# Some Fixed Point Theorems under Weak Semicompatibility

Authors: A. S. Saluja; Mukesh Kumar Jain
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

The aim of the present paper is to prove some fixed point theorems by using the recent notion “weak semicompatibility.” The new notion is proper generalization of semicompatibility and can be applicable on commuting and compatible maps. We used compatible and absorbing mappings to prove theorems which also include (E.A.) property.


---

## Body

## 1. Introduction and New Definitions

In 1995, Cho et al. [1] introduced the concept of semicompatibility and obtained the first result that established a situation in which a collection of mappings has a fixed point. They defined a pair of self-maps (
S
,
T
) to be semicompatible if(a)
S
x
=
T
x
⇒
S
T
x
=
T
S
x and
(b)
lim
⁡
n
→
∞
⁡
S
x
n
=
lim
⁡
n
→
∞
⁡
T
x
n
=
x. For some x
∈
X, implying lim
⁡
n
→
∞
⁡
S
T
x
n
=
T
x holds.Singh and Jain [2] observe that (b) implies (a). Hence they defined the semicompatibility by condition (b) only.Let(
X
,
d
) be a metric space and let f and g be two maps from (
X
,
d
) into itself. f and g are commuting maps if f
g
x
=
g
f
x for all x in X.To generalize the notion of commuting maps, Sessa [3] introduced the concept of weakly commuting maps. He defines f and g as being weakly commuting if d
(
f
g
x
,
g
f
x
)
≤
d
(
f
x
,
g
x
) for all x
∈
X. Obviously, commuting maps are weakly commuting but the converse is not true.In 1986, Jungck [4] gave more generalized commuting and weakly commuting maps called compatible maps. f and g are called compatible if
(1)
lim
⁡
n
→
∞
d
(
f
g
x
n
,
g
f
x
n
)
=
0
.
Whenever {
x
n
} is a sequence in X such that lim
⁡
n
→
∞
⁡
f
x
n
=
lim
⁡
n
→
∞
⁡
g
x
n
=
t for some t
∈
X, clearly, weakly commuting maps are compatible, but the implication is not reversible (see [4]).Afterwards, Jungck et al. [5] made another generalization of weakly commuting maps by introducing the concept of compatible maps of type (A). Previous f and g are said to be compatible of type (A) if in place of (1) we have the two following conditions:
(2)
lim
⁡
n
→
∞
d
(
f
g
x
n
,
g
g
x
n
)
=
0
,
lim
⁡
n
→
∞
d
(
g
f
x
n
,
f
f
x
n
)
=
0
.
It is clear to see that weakly commuting maps are compatible of type (A); from [5] it follows that the implication is not reversible.Two self-mapsf and g of metric space (
X
,
d
) are called g-compatible ([6] cited from [7]) if lim
⁡
d
(
g
f
x
n
,
f
f
x
n
)
=
0, whenever {
x
n
} is a sequence in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X. Similarly, two self-maps f and g of metric space (
X
,
d
) are called f-compatible ([6] cited from [7]) if lim
⁡
d
(
f
g
x
n
g
g
x
n
)
=
0, whenever {
x
n
} is a sequence in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X.Two self-mappingsf and g of a metric space (
X
,
d
) are called R weakly commuting [8] at a point x in X if d
(
f
g
x
,
g
f
x
)
≤
R
d
(
f
x
,
g
x
) for some R
>
0.The two self-mapsf and g of a metric space (
X
,
d
) are called R weakly commuting of type (
A
g
) [9] if there exists some positive real number R such that d
(
f
f
x
,
g
f
x
)
≤
R
d
(
f
x
,
g
x
) for all x in X.The two self-mapsf and g of a metric space (
X
,
d
) are called R weakly commuting of type (
A
f
) [9] if there exists some positive real number R such that d
(
f
g
x
,
g
g
x
)
≤
R
d
(
f
x
,
g
x
) for all x in X. It may be noted that compatible mappings f and g can be R weakly commuting of types (
A
f
) and (
A
g
).Letf and g
(
f
≠
g
) be two self-maps of metric space (
X
,
d
); then f will be called g-absorbing [10] if there exists a real number R
>
0 such that d
(
g
x
,
g
f
x
)
≤
R
d
(
f
x
,
g
x
) for all x
∈
X. Similarly, let f and g
(
f
≠
g
) be two self-maps of metric space (
X
,
d
); then g will be called f-absorbing [10] if there exists a real number R
>
0 such that d
(
f
x
,
f
g
x
)
≤
R
d
(
f
x
,
g
x
) for all x
∈
X.Letf and g be two self-mappings of metric space (
X
,
d
). The maps f and g satisfy the E.A. property [11] if there exists a sequence {
x
n
} in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t
∈
X.Two self-mapsf and g of metric space (
X
,
d
) are said to be f-compatible of type (
E
) [12] if lim
⁡
f
f
x
n
=
lim
⁡
f
g
x
n
=
g
t, whenever {
x
n
} is a sequence in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X. Similarly, two self-maps f and g of metric space (
X
,
d
) are said to be g-compatible of type (
E
) [12] if lim
⁡
g
g
x
n
=
lim
⁡
g
f
x
n
=
f
t, whenever {
x
n
} is a sequence in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X.Pant et al. [13] introduced a notion of weak reciprocal continuity as follows.Definition 1.
Two self-mappingsf and g of metric space (
X
,
d
) will be called weakly reciprocally continuous if lim
⁡
f
g
x
n
=
f
t or lim
⁡
g
f
x
n
=
g
t, whenever {
x
n
} is a sequence in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X.
Further Saluja et al. [14] introduced a notion of weak semicompatibility as follows.Definition 2.
Two self-mappingsf and g of a metric space (
X
,
d
) will be called weak semicompatible mappings if lim
⁡
n
→
∞
⁡
f
g
x
n
=
g
t or lim
⁡
n
→
∞
⁡
g
f
x
n
=
f
t, whenever {
x
n
} is a sequence in X such that lim
⁡
n
→
∞
⁡
f
x
n
=
lim
⁡
n
→
∞
⁡
g
x
n
=
t for some t in X.Example 3 (see [14]).
