Common Fixed Point Theorems in Uniform Spaces

International Journal of Latest Technology in Engineering, Management & Applied Science (IJLTEMAS)
Volume VI, Issue V, May 2017 | ISSN 2278-2540
www.ijltemas.in Page 4
Common Fixed Point Theorems in Uniform Spaces
Deepti Sharma
Department of Mathematics, Ujjain Engineering College, Ujjain (M.P.), 456010, India
Abstract: - In the process of generalization of metric spaces to
Topological spaces, a few aspects of metric spaces are lost.
Therefore, the requirement of generalization of metric spaces
leads to the theory of uniform spaces. Uniform spaces stand
somewhere in between metric spaces and general topological
spaces. Khan[6] extended fixed point theorems due to Hardy and
Rogers[2], Jungck[4] and Acharya[1] in uniform space by
obtaining some results on common fixed points for a pair of
commuting mappings defined on a sequentially complete
Hausdorff uniform space. Rhoades et. al.[7] generalized the
result of Khan[6] by establishing a general fixed point theorem
for four compatible maps in uniform space .
In this paper, a common fixed point theorem in
uniform spaces is proved which generalizes the result of Khan[6]
and Rhoades et al.[7] by employing the less restrictive condition
of weak compatibility for one pair and the condition of
compatibility for second pair, the result is proved for six self-
mappings.
AMS (2000) Subject Classification:- 54H25, 47H10.
Keywords: - Common fixed point, uniform space, compatible
mappings and weakly compatible mappings.
I. INTRODUCTION AND PRELIMINARY CONCEPTS
or the terminology, definition and basic properties of
uniform spaces, the reader can refer to Joshi[3].
Following Khan[6] and Rhoades et. al.[7], we
assume that throughout the paper, (X, U ) stands for a
sequentially complete Hausdorff uniform space and P be a
fixed family of pseudo-metrics on X which generates the
uniformity U . Following Kelley[5], they [6, 7] assumed :
(1.1) V(p, r) = { (x, y) : x, y  X, p(x, y)  r }.
(1.2) G =
i, i
n
(p r ) i i
i 1
V : V V : p P,r 0,i 1,2,...,n

 
    
 
and for  > 0,
(1.3) V =
i, i
n
(p r ) i i
i 1
V : p P,r 0,i 1,2,...,n .

 
   
 
Khan[6] and Rhoades et. al.[ 7] used the following
well-known lemmas taken from Acharya[1] in order to prove
their results.
Lemma 1.1. If V G and ,  > 0, then (V) = ()V.
Lemma 1.2. Let p be any pseudo-metric on X and ,  > 0.
If (x, y) 
   1 2p,r p,rV V  , then p(x, y) <  r1 + 
r2.
Lemma 1.3. If x, y  X, then, for every V in G there is a
positive number  such that (x, y)  V.
Lemma 1.4. For any arbitrary V  G there is a pseudo-metric
p on X
such that V = V(p, 1). This p is called a Minkowski pseudo-
metric of V.
Lemma 1.5. [6] Let { y
n
} be a sequence in a complete
metric space (X, p). If there exists k(0, 1) such that p(y
n+1
,
y
n
) ≤ k p(y
n
, y
n−1
) for all n, then {y
n
} converges to a point in
X.
Definition 1.1 [7] Let A and B be two self-maps of a uniform
space, p a pseudo-metric on X. A and B will be said to be
compatible on X if p(ABx
n
, BAx
n
) = 0, whenever {x
n
}
is a sequence in X such that {Ax
n
} and {Bx
n
} converge to the
same point t in X.
Definition 1.2. Let A and B be self-mappings of a uniform
space, p a pseudo-metric on X. Then the mappings A and B
are said to be weakly compatible if they commute at their
coincidence point, that is, Ax = Bx implies ABx = BAx for
some xX.
II. MAIN RESULT
Khan[6] extended fixed point theorems due to Hardy
and Rogers[2], Jungck[4] and Acharya[1] in uniform space by
obtaining some results on common fixed points for a pair of
commuting mappings defined on a sequentially complete
Hausdorff uniform space. Rhoades et. al.[7] generalized the
result of Khan[6] by establishing a general fixed point
theorem for four compatible maps in uniform space . In this
paper, a common fixed point theorem in uniform spaces is
proved which generalizes the result of Khan[6] and Rhoades
et al.[7] .
Theorem 2.1. Let A, B, S, T, P and Q be self-maps of X
satisfying the following conditions:
F

