American Journal of Mathematics and Statistics
p-ISSN: 2162-948X e-ISSN: 2162-8475
2013; 3(4): 190-193
doi:10.5923/j.ajms.20130304.02
1Department of Mathematics and Computer Science, Rani Durgawati University., Jabalpur, M.P., India
2Department of Mathematics and Computer Science, Govt. Model Science College, Jabalpur, M.P., India
Correspondence to: P. L. Powar, Department of Mathematics and Computer Science, Rani Durgawati University., Jabalpur, M.P., India.
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Copyright © 2012 Scientific & Academic Publishing. All Rights Reserved.
In the present paper, the authors have noticed that the condition, which has been imposed to compute the unique common fixed point for sequence of mappings used by Iseki[6] in 1974 and Babu G.V.R. et al.[2], can be replaced by another generalized condition, which significantly reduce large number of computational steps and established the same result. In addition, the generalized condition introduced in this paper includes several results on fixed point theory by considering special values of parameter. (see cf.[2],[21]).
Keywords: Common Fixed Point, Complete Metric Space, Subadditive Altering Distance Function
Cite this paper: P. L. Powar, G. R. K. Sahu, Common Fixed Point Theorem by Subadditive Altering Distance Function for Sequence of Mappings, American Journal of Mathematics and Statistics, Vol. 3 No. 4, 2013, pp. 190-193. doi: 10.5923/j.ajms.20130304.02.
is called a contraction mapping if there exists a real number k,
, such that d (Tx, Ty)
kd (x, y), for all x, y in X.The well-known Banach contraction theorem is given below:“If T is a mapping of a complete metric space X into itself such that d(Tx,Ty)
k d(x,y), for all x, yX and 0 < k <1. Then T has a unique fixed point.” Definition 2.2 A function
is called a subadditive altering distance function if the following properties are satisfied: (i)
is a continuous function (ii)
is a monotonically increasing function (iii)
(iv)
Lemma 2.1 (Lemma 1.3 of[14]). Let (M, d) be a metric space. Let {xn} be a sequence in M such that
. If {xn} is not a Cauchy sequence in M, then there exists an
for which, the subsequences
and
of {xn} may be obtained with m(k) > n(k) > k such that
and 
be a sequence of self maps of X. Suppose there are non-negative real numbers
such that for any x, y in X and i, j = 1, 2,…, n…
where
Then
has a unique fixed point. Sastry et al.[20] have initiated the following Theorem in 1999:Theorem 3.2 (Theorem 4.2 of[11]). Let (X, d) be a bounded complete metric space. Suppose
is a sequence of self maps of X such that TiTj = Tj Ti, for all i, j = 1, 2,…, n,…and satisfies the inequality: There exists k
(0, 1) and 
Then, the sequence
has a unique common fixed point. Babu et al.[2] had proved the following result in 2001:Theorem 3.3 (Theorem 4.5 of[11]). Let (X, d) be a complete metric space and
be a sequence of self maps of X. Suppose there is a
satisfying the following inequality:There exists k
[0, 1) such that
for all x, y
X and for all j = 1, 2,…n,…. Then the mappings
have a unique common fixed point in X.
be a sequence of self maps on a complete metric space (X, d) and
be an altering distance function satisfying the condition:![]() | (4.1) |
X, where 0 < a < ½ .Then
has a unique common fixed point in X. Remark 4.1 It may be noted that any specific choice of parameter ‘a’ should be bounded by 0 and ½.Proof. Let x be an arbitrary point in X and {xn} be a sequence of points of X. Consider ![]() | (4.2) |
![]() | (4.3) |
This implies that
Hence
. Thus, by induction, it follows
This implies that
is decreasing sequence of non-negative real numbers; hence, it converges to zero,
Since 
This implies that the sequence {
} is also a deceasing sequence of non-negative real number and hence
as
.Claim: The sequence {xn} is a Cauchy sequence. For this, it is sufficient to show that the subsequence {x2n} of {xn} is a Cauchy sequence, Let, if possible, {x2n} is not a Cauchy sequence, then there exists an
and monotonically increasing sequence of natural numbers {2m(k)} and {2n(k)} such that n(k) > m(k) , ![]() | (4.4) |
This implies that
Letting
, yields
(using lemma 2.1)which is a contradiction. Hence, {x2n} is a Cauchy sequence therefore {xn} is a Cauchy sequence in X. Since X is complete metric space, therefore {xn} converges to a point x
X.Claim: x is a common fixed point of sequence of mappings
. Consider
Applying limit as
, we get
This implies that
i.e. Tmx = x, for all m = 1, 2,… This show that x is fixed point of Tm,
m = 1, 2,… Thus, x is a common fixed point of the sequence of mappings
.Claim: x is unique. Let, if possible,
such that y is also a fixed point of Tn, for all n=1, 2,…i.e. Tny = y, for all n =1, 2, … Now 
This implies that
which is a contradiction. Thus, x is unique common fixed point of sequence of mappings
.
, the following may be noted :(i) Putting
and
where
and
in (4.1) leads to the Theorem 4.3 of[11].(ii) Putting
where 0 < k < 1 in (4.1) yields Theorem 4.2 of [11]. (iii) Putting
, where 
and 0 < k < 1 in (4.1) Theorem 4.5 of[11] can obtained.Note: (ii) and (iii) hold without sub-additive condition on
.