LetX
=
[
0,1
] and d be the usual metric on X. Define f
,
g
:
X
→
X by
(3)
f
x
=
x
∀
x
∈
[
0,1
]
,
g
x
=
{
0
,
x
∈
(
0,1
]
,
1
,
x
=
0
.
Taking {
x
n
}
=
1
/
n, since lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
0, we have
(4)
lim
⁡
f
g
x
n
=
lim
⁡
f
(
0
)
=
0
≠
g
(
0
)
,
lim
⁡
g
f
x
n
=
lim
⁡
g
(
x
n
)
=
lim
⁡
g
(
1
n
)
=
0
=
f
(
0
)
.
Therefore f and g are weak semicompatible.
Also here mapsf and g have no common fixed point.Now we give some more definitions to improve this result.Definition 4 (see [15]).
LetX be a set and f and g be self-maps of X. A point x in X is called coincidence point of f and g if and only if f
x
=
g
x. One will call w
=
f
x
=
g
x a point of coincidence of f and g.Definition 5 (see [16]).
Two self-mapsf and g of a set X are occasionally weakly compatible (owc) if and only if there is a point x in X which is coincidence point of f and g at which f and g commute.Lemma 6 (see [15]).
LetX be a set and let f and g be owc self-maps on X. If f and g have a unique point of coincidence, w
:
=
f
x
=
g
x, then w is the unique common fixed point of f and g.Definition 7 (see [17]).
Two self-mappingsf and g on a metric space (
X
,
d
) are called conditionally commuting if they commute on nonempty subset of the set of coincidence points whenever the set of their coincidence points is nonempty.
From the definition itself, it is clear that if two maps are weakly compatible or owc, then they are necessarily conditionally commuting; however, the conditionally commuting mappings are not necessarily weakly commuting or owc.Definition 8.
LetX be a set. A symmetric on X is a mapping d
:
X
×
X
→
[
0
,
∞
) such that
(5)
d
(
x
,
y
)
=
0
iff
x
=
y
,
d
(
x
,
y
)
=
d
(
y
,
x
)
for
x
,
y
∈
X
.Theorem 9.
LetX be a set with a symmetric d. Suppose that f and g are owc self-maps of X satisfying
(6)
(
a
)
d
(
f
x
,
f
y
)
≥
a
d
(
g
x
,
g
y
)
+
b
d
(
g
x
,
f
x
)
+
c
d
(
g
y
,
f
y
)
,
where a
,
c
>
1, b
∈
R, such that a
+
b
+
c
>
1. Then f and g have a unique common fixed point.Proof.
Since the maps are owc, there exists a pointx such that f
x
=
g
x and f
g
x
=
g
f
x. Substituting (a) with y
=
g
x gives
(7)
d
(
f
x
,
f
g
x
)
≥
a
d
(
g
x
,
g
f
x
)
+
b
d
(
g
x
,
f
x
)
+
c
d
(
g
f
x
,
f
g
x
)
=
a
d
(
g
x
,
f
g
x
)
.
Thus,0
≥
(
a
-
1
)
d
(
f
x
,
f
g
x
), which implies that f
x is a fixed point of f. But, since g
f
x
=
f
g
x, f
x is also a fixed point of g.
Suppose thatp and q are common fixed point of f and g. Substituting (1) with
(8)
d
(
p
,
q
)
=
d
(
f
p
,
f
q
)
≥
a
d
(
g
p
,
g
q
)
+
b
d
(
g
p
,
f
p
)
+
c
d
(
g
q
,
f
q
)
=
a
d
(
p
,
q
)
,
implies that p
=
q.Lemma 10 (see [18]).
Iffand g are compatible of type (A), then they are owc, but the converse is not true in general.
Now we give the following lemma with the fact of [18].Lemma 11.
Iffand g are either f-compatible or g-compatible, then they are owc, but the converse is not true in general. We give the following examples to prove this.Example 12.
LetX
=
[
0
,
∞
) with the usual metric. Define f
,
g
:
X
→
X by
(9)
f
x
=
{
3
,
if
x
∈
[
0,1
)
,
x
,
if
x
∈
[
1
,
∞
)
,
g
x
=
{
2
,
if
x
∈
[
0,1
)
,
1
x
,
if
x
∈
[
1
,
∞
)
.
We have f
(
1
)
=
g
(
1
)
=
1 and f
g
(
1
)
=
g
f
(
1
)
=
1; that is, f and g are owc. Now consider x
n
=
1
+
1
/
n for n
∈
{
1,2
,
3
…
}; we have lim
⁡
f
x
n
=
lim
⁡
f
(
1
+
1
/
n
)
=
1 and lim
⁡
g
x
n
=
lim
⁡
g
(
1
+
1
/
n
)
=
1. But lim
⁡
f
g
x
n
=
lim
⁡
f
g
(
1
+
1
/
n
)
=
3 and lim
⁡
g
g
x
n
=
lim
⁡
g
g
(
1
+
1
/
n
)
=
2; therefore lim
⁡
d
(
f
g
x
n
,
g
g
x
n
)
≠
0; hence f and g are not g-compatible.Example 13.
LetX
=
[
1,5
] with the usual metric. Define f
,
g
:
X
→
X by
(10)
f
x
=
{
1
,
if
x
=
1
,
5
,
if
x
∈
(
1,5
]
,
g
x
=
{
1
,
if
x
=
1
,
x
+
4
,
if
x
∈
(
1,5
]
.
We have f
(
1
)
=
g
(
1
)
=
1 and f
g
(
1
)
=
g
f
(
1
)
=
1; that is, f and g are owc. Now consider x
n
=
1
+
1
/
n for n
∈
{
1,2
,
3
…
}; we have lim
⁡
f
x
n
=
lim
⁡
f
(
1
+
1
/
n
)
=
5 and lim
⁡
g
x
n
=
lim
⁡
g
(
1
+
1
/
n
)
=
5. But lim
⁡
g
f
x
n
=
lim
⁡
g
f
(
1
+
1
/
n
)
=
g
(
5
)
=
9 and lim
⁡
f
f
x
n
=
lim
⁡
f
f
(
1
+
1
/
n
)
=
lim
⁡
f
(
5
)
=
5; therefore lim
⁡
d
(
g
f
x
n
,
f
f
x
n
)
≠
0; hence f and g are not f-compatible.Lemma 14.
