Faculty of Sciences and Mathematics, University of Niˇs, Serbia
Available at: http://www.pmf.ni.ac.yu/filomat
Filomat 21:2 (2007), 211–234
A COMMON FIXED POINT THEOREM
USING IMPLICIT RELATION
AND PROPERTY (E.A) IN METRIC SPACES
H. K. Pathak, RosanaRodr´ıguez-L´opez and R. K. Verma
Abstract
In this paper, we prove a common fixed point theorem for a quadru-
ple of mappings by using an implicit relation [6] and property (E.A)
[1] under weak compatibility.Our theorem improves and generalizes the
main Theorems of Popa [6] and Aamri and Moutawakil [1].Various ex-
amples verify the importance of weak compatibility and property (E.A)
in the existence of common fixed point and examples are also given to
the implicit relation and to validate our main Theorem. We also show
that prop erty (E.A) and Meir-Keeler type contractive condition are
independent to each other.
1. Introduction
The concept of weakly commuting mappings of Sessa [7] is sharpened by
Jungck [3] and further generalized by Jungck and Rhoades [4]. Similarly,
noncompatible mapping is generalized by Aamri and Moutawakil [1] called
property (E.A). Noncompatibility is also important to study the fixed point
theory. There may be pairs of mappings which are noncompatible but weakly
compatible (see Example 1 of Popa [6] p. 34, and Example 2.1below).
Let A and S be two self-maps of a metric space (X, d). Mappings A and
S are said to be weakly commuting[7] if
d(SAx, ASx) ≤ d(Ax, Sx), for all x ∈ X, (1.1)
compatible [3] if
lim
n→∞
d(ASx
n
, SAx
n
) = 0, (1.2)
whenever there exists a sequence {x
n
} in X such that lim
n→∞
Ax
n
=
lim
n→∞
Sx
n
= t, for some t ∈ X.
2000 Mathematics Subject Classification. 47H10, 54H25.
Key words and phrases. Compatible mappings, implicit relation, noncompatible map-
pings, property (E.A), weakly compatible mappings.
Received: March 30, 2007
212 H. K. Pathak, Rosana Rodr´ıguez-L´opez and R. K. Verma
noncompatible if there exists a sequence{x
n
} in X such that lim
n→∞
Ax
n
=
lim
n→∞
Sx
n
= t, for some t ∈ X and
lim
n→∞
d(ASx
n
, SAx
n
) is either nonzero or nonexistent, (1.3)
and weakly compatible if they commute at their coincidence points, i.e.,
ASu = SAu whenever Au = Su, for some u ∈ X. (1.4)
2. Preliminaries
Property (E.A) [1]. Let A and S be two self-maps of a metric space (X, d)
then they are said to satisfy property (E.A), if there exists a sequence {x
n
}
in X such that
lim
n→∞
Ax
n
= lim
n→∞
Sx
n
= t, for some t ∈ X. (2.1)
Notice that weakly compatible and property (E.A) are independent to each
other:
Example 2.1. Consider X = [0, 1] equipped with the usual metric d. De-
fine f, g : X → X by:
f(x) = 1 − x, if x ∈ [0,
1
2
] and f(x) = 0, if x ∈ (
1
2
, 1],
g(x) =
1
2
, if x ∈ [0,
1
2
] and g(x) =
3
4
, if x ∈ (
1
2
, 1].
Then, for the sequence {x
n
} = {
1
2
−
1
n
}, n ≥ 2, we have
lim
n→∞
f(
1
2
−
1
n
) = lim
n→∞
1
2
+
1
n
=
1
2
= lim
n→∞
g(
1
2
−
1
n
). Thus, the
pair (f, g) satisfies property (E.A). Further, f and g are weakly compatible
since x =
1
2
is their unique coincidence point and fg(1/2) = f(1/2) =
g(1/2) = gf(1/2). We further observe that lim
n→∞
d(fg(
1
2
−
1
n
), gf(
1
2
−
1
n
)) = lim
n→∞
d(f(
1
2
), g(
1
2
+
1
n
)) = d(
1
2
,
3
4
) 6= 0,showing that the pair (f, g)
is noncompatible.
Example 2.2. Let X =
R
+
and d be the usual metric on X. Define
f, g : X → X by:
fx = 0, if 0 < x ≤ 1 and f x = 1, if x > 1 or x = 0; and
gx = [x], the greatest integer that is less than or equal to x, ∀x ∈ X.
