A fixed point theorem for a pair of maps satisfying a general contractive condition of integral type

Abstract

We give a general condition which enables one to easily establish fixed point theorems for a pair of maps satisfying a contractive inequality of integral type.

Full text

Page 1

A FIXED POINT THEOREM FOR A PAIR OF MAPS
SATISFYING A GENERAL CONTRACTIVE
CONDITION OF INTEGRAL TYPE
P. VIJAYARAJU, B. E. RHOADES, AND R. MOHANRAJ
Received 18 October 2004 and in revised form 20 July 2005
We give a general condition which enables one to easily establish fixed point theorems for
a pair of maps satisfying a contractive inequality of integral type.
Branciari [1] obtained a fixed point result for a single mapping satisfying an analogue
of Banach’s contraction principle for an integral-type inequality. The second author [3]
proved two fixed point theorems involving more general contractive conditions. In this
paper, we establish a general principle, which makes it possible to prove many fixed point
theorems for a pair of maps of integral type.
Define Φ = {ϕ : ϕ : R+→ R} such that ϕ is nonnegative, Lebesgue integrable, and sat-
isfies
??
0ϕ(t)dt >0for each ? >0.
(1)
Let ψ : R+→ R+satisfy that
(i) ψ is nonnegative and nondecreasing on R+,
(ii) ψ(t)<t for each t >0,
(iii)?∞
Define Ψ = {ψ :ψ satisfies (i)–(iii)}.
Lemma 1. Let S and T be self-maps of a metric space (X,d). Suppose that there exists a
sequence {xn} ⊂ X with x0∈ X, x2n+1:= Sx2n, x2n+2:= Tx2n+1, such that {xn} is complete
and there exists a k ∈ [0,1) such that
?d(Sx,Ty)
0
n=1ψn(t)< ∞ for each fixed t >0.
ϕ(t)dt ≤ψ
??d(x,y)
0
ϕ(t)dt
?
(2)
for each distinct x,y ∈ {xn} satisfying either x =Ty or y = Sx, where ϕ ∈Φ, ψ ∈ Ψ.
Copyright© 2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:15 (2005) 2359–2364
DOI: 10.1155/IJMMS.2005.2359

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2360 Fixed point theorem for a pair of maps
Then, either
(a) S or T has a fixed point in {xn} or
(b) {xn} converges to some point p ∈X and
?d(xn,p)
0
ϕ(t)dt ≤
∞
?
i=n
ψi(d)for n>0,(3)
where
d :=
?d(x0,x1)
0
ϕ(t)dt.
(4)
Proof. Suppose that x2n+1= x2nfor some n. Then x2n= x2n+1= Sx2n, and x2nis a fixed
point of S. Similarly, if x2n+2=x2n+1for some n, then x2n+1is a fixed point of T.
Now assume that xn?=xn+1for each n. With x = x2n, y =x2n+1, (2) becomes
?d(x2n+1,x2n+2)
0
ϕ(t)dt ≤ ψ
??d(x2n,x2n+1)
0
ϕ(t)dt
?
.
(5)
Substituting x = x2n, y = x2n−1, (2) becomes
?d(x2n+1,x2n)
0
ϕ(t)dt ≤ψ
??d(x2n,x2n−1)
0
ϕ(t)dt
?
.
(6)
Therefore, for each n ≥0,
?d(xn,xn+1)
0
ϕ(t)dt ≤ψ
??d(xn−1,xn)
0
ϕ(t)dt
?
≤ ··· ≤ψn(d).
(7)
Let m,n ∈ N, m>n. Then, using the triangular inequality,
d?xn,xm
?≤
m−1
?
i=n
d?xi,xi+1
?.
(8)
It can be shown by induction that
?d(xn,xm)
0
ϕ(t)dt ≤
m−1
?
i=n
?d(xi,xi+1)
0
ϕ(t)dt.
(9)
Using (7) and (9),
?d(xn,xm)
0
ϕ(t)dt ≤
∞
?
i=n
ψi(d) ≤
∞
?
i=n
ψi(d).
(10)
Takingthelimitof(10)asm,n → ∞andusingcondition(iii)forψ,itfollowsthat {xn}
is Cauchy, hence convergent, since X is complete. Call the limit p. Taking the limit of (10)
as m → ∞ yields (3).
?

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P. Vijayaraju et al.2361
Theorem 2. Let (X,d) be a complete metric space, and let S, T be self-maps of X such that
for each distinct x,y ∈ X,
?d(Sx,Ty)
0
ϕ(t)dt ≤ ψ
??M(x,y)
0
ϕ(t)dt
?
,(11)
where k ∈ [0,1), ϕ ∈Φ, ψ ∈ Ψ, and
M(x,y):=max
?
d(x,y),d(x,Sx),d(y,Ty),
?d(x,Ty)+d(y,Sx)?
2
?
.
(12)
Then S and T have a unique common fixed point.
Proof. WewillfirstshowthatanyfixedpointofSisalsoafixedpointofT,andconversely.
Let p = Sp. Then
?
and (11) becomes
?d(p,Tp)
0
M(p,p) =max 0,0,d(p,Tp),d(p,Tp)
2
?
=d(p,Tp),(13)
ϕ(t)dt ≤ ψ
??d(p,Tp)
0
ϕ(t)dt
?
,(14)
which, from (1), implies that p = Tp.
Similarly, p = Tp implies that p =Sp.
We will now show that S and T satisfy (2).
M(x,Sx) = max
?
d(x,Sx),d(x,Sx),d(Sx,TSx),
?d(x,TSx)+0?
2
?
.
(15)
From the triangular inequality,
d(x,TSx)
2
≤
?d(x,Sx)+d(Sx,TSx)?
2
≤ max?d(x,Sx),d(Sx,TSx)?.
(16)
Thus, (11) becomes
?d(Sx,TSx)
0
ϕ(t)dt ≤k
?d(Sx,TSx)
0
ϕ(t)dt, (17)
a contradiction to (1).
Therefore, for all x ∈ X, M(x,Sx) = d(x,Sx), and (2) is satisfied. If condition (a) of
Lemma 1 is true, then S or T has a fixed point. But it has already been shown that any
fixed point of S is also a fixed point of T, and conversely. Thus S and T have a common
fixed point.
Suppose that conclusion (b) of Lemma 1 is true. Then, from (3),
?d(Sx2n,Tp)
0
ϕ(t)dt ≤ψ
??d(x2n,p)
0
ϕ(t)dt
?
, (18)
which implies, since X is complete, that limd(Sx2n,Tp) = 0.