(2.1) (PQx, Ax)  V1, (STy, By)  V2, (PQx, By)  V3 (STy,
Ax)  V4,
(PQx, STy)  V5 implies that (Ax, By)  
1
V
1
 
2
V
2


3
V
3


4
V
4
 
5
V
5
, where 
i
= 
i
(x, y) are non-negative
functions from X  X→ [0, 1) satisfying
x,y X
sup

5
i
i 1
1

 
and 
3
= 
4
;
(2.2) A(X)  ST(X), B(X)  PQ(X);
(2.3) either A or PQ is continuous;
(2.4) (A, PQ) is compatible and (B, ST) is weakly
compatible; and
(2.5) PQ = QP, ST = TS, AQ = QA and BT = TB,
Then A, B, S, T, P and Q have a unique common fixed point
in X.
Proof. Let V  G be arbitrary and p the Minkowski pseudo-
metric of V. For x, y X, let p(PQx, Ax) = r
1
, p(STy, By) = r
2
,
p(PQx, By) = r
3
, p( STy, Ax) = r
4
, p(PQx, STy) = r
5
.
For any Ɛ > 0, (PQx, Ax)  (r
1
+ Ɛ)V, (STy, By)  ( r
2
+ Ɛ)V ,
(PQx, By)  (r
3
+ Ɛ)V, (STy, Ax)  (r
4
+ Ɛ)V, (PQx, STy) 
(r
5
+ Ɛ)V.
From (2.1),
(Ax, By)  
1
( r
1
+ Ɛ)V  
2
( r
2
+ Ɛ)V  
3
( r
3
+ Ɛ)V  
4
( r
4
+
Ɛ)V  
5
(r
5
+ Ɛ)V.
where 
i
= 
i
(x, y). Using Lemmas 1.1−1.3, we get
p(Ax, By)  
1
( r
1
+ Ɛ) + 
2
( r
2
+ Ɛ)+ 
3
( r
3
+ Ɛ) + 
4
( r
4
+ Ɛ)+

5
( r
5
+ Ɛ).
Since Ɛ is arbitrary, we have
(2.6) p(Ax, By) ≤ 
1
p(PQx, Ax) + 
2
p(STy, By)
+ 
3
p(PQx, By) + 
4
p(STy, Ax) + 
5
p(PQx, STy).
Now, let x
0
be an arbitary point in X. As A(X)  ST(X) and
B(X)  PQ(X), then there exists x
1
, x
2
 X such that Ax
0
=
STx
1
= y
0
and Bx
1
= PQx
2
= y
1
.In general construct a
sequences {y
n
} in X such that y
2n
= STx
2n+1
= Ax
2n
and
y
2n+1
= Bx
2n+1
= PQx
2n+2
for n = 0, 1, 2,… .Now, we show that
{y
n
} is a Cauchy sequence in X. From (2.6), we have
p (y
2n
, y
2n+1
) = p(Ax
2n
, Bx
2n+1
)
≤ 1 p(PQx
2n
, Ax
2n
) + 2 p(STx
2n+1
, Bx
2n+1
)
+ 3 p(PQx
2n
, Bx
2n+1
)+ 4 p(STx
2n+1
, Ax
2n
)
+ 5 p(PQx
2n
, STx
2n+1
).
p (y
2n
, y
2n+1
) ≤ 1 3 5
2 31
    
   
p(y
2n−1
, y
2n
).
=  p(y
2n−1
, y
2n
).
In general p(y
n
, y
n+1
) ≤  p(y
n−1
, y
n
).From lemma (2.1), {y
n
}
converges to some point z in X. Thus, the subsequences
{Ax
2n
}, {Bx
2n+1
}, {STx
2n+1
} and {PQx
2n+2
} of sequence {y
n
}
also converges to z in X.
Case I. Suppose A is continuous, we have A2
x
2n
→ Az and
A(PQ)x
2n
→ Az. The compatibility of the pair (A, PQ) gives
that (PQ)Ax
2n
→ Az.
Step 1. Putting x = Ax
2n
and y = x
2n+1
in (2.6), we have
p(AAx
2n
, Bx
2n+1
) ≤ 
1
p(PQAx
2n
, AAx
2n
)
+ 
2
p(STx
2n+1
, Bx
2n+1
) + 
3
p(PQAx
2n
, Bx
2n+1
)
+ 
4
p(STx
2n+1
, AAx
2n
)+ 
5
p(PQAx
2n
, STx
2n+1
).
Letting n→∞ and using above results, we get
p(Az, z) ≤ (
3
+ 
4
+ 
5
) p(Az, z). So that Az = z.
Step 2. Since A(X)  ST(X), there exists uX such that
z = Az = STu. Putting x = x
2n
and y = u in (2.6), we get
p(Ax
2n
, Bu)≤ 
1
p(PQx
2n
, Ax
2n
) + 
2
p(STu, Bu)
+ 
3
p(PQx
2n
, Bu)+ 
4
p(STu, Ax
2n
) + 
5
p(PQx
2n
, STu).
p(z, Bu) ≤ (
2
+ 
3
) p(Bu, z). So that z = Bu. The weak
compatibility of the pair (B, ST) gives that STBu = BSTu.
Hence STz = Bz.
Step 3. Putting x = x
2n
and y = z in (2.6), we get