Consider a sequence {x
n
} = {1 +
1
n
}
n≥2
in (1, 2), then we have
lim
n→∞
fx
n
= 1 = lim
n→∞
gx
n
. Similarly for the sequence {y
n
} = {1 −
1
n
}
n≥2
in (0, 1), we have lim
n→∞
fy
n
= 0 = lim
n→∞
gy
n
. Thus the pair
(f, g) satisfies(E.A). However, f and g are not weakly compatible; as each
u
1
∈ (0, 1) and u
2
∈ (1, 2) are coincidence points of f and g, where they do
not commute. Moreover, they commute at x = 0, 1, 2, ... but none of these
points are coincidence points of f and g. Further, (f, g) is noncompatible.
Hence, (E.A) 6=⇒ weak compatibility.
Common fixed point theorem using implicit relation and property... 213
3. Implicit relation
Let
R
and
R
+
denote the set of real and non-negative real numbers,
respectively. We now state an implicit relation [6] as follows:
Let F be the set of all continuous functions F with F (t
1
, ..., t
6
) : R
+
6
→
R
satisfying the following conditions:
(F
1
) : F (u, 0, u, 0, 0, u) ≤ 0 =⇒ u = 0, (3.1)
(F
2
) : F (u, 0, 0, u, u, 0) ≤ 0 =⇒ u = 0. (3.2)
The function F (t
1
, ..., t
6
) : R
+
6
→
R
is said to satisfy condition (F
u
) if:
(F
u
) : F (u, u, 0, 0, u, u) ≥ 0, ∀u > 0. (3.3)
The following are examples of the implicit relation defined above. Another
examples can be found in [5-6].
Example 3.1. Let F (t
1
, ..., t
6
) = pt
1
− qt
2
− r(t
3
+ t
4
) − s(t
5
+ t
6
), where
p, q, r, s ≥ 0, 0 ≤ r + s < p and 0 ≤ q + 2s ≤ p, then:
(F
1
) : F (u, 0, u, 0, 0, u) = u(p − r − s) ≤ 0 implies u = 0;
(F
2
) : F (u, 0, 0, u, u, 0) = u(p − r − s) ≤ 0 implies u = 0 and
(F
u
) : F (u, u, 0, 0, u, u) = u(p − q − 2s) ≥ 0, ∀u > 0.
Example 3.2. Let F (t
1
, ..., t
6
) = pt
1
− max{qt
2
, (t
3
+ t
4
)/2, s(t
5
+ t
6
)/2},
where 0 ≤ s < q < 1/2 < p, then:
(F
1
) : F (u, 0, u, 0, 0, u) = pu − max{u/2, su/2} = u(p − 1/2) ≤ 0 ⇒ u = 0;
(F
2
) : F (u, 0, 0, u, u, 0) = pu − max{u/2, su/2} = u(p − 1/2) ≤ 0 ⇒ u = 0;
(F
u
) : F (u, u, 0, 0, u, u) = pu − max{qu, 0, su} = u(p − q) ≥ 0, ∀u > 0.
Example 3.3. Let F (t
1
, ..., t
6
) = t
1
− max{qt
2
, r(t
3
+ t
4
)/2, (t
5
+ t
6
)/2},
where 0 ≤ q < 1 ≤ r < 2, then:
(F
1
) : F (u, 0, u, 0, 0, u) = u− max{0, ru/2, u/2} = u(1− r/2) ≤ 0 ⇒ u = 0;
(F
2
) : F (u, 0, 0, u, u, 0) = u− max{0, ru/2, u/2} = u(1− r/2) ≤ 0 ⇒ u = 0;
(F
u
) : F (u, u, 0, 0, u, u) = u − max{qu, 0, u} = u − u = 0, ∀u > 0.
Example 3.4. Let F (t
1
, ..., t
6
) = t
1
− h max{t
2
, t
3
, t
4
, t
5
, t
6
}, where 0 ≤
h < 1, then:
(F
1
) : F (u, 0, u, 0, 0, u) = u − h max{0, u, 0, 0, u} = u(1 − h) ≤ 0 ⇒ u = 0;
(F
2
) : F (u, 0, 0, u, u, 0) = u − h max{0, 0, u, u, 0} = u(1 − h) ≤ 0 ⇒ u = 0;
(F
u
) : F (u, u, 0, 0, u, u) = u − h max{u, 0, 0, u, u} = u(1 − h) ≥ 0, ∀u > 0.
Example 3.5. Let F (t
1
, ..., t
6
) = t
2
1
− at
2
2
− t
3
t
4
− bt
2
5
− ct
2
6
where a, b, c ≥ 0
214 H. K. Pathak, Rosana Rodr´ıguez-L´opez and R. K. Verma
and 0 < a + b + c < 1 then:
(F
1
) : F (u, 0, u, 0, 0, u) = u
2
(1 − c) ≤ 0 ⇒ u = 0;
(F
2
) : F (u, 0, 0, u, u, 0) = u
2
(1 − b) ≤ 0 ⇒ u = 0 and
(F
u
) : F (u, u, 0, 0, u, u) = u
2
(1 − a − b − c) ≥ 0, ∀u > 0.
