bz = z.
Proof of (T.2.g) :
If possible let z is not unique, let w be another fixed point.
Then, z = Az = Bz = Sz = Tz and
w = Aw = Bw = Sw = Tw ( VI )
Consider,
d(Aw, Bz, a) [max{ d(Sw, Tz, a), d(Sw, Aw, a), d(Sw, Bz, a),
d(Tz, Bz, a), d(Tz, Aw, a),
(d(Sw, Bw, a) d(Tz, Az, a))}] [by condition (2.3)]
Therefore, d(w, z, a) [max{d(w, z, a), d(w, w, a), d(w, z, a), d(z, z, a), d(z, w, a),
(d(w, w, a) d(z, z, a))}] [by ( VI )]
= [max{d(w, z, a), 0, d(w, z, a), 0, d(z, w, a), 0}] [by Definition 1.]
= [d(z, w, a)]
d(w, z, a) [d(z, w, a)] < d(z, w, a) [since (t)<t] This is a contradiction.
This implies, d(w, z, a) = 0 aX. Hence w = z. Therefore, z is the unique common fixed point of S, T, A and B.
CONCLUSION
While observing both the theorems we analyse that the property of a selfmap, the continuity nature of the map is clearly exposed. Further commutativity plays an essential role in obtaining fixed point and its part is more dominant compared to the other two. In other words the concept of continuity and the property of commutativity gets binded to obtain a fixed point. To conclude we can say that the notion of continuity and commutativity is corelated and both plays an effective part for the existence of the fixed point.
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