p(Ax
2n
, Bz) ≤ 
1
p(PQx
2n
, Ax
2n
) + 
2
p(STz, Bz)
+ 
3
p(PQx
2n
, Bz)+ 
4
p(STz, Ax
2n
) + 
5
p(PQx
2n
, STz).
p(z, Bz) ≤ (
3
+ 
4
+ 
5
) p(Bz, z). So that z = Bz = STz.
Step 4. Putting x = x
2n
and y = Tz in (2.6), we get
p(Ax
2n
, BTz)≤ 
1
p(PQx
2n
, Ax
2n
) + 
2
p(STTz, BTz)
+ 
3
p(PQx
2n
, BTz)+ 
4
p(STTz, Ax
2n
) + 
5
p(PQx
2n
, STTz).
Since BT = TB and ST = TS. We have BTz = Tz and
ST(Tz) = Tz.
Letting n→∞ and using above results, we get z = Tz . Now
STz = z, which implies that Sz = z. Hence Sz = Tz = Bz = z.
Step 5. As B(X)  PQ(X), there exists vX such that
z = Bz = PQv. Putting x = v and y = x
2n+1
in (2.6), letting
n→∞ and using above results, we get p(Av, z) ≤ (
1
+ 
4
)p(Av, z). So that Av = z = PQv. As the pair (A, PQ) is
compatible implies weakly compatible. Therefore APQv =
PQAv implies Az = PQz. Hence PQz = Az = z.
Step 6. Putting x = Qz and y = z in (2.6), As AQ = QA and
PQ = QP.We have AQz = Qz and PQ(Qz) = Qz. Using above
results, we get p(Qz, z) ≤ (
3
+ 
4
+ 
5
) p(z, Qz).So that Qz =
z. Therefore PQz = z, which implies that Pz = z.
Hence Az = Bz = Sz = Tz = Pz = Qz = z.
Thus, z is a common fixed point of A, B, S, T, P and Q.
Case II. Similarly by taking PQ is continuous, it can be
proved that z is a common fixed point of A, B, S, T, P and Q
Uniqueness.
Let w is also a common fixed point of A, B, S, T, P and Q,
Then w = Aw = Bw = Sw = Tw = Pw = Qw.
Putting x = z and y = w in (2.6), we get p(z, w) ≤ (
3
+ 
4
+

5
) p(z, w). So that z = w. Therefore, z is unique common
fixed point of A, B, S, T, P and Q.
Corollary 2.1. Let A, B, S, T, P and Q be self-maps of X
satisfying the conditions (2.1), (2.2), (2.3), (2.5) of theorem
(2.1) and the pairs (A, PQ) and (B, ST) are compatible.
Then A, B, S, T, P and Q have a unique common fixed point
in X.
Proof. As compatibility implies weak compatibility, the proof
follows from theorem 2.1.
REFERENCES
[1]. Acharya, S.P.,Some Results on fixed points in uniform spaces,
Yokohama Math. J. 22(1974), 105-116.
[2]. Hardy, G.E. and Rogers, T.D., A generalization of a fixed point
theorem of Reich, Bull. Cand. Math. Soc., 16 (1973), 201-206.
[3]. Joshi, K.D., Introduction to General Topology, Wiley Eastern,
1992.
[4]. Jungck, G., Commuting mappings and fixed points, Amer. Math.
Monthly, 83 (1976), 261-263.
[5]. Kelley, J. L., General topology, Van Nostrand, 1955.
[6]. Khan, M.S., Commuting maps and fixed points in uniform spaces,
Bull. Cal. Math. Soc. 29(1981), 499-507.
[7]. Rhoades, B. E., Singh, B. and Chauhan, M.S., Fixed points of
compatible mappings in uniform spaces, Ranchi Univ. Math. J.,
25(1994), 83-87.

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