Example 3.6. Let F (t
1
, ..., t
6
) = t
3
1
− k(t
3
2
+ t
3
3
+ t
3
4
+ t
3
5
+ t
3
6
), where
0 ≤ k ≤ 1/3,then:
(F
1
) : F (u, 0, u, 0, 0, u) = u
3
(1 − 2k) ≤ 0 ⇒ u = 0;
(F
2
) : F (u, 0, 0, u, u, 0) = u
3
(1 − 2k) ≤ 0 ⇒ u = 0 and
(F
u
) : F (u, u, 0, 0, u, u) = u
3
(1 − 3k) ≥ 0, ∀u > 0.
The following lemma is useful to prove the existence of common fixed point.
Lemma 3.7 [6]. Let (X, d) be a metric space and A, B, S and T be four
self-mappings on Xsatisfying:
F (d(Ax, By), d(Sx, T y), d(Ax, Sx), d(By, T y), d(By, Sx), d(Ax, T y)) < 0,
for all x, y ∈ X, where F satisfies property (F
u
). Then A, B, S and T have
at most one common fixed point.
The following theorem was proved by Popa [6] for a Meir-Keeler type
contractive condition using the implicit relation:
Theorem A. ([6]) Let A, B, S and T be self-mappings of a metric space
(X, d) such that
(a) A(X) ⊆ T (X), B(X) ⊆ S(X),
(b) given ² > 0, there exists δ > 0 such that
² ≤ max {d(Sx, T y), d(Ax, Sx), d(By, T y),
1
2
[d(By, Sx) + d(Ax, T y))]} <
² + δ =⇒ d(Ax, By) < ²,
(c) there exists F ∈ F such that the inequality:
F (d(Ax, By), d(Sx, T y), d(Ax, Sx), d(By, T y), d(By, Sx), d(Ax, T y)) < 0,
holds for all x, y ∈ X.
If one of A(X), B(X), S(X) or T (X) is a complete subspace of X then,
(d) A and S have a coincidence point.
(e) B and T have a coincidence point.
Moreover, if the pairs (A, S) and (B, T ) are weakly compatible, then A,
B, S and T have a unique common fixed point.
Common fixed point theorem using implicit relation and property... 215
The following theorem was proved by Aamri and Moutawakil [1] under
property (E.A)using a contractive condition:
Theorem B. ([1]) Let A, B, S and T be self-mappings of a metric space
(X, d) such that
(a) d(Ax, By) ≤ φ(max {d(Sx, T y), d(By, Sx), d(By, T y)}, ∀(x, y) ∈
X
2
,
where φ :
R
+
→
R
+
is a non-decreasing function on
R
+
such that 0 < φ(t) <
t, for each t ∈ (0, ∞).
(b) (A, S) and (B, T ) are weakly compatible,
(c) (A, S) or (B, T ) satisfy property (E.A),
(d) A(X) ⊆ T (X), B(X) ⊆ S(X).
If the range of one of the mappings is a complete subspace of X, then A,
B, Sand T have a unique common fixed point.
In this paper, we intend to unify Theorem A and Theorem B by imposing
property (E.A).Theorem A uses the Meir-Keeler type contractive condition
which is to be removed by an independent notion viz. property (E.A). Sim-
ilarly, Theorem B uses a φ-contractive condition which is to be removed by
its generalized condition viz. implicit relation. Thus we will use property
(E.A) of [1] and implicit relation of [6] to unify under property(E.A) and
implicit relation.
The following two examples show that Meir-Keeler type contractive con-
dition and property (E.A) are independent to each other.
Example 3.8. Let A, B, S and T be four self-mappings of the metric space
([0, 1], d) with the usual metric d defined by
Ax = Bx = 0, if x = 0 or x = 1, Ax = Bx =
1
n+1
,if
1
n+1
≤ x <
1
n
, n ∈ N;
and
Sx = T x = x, ∀x ∈ X.
The Meir-Keeler type contractive condition is defined by:
given ² > 0, there exists a δ > 0 such that for all x, y ∈ X,
² ≤ M (x, y) < ² + δ =⇒ d(Ax, By) < ². (MKC)
where
M(x, y) = max {d(Sx, T y), d(Ax, Sx), d(By, T y),
1
2
[d(By, Sx)+d(Ax, T y))]}.
Let us discuss property (E.A) and MKC condition for various cases.
Here, (B, T ) satisfies property (E.A). Indeed, taking {
1
n+1
} ⊆ [0, 1], we
getlim
n→∞
B(
1
n+1
) = lim
n→∞
1
n+1
= 0 = lim
n→∞
T (
1
n+1
) = lim
n→∞
1
n+1
.
Similarly, (A,S) satisfies property (E.A).
Next, we check that property MKC is not valid.