Iff and g are f-compatible of type (E) or g-compatible of type (E), then they are owc, but the converse is not true in general. Here we prove the converse condition.Proof.
IfX
=
[
1
,
∞
) with the usual metric, define f and g by
(11)
f
x
=
{
1
x
,
x
∈
[
1,2
)
,
1
+
x
,
x
∈
[
2
,
∞
)
,
g
x
=
{
1
x
2
,
x
∈
[
1,2
)
,
5
-
x
,
x
∈
[
2
,
∞
)
.
We have f
(
1
)
=
g
(
1
), f
(
2
)
=
g
(
2
), and also f
g
(
1
)
=
g
f
(
1
), but f
g
(
2
)
≠
g
f
(
2
). This implies that f and g are owc.
If we take sequencex
n
=
2
+
1
/
n
,
then
(12)
lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
3
,
lim
⁡
f
f
x
n
=
lim
⁡
f
f
(
2
+
1
n
)
=
lim
⁡
f
(
3
+
1
n
)
=
4
≠
g
(
3
)
,
lim
⁡
f
g
x
n
=
lim
⁡
f
g
(
2
+
1
n
)
=
lim
⁡
f
(
3
-
1
n
)
=
4
≠
g
(
3
)
.
This implies that f and g are not f-compatible of type (E).


## 2. Main Results

Theorem 15.
Letf and g be weak semicompatible, R weakly commuting type of A
f, and self-mappings of a complete metric space (
X
,
d
) such that (a)
f
(
X
)
⊆
g
(
X
);
(b)
d
(
f
x
,
f
y
)
≥
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
);
a
,
c
>
1
(
a
+
c
>
0
) and b
∈
R such that a
+
b
+
c
>
1;
(c)
f and g are either f-compatible of type (E) or g-compatible of type (E).If fand g are conditional commuting, then f and g have a common fixed point in X.Proof.
Letx
0 be any point in X. Since f
(
X
)
⊆
g
(
X
), there exist x
1
∈
X such that f
x
1
=
g
x
0
=
y
0. Similarly, we can have a sequence
(13)
f
x
n
+
1
=
g
x
n
=
y
n
.
Now by (b), we have
(14)
d
(
f
x
n
,
f
x
n
+
1
)
≥
a
d
(
g
x
n
,
g
x
n
+
1
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
x
n
+
1
,
g
x
n
+
1
)
,
d
(
y
n
-
1
,
y
n
)
≥
a
d
(
y
n
,
y
n
+
1
)
+
b
d
(
y
n
-
1
,
y
n
)
+
c
d
(
y
n
,
y
n
+
1
)
,
d
(
y
n
,
y
n
+
1
)
≤
1
-
b
a
+
c
d
(
y
n
-
1
,
y
n
)
.
Since (
1
-
b
)
/
(
a
+
c
)
=
k
<
1
(
a
+
b
+
c
>
1
),
(15)
d
(
y
n
,
y
n
+
1
)
≤
k
d
(
y
n
-
1
,
y
n
)
.
Similarly, we can obtain d
(
y
n
-
1
,
y
n
)
≤
k
d
(
y
n
-
2
,
y
n
-
1
). Therefore by (15) we have d
(
y
n
,
y
n
+
1
)
≤
k
2
d
(
y
n
-
2
,
y
n
-
1
). By the same way, we can have
(16)
d
(
y
n
,
y
n
+
1
)
≤
k
n
d
(
y
0
,
y
1
)
.
Now we will show that {
y
n
} is a Cauchy sequence. For any integer p
>
0, we get
(17)
d
(
y
n
,
y
n
+
p
)
≤
d
(
y
n
,
y
n
+
1
)
+
d
(
y
n
+
1
,
y
n
+
2
)
+
⋯
+
d
(
y
n
+
p
-
1
,
y
n
+
p
)
,
now
by
(
16
)
d
(
y
n
,
y
n
+
p
)
≤
k
n
d
(
y
0
,
y
1
)
+
k
n
+
1
d
(
y
0
,
y
1
)
+
⋯
+
k
n
+
p
-
1
d
(
y
0
,
y
1
)
,
d
(
y
n
,
y
n
+
p
)
≤
k
n
[
1
+
k
+
k
2
+
k
3
+
⋯
+
k
p
-
1
]
d
(
y
0
,
y
1
)
,
d
(
y
n
,
y
n
+
p
)
≤
k
n
1
-
k
d
(
y
0
,
y
1
)
.
Since k
<
1, then taking limit n
→
∞ we have d
(
y
n
,
y
n
+
p
)
→
0. Therefore {
y
n
} is a Cauchy sequence. Since X is complete, there exists a point t in X such that {
y
n
}
→
t as n
→
∞; moreover y
n
=
f
x
n
+
1
=
g
x
n
→
t for t
∈
X.Case 1 (f and g are f-compatible of type (E)).
Sincef and g are weak semicompatible, this yields either lim
⁡
f
g
x
n
=
g
t or lim
⁡
g
f
x
n
=
f
t. First we take lim
⁡
g
f
x
n
=
f
t.
Sincef and g are f-compatible of type (E), this yields lim
⁡
f
f
x
n
=
lim
⁡
f
g
x
n
=
g
t. Now by (b), we have
(18)
d
(
f
f
x
n
,
f
t
)
≥
a
d
(
g
f
x
n
,
g
t
)
+
b
d
(
f
f
x
n
,
g
f
x
n
)
+
c
d
(
f
t
,
g
t
)
,
Now limiting n
→
∞ yields
(19)
d
(
g
t
,
f
t
)
≥
a
d
(
f
t
,
g
t
)
+
b
d
(
g
t
,
f
t
)
+
c
d
(
f
t
,
g
t
)
,
Since a
+
b
+
c
>
1, this yields f
t
=
g
t.