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2362 Fixed point theorem for a pair of maps
Therefore,
d(p,Tp) ≤ d?p,Sx2n
?+d?Sx2n,Tp?− →0,(19)
and p isafixedpointofT,henceafixedpointofS.Condition(11)clearlyimpliesunique-
ness of the fixed point.
?
Every contractive condition of integral type automatically includes a corresponding
contractive condition not involving integrals, by setting ϕ(t) ≡1 over R+.
There are many contractive conditions of integral type which satisfy (2). Included
among these are the analogues of the many contractive conditions involving rational ex-
pressions and/or products of distances. We conclude this paper with one such example.
Corollary 3. Let (X,d) be a complete metric space, S and T self-maps of X such that, for
each distinct x,y ∈ X,
?d(Sx,Ty)
0
ϕ(t)dt ≤k
?n(x,y)
0
ϕ(t)dt,(20)
where ϕ ∈ Φ, k ∈ [0,1), and
n(x,y):= max
?d(y,Ty)?1+d(x,Sx)?
1+d(x,y)
,d(x,y)
?
.
(21)
Then S and T have a unique common fixed point.
Proof.
n(x,Sx) = max?d(Sx,TSx),d(x,Sx)?.
(22)
As in the proof of Theorem 2, it is easy to show that any fixed point of S is also a fixed
point of T, and conversely.
If n(x,Sx) = d(Sx,TSx), then an argument similar to that of Theorem 2 leads to a con-
tradiction. Therefore n(x,Sx) = d(x,Sx), and either S or T has a common fixed point, or
(3) is satisfied. In the latter case, with limxn= p, n(p,p) = 0, so that, from (20), p is a
fixed point of S, hence of T. Uniqueness of p is easily established.
Corollary 3 is also a consequence of Lemma 1.
We now provide an example, kindly supplied by one of the referees, to show that
Lemma 1 is more general than [2, Theorem 3.1].
Example4. LetX := {1/n:n ∈ N∪{0}}withtheEuclideanmetricandS,T areself-maps
of X defined by

n+1
1
n+2
S
?1
n
?
=


1
if n is odd,
if n is even,
0if n = ∞,
T
?1
n
?
=



1
n+1
1
n+2
0
if n is even,
if n is odd,
if n = ∞.
(23)

Page 5

P. Vijayaraju et al.2363
For each n, define x2n+1= Sx2n, x2n+2= Tx2n+1. With x0= 1, let O(1) denote the orbit
of x0= 1; that is, O(1) = {1,1/2,1/3,...} and O(1) = O(1) ∪ {0} = X. For x,y ∈ O(1),
y = 1/m, m even and x = 1/n =Ty = 1/(m+1), Sx = 1/(m+2), so that
????
d(x,y) =
m
d(Sx,Ty) =
1
m+1−
????1
1
m+1
????
????=
1
m+1−
????=1
1
m+2=
1
(m+1)(m+2),
1
m(m+1).
n−1
????=
1
m+1−1
nm−
1
m+1=
(24)
Thus
d(Sx,Ty)
d(x,y)
=
m
m+2≤1.
(25)
Also
sup
n∈N
d(Sx,Ty)
d(x,y)
= 1,(26)
so that there is no number c ∈ [0,1) such that d(Sx,Ty) ≤ cd(x,y) for x,y ∈ O(1) and
x = Ty. Therefore, [2, Theorem 3.1] cannot be used. On the other hand, the hypotheses
of Lemma 1 are satisfied. To see this, it will be shown that condition (2) is satisfied for
some ϕ ∈ Φ.
We will first show that for any x = 1/n, y = 1/m ∈ O(1) satisfying either x = Ty or
y = Sx,
????
There are four cases.
d(Sx,Ty) ≤
1
n+1−
1
m+1
????.
(27)
Case 1. y = 1/m, m even, x =1/n =Ty =1/(m+1), and Sx =1/(m+2). Then
????
Case 2. y = 1/m, m odd, x = 1/n =Ty = 1/(m+2), and Sx = 1/(m+3). Then
????
≤
d(Sx,Ty) =
1
m+2−
1
m+1
????=
????
1
n+1−
1
m+1
????.
(28)
d(Sx,Ty) =
1
m+3−
1
m+1−
1
m+2
1
m+3=
????=
1
m+2−
1
n+1−
1
m+3
1
m+1
????
????.
(29)
Case 3. x = 1/n, n even, y = 1/m =Sx = 1/(n+2), and Ty = 1/(n+3). Then
????
≤
d(Sx,Ty) =
1
n+2−
1
n+1−
1
n+3
1
n+3=
????=
1
n+2−
1
n+1−
1
n+3
1
n+3
????
????.
(30)