A common fixed point theorem using implicit relation and property (E.A) in metric spaces

TL;DR: In this paper, a common fixed point theorem for a quadruple of mappings with weak compatibility was proved for weak compatibility and property (E.A) in the existence of common fixed points.

Abstract: In this paper, we prove a common fixed point theorem for a quadruple 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 examples 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 property (E.A) and Meir-Keeler type contractive condition are independent to each other. .

Summary

1. Introduction

• 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).
• Received: March 30, 2007 noncompatible if there exists a sequence{xn} in X such that limn→.

3. Implicit relation

• Let R and R+ denote the set of real and non-negative real numbers, respectively.
• Let us discuss property (E.A) and MKC condition for various cases.
• For every sequence {xn} in X, the authors get limn→.
• In their main Theorem the authors will apply property (E.A) and implicit relation.
• Meir-Keeler type contractive condition of Theorem A, the authors will impose property (E.A).

4. Main Results

• In this section the authors state and prove their main result.
• Assume that one of the following conditions hold: (v) {Byn} is a bounded sequence for every {yn} ⊆ X such that{Tyn} is convergent (in case (A,S) satisfies property (E.A)), and {Ayn} is a bounded sequence for every {yn} ⊆ X such that{Syn} is convergent (in case (B, T ) satisfies property (E.A)).

B, S and T have a unique common fixed point.

• Suppose that (B, T ) satisfies property (E.A) then, by definition, there exists a sequence {xn}in X such that limn→.
• Hence, in both cases, {Ayn} is a bounded sequence and {d(Ayn, Bxn)} is bounded,hencelimsupn→∞d(Ayn, Bxn) is a finite number.
• Indeed, in this case, limsupn→∞pn = limsupn→∞qn.
• The weak compatibility of A and S implies that ASu = SAu so that AAu = ASu = SAu = SSu.
• The authors can relax continuity ofF , extending this result to the more general case whereF is continuous at certain points of the boundary of (R+)6.

X, then A, B and S have a unique common fixed point.

• It is clear that two non compatible self-maps also satisfy property(E.A).
• The following remarks and examples validate their main Theorem 4.1. Remark 4.10.
• A similar reasoning provides the validity of (ĈS) and (H1).
• Thus both pairs satisfy property (E.A) and condition (iv) is satisfied for both pairs.
• The authors thank the referee’s comments on the paper.

Full text

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.

Frequently asked questions

Q1. What are the contributions mentioned in the paper "A common fixed point theorem using implicit relation and property (e.a) in metric spaces" ?

In this paper, the authors prove a common fixed point theorem for a quadruple of mappings by using an implicit relation [ 6 ] and property ( E. A ) [ 1 ] under weak compatibility. The authors also show that property ( E. A ) and Meir-Keeler type contractive condition are independent to each other. 

Q2. What is the condition that satisfies the property (E.A)?

If condition (vi) holds, and the authors suppose that {Ayn} is not bounded, then there exists a subsequence {nk} of nonnegative integer numbers such that d(Aynk , Bxnk) → +∞, as k → ∞. 

Q3. what is the irrational number of x and y?

If x is an irrational number of X and y ∈ X, the authors have:F (0, |3−y|, 1, |y−2|, 1, |y−2|) = 0−hmax{2|3−y|, 1, |y−2|, 1, |y−2|} ≤ 0,as 0 ≤ h < 12 . 

Q4. what is the irrational number of x?

If x is a rational number of X and y ∈ X, then the authors have: F (0, |2− y|, 0, |2− y|, 0, |2− y|) = 0− hmax{2|2− y|, 0, |2− y|, 0, |2− y|} = −2h|2− y| ≤ 0, as 0 ≤ h < 12 . 

Q5. What is the meaning of a weakly compatible mapping?

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 {xn} in X such that limn→ 

Q6. What is the meaning of weakly compatible mappings?

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] iflimn→∞d(ASxn, SAxn) = 0, (1.2) whenever there exists a sequence {xn} in X such that limn→ 

Q7. what is the case for x and y?

Let us define a function F ∈ F such that F : R+6 → R where F (t1, ..., t6) = t1 − hmax{2t2, t3, t4, t5, t6}, for each t1, ..., t6 ≥ 0, where 0 ≤ h < 12 , then the authors observe that: (i) A(X) = {2} ⊆ T (X) = X and B(X) = {2} ⊆ S(X) = {2, 3}. (ii) Let us discuss two cases for the elements x, y ∈ X and obtain the implicit relation: Case I. 

Q8. what is the proof of the main theorem 4.1?

From the proof of Theorem 4.1, the authors deduce that the strict ’<’ sign can be replaced by ’≤’if the authors admit that the inequality in condition (Fu) is also strict, that is, F (u, u, 0, 0, u, u) > 0, ∀u > 0. 

Q9. what is the simplest way to describe a metric space?

Let A and S be two weakly compatible self-mappings of a metric space (X, d) suchthat (i) A and S satisfy property (E.A), (ii) there exists a continuous (see Remark 4.3) function F : R+6 → R inF such thatF (d(Ax,Ay), d(Sx, Sy), d(Ax, Sx), d(Ay, Sy), d(Ay, Sx), d(Ax, Sy)) < 0,for all x, y ∈ X, where F satisfies all the conditions of implicit relation. 

Q10. what is the simplest example of a metric space?

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, Ty), d(Ax, Sx), d(By, Ty), d(By, Sx), d(Ax, Ty)) < 0,for all x, y ∈ X, where F satisfies property (Fu). 

Q11. if x = 0 and y = 1n, then d( 1?

² ≤ M(x, y) < ² + δ yields² ≤ max { d ( 0,1 n) , d(0, 0), d ( 1 n , 1 n ) , 1 2 ( d ( 1 n , 0 ) + d ( 1 n , 0 ))}= 1 n < ² + δ,showing that ² ≤ d(Ax,By) = 1n , contradicting MKC condition. 

Q12. ii) If x = 0 and y = 1n+1, then?

On the other hand, for a given ² > 0, there exists a δ > 0 such that for x = xn = yn = y ∈ [ 1n+1 , 1n), the authors have that ² ≤ M(x, y) < ² + δ yields² ≤ max { 0, ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ , ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ , 1 2 [∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ + ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ ]}= ∣∣∣∣ 1 n + 1 − xn ∣∣∣∣ < ² + δ,implies d(Axn, Byn) = d ( 1 n+1 , 1 n+1 ) = 0 < ². 

Q13. What is the main result of the theorem?

Assume that one of the following conditions hold: (v) {Byn} is a bounded sequence for every {yn} ⊆ X such that{Tyn} is convergent (in case (A,S) satisfies property (E.A)), and {Ayn} is a bounded sequence for every {yn} ⊆ X such that{Syn} is convergent (in case (B, T ) satisfies property (E.A)). 

Q14. what is the proof of a common fixed point of A and S?

Thus AAu = Au = SAu, and Au is a common fixed point of A and S. Similarly the authors can prove that Bv is a common fixed point of B and T .