Now the conditional commutativity off and g implies f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we have
(20)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Since a
>
1, this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.
Now we takelim
⁡
f
g
x
n
=
g
t.
Sincef and g are f-compatible of type (E), this yields lim
⁡
f
f
x
n
=
lim
⁡
f
g
x
n
=
g
t.
Alsof and g are R-weakly commuting type of A
f and this implies d
(
f
g
x
n
,
g
g
x
n
)
≤
R
d
(
f
x
n
,
g
x
n
).
Now limitingn
→
∞ yields lim
⁡
g
g
x
n
=
g
t. Again by (b), we have
(21)
d
(
f
x
n
,
f
g
x
n
)
≥
a
d
(
g
x
n
,
g
g
x
n
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
g
x
n
,
g
g
x
n
)
.
Now limitingn
→
∞ yields
(22)
d
(
t
,
g
t
)
≥
a
d
(
t
,
g
t
)
+
b
d
(
t
,
t
)
+
c
d
(
g
t
,
g
t
)
.
Sincea
>
1 implies g
t
=
t, now by (b), we get
(23)
d
(
f
x
n
,
f
t
)
≥
a
d
(
g
x
n
,
g
t
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
t
,
g
t
)
.
Now limitingn
→
∞ with c
>
1 yields f
t
=
t, and hence f
t
=
g
t
=
t and t is common fixed point of f and g.Case 2 (fand g are g-compatible of type (E)).
Sincef and g are weak semicompatible, this yields either lim
⁡
f
g
x
n
=
g
t or lim
⁡
g
f
x
n
=
f
t.
First we takelim
⁡
g
f
x
n
=
f
t; since f and g are g-compatible of type (E), this yields lim
⁡
g
g
x
n
=
lim
⁡
g
f
x
n
=
f
t. Also f and g are R-weakly commuting type of A
f and this implies d
(
f
g
x
n
,
g
g
x
n
)
≤
R
d
(
f
x
n
,
g
x
n
). Now limiting n
→
∞, we get lim
⁡
f
g
x
n
=
f
t. Now by (b), we get
(24)
d
(
f
g
x
n
,
f
t
)
≥
a
d
(
g
g
x
n
,
g
t
)
+
b
d
(
f
g
x
n
,
g
g
x
n
)
+
c
d
(
f
t
,
g
t
)
.
Now limiting n
→
∞, we get d
(
f
t
,
f
t
)
≥
a
d
(
f
t
,
g
t
)
+
b
d
(
f
t
,
f
t
)
+
c
d
(
f
t
,
g
t
), Since a
+
c
>
0, this yields f
t
=
g
t.
Now the conditional commutativity off and g implies f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we get
(25)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Since a
>
1, this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.
Now we takelim
⁡
f
g
x
n
=
g
t.
Sincef and g are g-compatible of type (E), this yields lim
⁡
g
g
x
n
=
lim
⁡
g
f
x
n
=
f
t. Now by (b) d
(
f
g
x
n
,
f
t
)
≥
a
d
(
g
g
x
n
,
g
t
)
+
b
d
(
f
g
x
n
,
g
g
x
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
g
t
,
f
t
)
≥
a
d
(
f
t
,
g
t
)
+
b
d
(
g
t
,
f
t
)
+
c
d
(
f
t
,
g
t
), since a
+
b
+
c
>
1 yields f
t
=
g
t.
Now the conditional commutativity off and g implies f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we get
(26)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Since a
>
1
,
this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.Example 16.
Letx
,
y
∈
X
(
x
≥
y
) and X
=
[
1,7
] and d be the usual metric on X. Define f
,
g
:
X
→
X as follows:
(27)
f
x
=
1
if
x
=
1
,
f
x
=
x
+
5
2
if
x
>
5
,
f
x
=
7
if
1
<
x
≤
5
,
g
x
=
1
if
x
=
1
,
g
x
=
x
if
x
≥
5
,
g
x
=
6
if
1
<
x
<
5
.
When we take constant sequence {
x
n
}
=
1, then lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
1, lim
⁡
f
g
x
n
=
1
=
g
(
1
), and lim
⁡
g
f
x
n
=
1
=
f
(
1
), but if we take sequence x
n
=
5
+
1
/
n, lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
5, and also
(28)
lim
⁡
f
g
x
n
=
lim
⁡
f
g
(
5
+
1
n
)
=
lim
⁡
f
(
5
+
1
n
)
=
lim
⁡
(
5
+
1
/
n
+
5
2
)
=
5
=
g
(
5
)
,
(29)
lim
⁡
g
f
x
n
=
lim
⁡
g
f
(
5
+
1
n
)
=
lim
⁡
g
(
5
+
1
/
n
+
5
2
)
=
lim
⁡
g
(
5
+
1
2
n
)
=
5
≠
f
(
5
)
.
Therefore maps f and g are weak semicompatible.It can easily be shown that mapsf and g satisfy conditional commutativity at x
=
1. Moreover if we take sequence x
n
=
5
+
1
/
n, lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
5.Also,lim
⁡
f
f
x
n
=
lim
⁡
f
f
(
5
+
1
/
n
)
=
lim
⁡
f
(
5
+
1
/
2
n
)
=
5
=
g
(
5
) and lim
⁡
f
g
x
n
=
lim
⁡
f
g
(
5
+
1
/
n
)
=
lim
⁡
f
(
5
+
1
/
n
)
=
5
=
g
(
5
). Hence maps f and g are f-compatible of type (E). The mappings f and g are R-weakly commuting of type A
f; that is,
(30)
d
(
f
g
x
,
g
g
x
)
≤
R
d
(
f
x
,
g
x
)
with
R
=
1
∀
x
∈
X
.Moreover the pair(
f
,
g
) satisfies equally the condition of (b) if x
,
y
=
1. Again the pair (
f
,
g
) satisfies equally the condition of (b) if x
,
y
∈
(
1,5
) and x
,
y
=
5 for a
=
2, b
=
-
3, and c
=
3.Now finally iff
x
=
(
x
+
5
)
/
2 and g
x
=
x for x
,
y
>
5, the left hand side of given inequality yields d
(
f
x
,
f
y
)
=
|
f
x
-
f
y
|
=
(
x
-
y
)
/
2 and right hand side gives
(31)
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
)
2
|
x
-
y
|
-
6
|
x
+
5
2
-
x
|
+
6
|
y
+
5
2
-
y
|
⟹
2
(
x
-
y
)
-
6
(
x
-
x
+
5
2
)
+
6
(
y
-
y
+
5
2
)
=
-
(
x
-
y
)
.
Therefore the pair (
f
,
g
) is satisfied greater than the condition of (b) when a
=
2, b
=
-
6, and c
=
6, and 1 is a common fixed points of f and g.Corollary 17.
LetX be a set and let d be the symmetric on X. Let maps f and g satisfy all the conditions of Theorem 15. Since either f and g are f-compatible of type (E) or f and g are g-compatible of type (E), then by Lemma 14 the pair (
f
,
g
) will be owc and therefore the conclusion of Theorem 15 follows from Theorem 9.Theorem 18.
Letf and g be weak semicompatible and self-mappings of a metric space (
X
,
d
) such that(a)
f
(
X
)
⊆
g
(
X
);
(b)
d
(
f
x
,
f
y
)
≥
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
);
c
>
1 and a
,
b
∈
R such that a
+
c
>
0
&
a
+
b
+
c
>
1;
(c)
f and g satisfying E.A. property.
Iff is g-absorbing or g is f-absorbing, then f and g have a common fixed point in X.Proof.
Sincef and g satisfy E.A. property, then there exists a sequence {
x
n
} in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X.Case 1 (suppose thatg is f-absorbing).
Sincef and g are weak semicompatible, this yields either lim
⁡
f
g
x
n
=
g
t or lim
⁡
g
f
x
n
=
f
t.
First we takelim
⁡
g
f
x
n
=
f
t.
Sincef
(
X
)
⊆
g
(
X
), for each sequence, {
x
n
} there exists a sequence {
y
n
} in X such that lim
⁡
f
x
n
=
lim
⁡
g
y
n. Therefore, lim
⁡
f
x
n
=
lim
⁡
g
y
n
=
t and lim
⁡
g
x
n
=
t. By using (b), d
(
f
x
n
,
f
y
n
)
≥
a
d
(
g
x
n
,
g
y
n
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
y
n
,
g
y
n
). Limiting n
→
∞ with c
>
1 yields
(32)
lim
⁡
f
y
n
=
t
,
and
so
lim
⁡
f
x
n
=
lim
⁡
g
y
n
=
t
,
lim
⁡
g
x
n
=
t
,
lim
⁡
f
y
n
=
t
.
Since g is f-absorbing, this yields d
(
f
y
n
,
f
g
y
n
)
≤
R
d
(
f
y
n
,
g
y
n
). Limiting n
→
∞ yields lim
⁡
f
g
y
n
=
t. By (32) it is easy to see that lim
⁡
f
f
x
n
=
t. Now by (b), d
(
f
x
n
,
f
f
x
n
)
≥
a
d
(
g
x
n
,
g
f
x
n
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
f
x
n
,
g
f
x
n
). Limiting n
→
∞ yields d
(
t
,
t
)
≥
a
d
(
t
,
f
t
)
+
b
d
(
t
,
t
)
+
c
d
(
t
,
f
t
). Since a
+
c
>
0, it gives f
t
=
t. Since lim
⁡
f
f
x
n
=
t, therefore lim
⁡
f
f
x
n
=
f
t.
Sincef
(
X
)
⊆
g
(
X
), then there exists a point u in X such that f
t
=
g
u. Now by (b), d
(
f
f
x
n
,
f
u
)
≥
a
d
(
g
f
x
n
,
g
u
)
+
b
d
(
f
f
x
n
,
g
f
x
n
)
+
c
d
(
f
u
,
g
u
). Limiting n
→
∞ yields
(33)
d
(
f
t
,
f
u
)
≥
a
d
(
f
t
,
g
u
)
+
b
d
(
f
t
,
f
t
)
+
c
d
(
f
u
,
g
u
)
,
d
(
g
u
,
f
u
)
≥
a
d
(
g
u
,
g
u
)
+
c
d
(
f
u
,
g
u
)
.
Since c
>
1, this yields f
u
=
g
u. Since g is f-absorbing, this yields d
(
f
u
,
f
g
u
)
≤
R
d
(
f
u
,
g
u
) implying f
g
u
=
f
u or f
f
u
=
f
u. Again by (b), we have
(34)
d
(
f
u
,
f
f
u
)
≥
a
d
(
g
u
,
g
f
u
)
+
b
d
(
f
u
,
g
u
)
+
c
d
(
f
f
u
,
g
f
u
)
,
d
(
f
u
,
f
u
)
≥
a
d
(
f
u
,
g
f
u
)
+
b
d
(
f
u
,
f
u
)
+
c
d
(
f
u
,
g
f
u
)
.
Since a
+
c
>
0, this yields g
f
u
=
f
u or g
f
u
=
f
f
u
=
f
u. Hence f
u is common fixed point of f and g.
Now we takelim
⁡
f
g
x
n
=
g
t.
Sinceg is f-absorbing, this yields d
(
f
x
n
,
f
g
x
n
)
≤
R
d
(
f
x
n
,
g
x
n
). Limiting n
→
∞ yields d
(
t
,
g
t
)
≤
R
d
(
t
,
t
), which implies g
t
=
t. Now by (b), d
(
f
x
n
,
f
t
)
≥
a
d
(
g
x
n
,
g
t
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
t
,
f
t
)
≥
a
d
(
t
,
t
)
+
b
d
(
t
,
t
)
+
c
d
(
f
t
,
t
). Since c
>
1, this yields f
t
=
t. Hence f
t
=
g
t
=
t or t is common fixed point of f and g.Case 2 (supposef is g-absorbing).
By virtue of (32) it is easy to see that lim
⁡
f
y
n
=
lim
⁡
g
y
n
=
t.
Sincef and g are weak semicompatible mappings, this yields lim
⁡
f
g
y
n
=
g
t & lim
⁡
g
f
y
n
=
f
t.
First we takelim
⁡
g
f
y
n
=
f
t.
Sincef is g-absorbing, this yields d
(
g
y
n
,
g
f
y
n
)
≤
R
d
(
f
y
n
,
g
y
n
). Limiting n
→
∞ yields lim
⁡
g
f
y
n
=
t. This implies f
t
=
t. Now by (b), d
(
f
y
n
,
f
f
y
n
)
≥
a
d
(
g
y
n
,
g
f
y
n
)
+
b
d
(
f
y
n
,
g
y
n
)
+
c
d
(
f
f
y
n
,
g
f
y
n
). Limiting n
→
∞ yields d
(
t
,
f
f
y
n
)
≥
a
d
(
t
,
t
)
+
b
d
(
t
,
t
)
+
c
d
(
f
f
y
n
,
t
). Since c
>
1, this yields lim
⁡
f
f
y
n
=
t or lim
⁡
f
f
y
n
=
f
t.
Sincef
(
X
)
⊆
g
(
X
), then there exists a point u in X such that f
t
=
g
u. Now by (b), d
(
f
f
y
n
,
f
u
)
≥
a
d
(
g
f
y
n
,
g
u
)
+
b
d
(
f
f
y
n
,
g
f
y
n
)
+
c
d
(
f
u
,
g
u
). Limiting n
→
∞ yields
(35)
d
(
f
t
,
f
u
)
≥
a
d
(
f
t
,
f
t
)
+
b
d
(
f
t
,
f
t
)
+
c
d
(
f
u
,
g
u
)
,
d
(
g
u
,
f
u
)
≥
c
d
(
f
u
,
g
u
)
.
Since c
>
1, this yields f
u
=
g
u. Since f is g-absorbing, this yields d
(
g
u
,
g
f
u
)
≤
R
d
(
f
u
,
g
u
) implying g
f
u
=
f
u. Again by (b), we have
(36)
d
(
f
u
,
f
g
u
)
≥
a
d
(
g
u
,
g
g
u
)
+
b
d
(
f
u
,
g
u
)
+
c
d
(
f
g
u
,
g
g
u
)
,
d
(
f
u
,
f
f
u
)
≥
a
d
(
f
u
,
f
u
)
+
b
d
(
f
u
,
f
u
)
+
c
d
(
f
f
u
,
f
u
)
.
Sincec
>
1, this yields f
f
u
=
f
u or g
f
u
=
f
f
u
=
f
u. Hence f
u is common fixed point of f and g.
Now we takelim
⁡
f
g
y
n
=
g
t.
Sincef is g-absorbing, this yields d
(
g
x
n
,
g
f
x
n
)
≤
R
d
(
g
x
n
,
f
x
n
) and d
(
g
y
n
,
g
f
y
n
)
≤
R
d
(
g
y
n
,
f
y
n
). Limiting n
→
∞ yields lim
⁡
g
f
x
n
=
t
&
lim
⁡
g
f
y
n
=
t. By virtue of (32)   lim
⁡
g
f
x
n
=
lim
⁡
g
g
y
n
=
t.
Now by (b),d
(
f
y
n
,
f
g
y
n
)
≥
a
d
(
g
y
n
,
g
f
y
n
)
+
b
d
(
f
y
n
,
g
y
n
)
+
c
d
(
f
g
y
n
,
g
g
y
n
). Limiting n
→
∞ yields d
(
t
,
g
t
)
≥
a
d
(
t
,
t
)
+
b
d
(
t
,
t
)
+
c
d
(
g
t
,
t
). Since c
>
1, this yields g
t
=
t.
Now by (b),d
(
f
y
n
,
f
t
)
≥
a
d
(
g
y
n
,
g
t
)
+
b
d
(
f
y
n
,
g
y
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
t
,
f
t
)
≥
a
d
(
t
,
t
)
+
b
d
(
t
,
t
)
+
c
d
(
f
t
,
t
). Since c
>
1, this yields f
t
=
t. Hence f
t
=
g
t
=
t or t is common fixed point of f and g.Example 19.
Letx
,
y
∈
X
(
x
≥
y
) and X
=
[
1,10
] and d be the usual metric on X. Define f
,
g
:
X
→
X as follows:
(37)
f
x
=
1
if
x
=
1
,
f
x
=
5
if
x
>
5
,
f
x
=
6
if
1
<
x
≤
5
,
g
x
=
1
if
x
=
1
,
g
x
=
x
if
x
>
5
,
g
x
=
5
if
1
<
x
≤
5
.
When we take constant sequence {
x
n
}
=
1, then lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
1,  lim
⁡
f
g
x
n
=
1
=
g
(
1
), and lim
⁡
g
f
x
n
=
1
=
f
(
1
). But if we take sequence x
n
=
5
+
1
/
n, lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
5, and also
(38)
lim
⁡
f
g
x
n
=
lim
⁡
f
g
(
5
+
1
n
)
=
lim
⁡
f
(
5
+
1
n
)
=
5
=
g
(
5
)
,
lim
⁡
g
f
x
n
=
lim
⁡
g
f
(
5
+
1
n
)
=
lim
⁡
g
(
5
)
=
5
≠
f
(
5
)
.Therefore mapsf and g are weak semicompatible.It can be easily seen that mapsf and g satisfy E.A. property. Moreover map g is f-absorbing at x
=
1. Map g is f-absorbing at 1
<
x
≤
5 for R
>
0 and map g is f-absorbing at x
>
5 for R
>
0. Moreover the pair (
f
,
g
) satisfies equally the condition of (b) if x
,
y
=
1 and the pair (
f
,
g
) satisfies equally the condition of (b) for x
,
y
∈
(
1,5
] when a
=
2
,
b
=
-
3, and c
=
3. Now finally if f
x
=
5 and g
x
=
x for x
,
y
>
5, the left hand side of the given inequality yields d
(
f
x
,
f
y
)
=
|
f
x
-
f
y
|
=
0 and right hand side gives
(39)
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
)
,
2
|
x
-
y
|
-
6
|
5
-
x
|
+
6
|
5
-
y
|
⟹
2
(
x
-
y
)
-
6
(
x
-
5
)
+
6
(
y
-
5
)
=
4
(
-
x
+
y
)
(
negative
or
zero
)
.Therefore the pair(
f
,
g
) is satisfied greater than or is equal to the condition of (b) when a
=
2, b
=
-
6, and c
=
6, and 1 is a common fixed points of f and g.Theorem 20.
Letf and g be weak semicompatible and self-mappings of a metric space (
X
,
d
) such that (a)
f
(
X
)
⊆
g
(
X
);
(b)
d
(
f
x
,
f
y
)
≥
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
);
a
,
c
>
1
(
a
+
c
>
0
) and b
∈
R such that a
+
b
+
c
>
1;
(c)
f and g satisfying E.A. property.
Iff and g are f-compatible or fand g are g-compatible, then f and g have a common fixed point in X.Proof.
Sincef and g satisfy E.A. property, then there exists a sequence {
x
n
} in X such that lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
t for some t in X.Case 1 (supposef and g are f-compatible).
Sincef and g are weak semicompatible, this yields either lim
⁡
f
g
x
n
=
g
t or lim
⁡
g
f
x
n
=
f
t.
First we takelim
⁡
g
f
x
n
=
f
t.
Sincef
(
X
)
⊆
g
(
X
), for each sequence, {
x
n
} there exists a sequence {
y
n
} in X such that lim
⁡
f
x
n
=
lim
⁡
g
y
n. Therefore, lim
⁡
f
x
n
=
lim
⁡
g
y
n
=
t, and lim
⁡
g
x
n
=
t. By using (b), d
(
f
x
n
,
f
y
n
)
≥
a
d
(
g
x
n
,
g
y
n
)
+
b
d
(
f
x
n
,
g
x
n
)
+
c
d
(
f
y
n
,
g
y
n
). Limiting n
→
∞ with c
>
1 yields lim
⁡
f
y
n
=
t, so
(40)
lim
⁡
f
x
n
=
lim
⁡
g
y
n
=
t
,
lim
⁡
g
x
n
=
t
,
lim
⁡
f
y
n
=
t
.
Sincef and g are f-compatible, this yields lim
⁡
d
(
f
g
y
n
,
g
g
y
n
)
=
0 or lim
⁡
f
g
y
n
=
lim
⁡
g
g
y
n. By virtue of (40)   lim
⁡
g
f
x
n
=
lim
⁡
g
g
y
n
=
f
t. Therefore, lim
⁡
f
g
y
n
=
f
t. Now by (b), d
(
f
g
y
n
,
f
t
)
≥
a
d
(
g
g
y
n
,
g
t
)
+
b
d
(
f
g
y
n
,
g
g
y
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
f
t
,
f
t
)
≥
a
d
(
f
t
,
g
t
)
+
b
d
(
f
t
,
f
t
)
+
c
d
(
f
t
,
g
t
). Since a
+
c
>
0, this yields f
t
=
g
t.
Sincef-compatibility of f and g implies commutativity at coincidence point, this gives f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we get
(41)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Since a
>
1, this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.
Now we takelim
⁡
f
g
x
n
=
g
t.
Sincef and g are f-compatible, this yields lim
⁡
d
(
f
g
x
n
,
g
g
x
n
)
=
0 or lim
⁡
f
g
x
n
=
lim
⁡
g
g
x
n. Therefore, lim
⁡
g
g
x
n
=
g
t. Now by (b), d
(
f
g
x
n
,
f
t
)
≥
a
d
(
g
g
x
n
,
g
t
)
+
b
d
(
f
g
x
n
,
g
g
x
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
g
t
,
f
t
)
≥
a
d
(
g
t
,
g
t
)
+
b
d
(
g
t
,
g
t
)
+
c
d
(
f
t
,
g
t
). Since c
>
1, this implies f
t
=
g
t.
Sincef-compatibility of f and g implies commutativity at coincidence point, this gives f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we get
(42)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Since a
>
1, this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.Case 2 (supposef and g are g-compatible).
Sincef and g are weak semicompatible, this yields either lim
⁡
f
g
x
n
=
g
t or lim
⁡
g
f
x
n
=
f
t.
First we takelim
⁡
g
f
x
n
=
f
t.
Sincef and g are g-compatible, this yields lim
⁡
d
(
g
f
x
n
,
f
f
x
n
)
=
0 or lim
⁡
g
f
x
n
=
lim
⁡
f
f
x
n. Therefore lim
⁡
f
f
x
n
=
f
t. Now by (b), d
(
f
f
x
n
,
f
t
)
≥
a
d
(
g
f
x
n
,
g
t
)
+
b
d
(
f
f
x
n
,
g
f
x
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
f
t
,
f
t
)
≥
a
d
(
f
t
,
g
t
)
+
b
d
(
f
t
,
f
t
)
+
c
d
(
f
t
,
g
t
). Since a
+
c
>
0, this implies f
t
=
g
t.
Sinceg-compatibility of f and g implies commutativity at coincidence point, this gives f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we have
(43)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Since a
>
1
,
this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.
Now we takelim
⁡
f
g
x
n
=
g
t.
Sincefand g are g-compatible, this yields
(44)
lim
⁡
d
(
g
f
x
n
,
f
f
x
n
)
=
0
or
lim
⁡
g
f
x
n
=
lim
⁡
f
f
x
n
.
Sincef and g satisfy E.A. property, this yields with (44)   lim
⁡
g
g
x
n
=
lim
⁡
f
g
x
n
=
g
t. Now by (b), d
(
f
g
x
n
,
f
t
)
≥
a
d
(
g
g
x
n
,
g
t
)
+
b
d
(
f
g
x
n
,
g
g
x
n
)
+
c
d
(
f
t
,
g
t
). Limiting n
→
∞ yields d
(
g
t
,
f
t
)
≥
a
d
(
g
t
,
g
t
)
+
b
d
(
g
t
,
g
t
)
+
c
d
(
f
t
,
g
t
). Since c
>
1, this implies f
t
=
g
t.
Sinceg-compatibility of f and g implies commutativity at coincidence point, this gives f
g
t
=
g
f
t or f
g
t
=
g
f
t
=
f
f
t
=
g
g
t. Now by (b), we have
(45)
d
(
f
f
t
,
f
t
)
≥
a
d
(
g
f
t
,
g
t
)
+
b
d
(
f
f
t
,
g
f
t
)
+
c
d
(
f
t
,
g
t
)
,
d
(
f
f
t
,
f
t
)
≥
a
d
(
f
f
t
,
f
t
)
.
Sincea
>
1, this implies f
f
t
=
f
t or f
f
t
=
g
f
t
=
f
t. Hence f
t is common fixed point of f and g.Example 21.
Letx
,
y
∈
X
(
x
≥
y
) and X
=
[
1,25
] and d be the usual metric on X. Define f
,
g
:
X
→
X as follows:
(46)
f
x
=
1
if
x
=
1
,
x
≥
5
,
f
x
=
5
if
1
<
x
<
5
,
g
x
=
1
if
x
=
1
,
g
x
=
x
+
4
if
1
<
x
<
5
,
g
x
=
x
5
if
x
≥
5
,
if we take constant sequence x
n
=
1, lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
1, lim
⁡
f
g
x
n
=
lim
⁡
f
(
1
)
=
1
=
g
(
1
), and lim
⁡
g
f
x
n
=
lim
⁡
g
(
1
)
=
1
=
f
(
1
), but when we take sequence x
n
=
5
+
1
/
n, then lim
⁡
f
x
n
=
1 and lim
⁡
g
x
n
=
lim
⁡
(
(
5
+
1
/
n
)
/
5
)
=
1 or lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
1. Moreover lim
⁡
f
g
x
n
=
lim
⁡
f
(
(
5
+
1
/
n
)
/
5
)
=
lim
⁡
f
(
1
+
1
/
5
n
)
=
5
≠
g
(
1
) and lim
⁡
g
f
x
n
=
lim
⁡
g
(
1
)
=
1
=
f
(
1
).Thereforef and g are weak semicompatible. It can be easily shown that mappings f and g satisfy E.A. property. When we take sequence x
n
=
1, it is easy to show that maps f and g are f-compatible. When we take sequence x
n
=
5
+
1
/
n, then lim
⁡
f
x
n
=
lim
⁡
f
(
5
+
1
/
n
)
=
1 and lim
⁡
g
x
n
=
lim
⁡
g
(
5
+
1
/
n
)
=
l; therefore lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
1.Nowlim
⁡
f
g
x
n
=
lim
⁡
f
g
(
5
+
1
/
n
)
=
5 and lim
⁡
g
g
x
n
=
lim
⁡
g
g
(
5
+
1
/
n
)
=
lim
⁡
g
(
1
+
1
/
5
n
)
=
5. Thus lim
⁡
f
g
x
n
=
lim
⁡
g
g
x
n. Therefore maps f and g are f-compatible. Again if we take sequence x
n
=
1
+
1
/
n, then lim
⁡
f
x
n
=
lim
⁡
f
(
1
+
1
/
n
)
=
5 and lim
⁡
g
x
n
=
lim
⁡
g
(
1
+
1
/
n
)
=
5; thus lim
⁡
f
x
n
=
lim
⁡
g
x
n
=
1. Now lim
⁡
f
g
x
n
=
lim
⁡
f
g
(
1
+
1
/
n
)
=
lim
⁡
f
(
5
+
1
/
n
)
=
1 and lim
⁡
g
g
x
n
=
g
g
(
1
+
1
/
n
)
=
g
(
5
+
1
/
n
)
=
1. Thus lim
⁡
f
g
x
n
=
lim
⁡
g
g
x
n. Therefore maps f and g are f-compatible.Moreover the pair(
f
,
g
) is satisfying equally the condition of (b) if x
,
y
=
1and if we take f
x
=
1 and g
x
=
x
/
5, for x
,
y
≥
5, left hand side gives d
(
f
x
,
f
y
)
=
|
f
x
-
f
y
|
=
0 and right hand side gives
(47)
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
)
,
4
|
x
5
-
y
5
|
-
4
|
1
-
x
5
|
+
3
|
1
-
y
5
|
⟹
4
(
x
5
-
y
5
)
-
4
(
x
5
-
1
)
+
3
(
y
5
-
1
)
=
1
-
y
5
(
negative
or
zero
)
.
Therefore the pair (
f
,
g
) is satisfied greater than or is equal to the condition of (b) when a
=
4, b
=
-
4, and c
=
3.Now finally iff
x
=
5 and g
x
=
x
+
4 for 1
<
x
<
5. The left hand side of given inequality yields d
(
f
x
,
f
y
)
=
|
f
x
-
f
y
|
=
0 and right hand side gives
(48)
a
d
(
g
x
,
g
y
)
+
b
d
(
f
x
,
g
x
)
+
c
d
(
f
y
,
g
y
)
,
4
|
(
x
-
4
)
-
(
y
-
4
)
|
-
4
|
5
-
(
x
-
4
)
|
+
3
|
5
-
(
y
-
4
)
|
⟹
4
(
x
-
y
)
-
4
(
x
-
1
)
+
3
(
y
-
1
)
=
1
-
y
(
negative
)
.
Therefore the pair (
f
,
g
) is satisfied greater than the condition of (b) when a
=
4, b
=
-
4, and c
=
3, and 1 is a common fixed points of f and g.Corollary 22.
LetX be a set and let d be the symmetric on X. Let maps f and g satisfy all the conditions of Theorem 20. Since f and g are either f-compatible or g-compatible, then, by Lemma 11, pair (
f
,
g
) will be owc and therefore the conclusion of Theorem 20 follows from Theorem 9.



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*Source: 618965-2014-04-27.xml*