3. Main Results
Our first new result is the following.
Theorem 8.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
S
,
T
,
R
:
X
→
X
be three selfmappings satisfying the following condition:
(6)
G
S
x
,
T
y
,
R
z
≤
A
·
G
x
,
S
x
,
T
y
G
y
,
T
y
,
R
z
+
G
x
,
y
,
z
2
+
G
x
,
S
x
,
T
y
G
x
,
y
,
z
G
x
,
S
x
,
T
y
+
G
x
,
y
,
z
+
G
y
,
T
y
,
R
z

1
+
B
·
G
y
,
T
y
,
R
z
1
+
G
x
,
S
x
,
T
y
1
+
G
x
,
y
,
z

1
+
C
·
G
x
,
y
,
z
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
,
A
,
B
,
C
≥
0
with
0
≤
A
+
B
+
C
<
1
,
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
≠
0
. Then,
S
,
T
, and
R
have a common fixed point. Further, if
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
=
0
, then
S
,
T
, and
R
have a unique common fixed point in
X
.
Proof.
Let
x
0
be arbitrary in
X
; we define a sequence
x
n
by the following rules:
(7)
x
3
n
+
1
=
S
x
3
n
,
x
3
n
+
2
=
T
x
3
n
+
1
,
x
3
n
+
3
=
R
x
3
n
+
2
,
∀
n
∈
X
.
Now, we have to show that
x
n
is a
G
Cauchy sequence in
X
. Consider
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
≠
0
; from (6), we have
(8)
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
=
G
S
x
3
n
,
T
x
3
n
+
1
,
R
x
3
n
+
2
≤
A
·
G
x
3
n
,
S
x
3
n
,
T
x
3
n
+
1
G
x
3
n
+
1
,
T
x
3
n
+
1
,
R
x
3
n
+
2
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
2
+
G
x
3
n
,
S
x
3
n
,
T
x
3
n
+
1
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
G
x
3
n
,
S
x
3
n
,
T
x
3
n
+
1
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
+
1
,
T
x
3
n
+
1
,
R
x
3
n
+
2

1
+
B
·
G
x
3
n
+
1
,
T
x
3
n
+
1
,
R
x
3
n
+
2
1
+
G
x
3
n
,
S
x
3
n
,
T
x
3
n
+
1
1
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2

1
+
C
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
=
A
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
2
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3

1
+
B
·
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
1
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
1
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2

1
+
C
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
=
A
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3

1
+
B
·
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
1
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
1
+
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2

1
+
C
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
=
A
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
B
·
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
+
C
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
=
A
+
C
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
+
B
·
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
,
which implies that
(9)
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
≤
h
·
G
x
3
n
,
x
3
n
+
1
,
x
3
n
+
2
,
where
h
=
(
A
+
C
)
/
(
1

B
)
.
Similarly,
(10)
G
x
3
n
+
3
,
x
3
n
+
4
,
x
3
n
+
5
≤
h
·
G
x
3
n
+
2
,
x
3
n
+
3
,
x
3
n
+
4
.
Therefore, for all
n
, we have
(11)
G
x
n
+
1
,
x
n
+
2
,
x
n
+
3
≤
h
·
G
x
n
,
x
n
+
1
,
x
n
+
2
≤
⋯
≤
h
n
+
1
·
G
x
0
,
x
1
,
x
2
.
Now, for all
l
,
m
,
n
, with
l
>
m
>
n
, using rectangular inequality, the second axiom of the
G
metric, and (11), we have
(12)
G
x
n
,
x
m
,
x
l
≤
G
x
n
,
x
n
+
1
,
x
n
+
1
+
G
x
n
+
1
,
x
n
+
2
,
x
n
+
2
+
⋯
+
G
x
l

2
,
x
l

1
,
x
l
≤
G
x
n
,
x
n
+
1
,
x
n
+
2
+
G
x
n
+
1
,
x
n
+
2
,
x
n
+
3
+
⋯
+
G
x
l

2
,
x
l

1
,
x
l
≤
h
n
+
h
n
+
1
+
⋯
+
h
l

2
·
G
x
0
,
x
1
,
x
2
=
h
n
1

h
·
G
x
0
,
x
1
,
x
2
,
where
G
(
x
n
,
x
m
,
x
l
)
→
0
as
n
,
m
,
l
→
∞
.
This shows that
x
n
is a
G
Cauchy sequence. But
(
X
,
G
)
is
G
complete
G
metric space so there exists
w
in
X
such that
x
n
→
w
as
n
tends to infinity.
Now, we assume that
s
w
≠
w
. Using condition (6), we have
(13)
G
S
w
,
x
3
n
+
2
,
x
3
n
+
3
=
G
S
w
,
T
x
3
n
+
1
,
R
x
3
n
+
2
≤
A
·
G
w
,
S
w
,
T
x
3
n
+
1
G
x
3
n
+
1
,
T
x
3
n
+
1
,
R
x
3
n
+
2
+
G
w
,
x
3
n
+
1
,
x
3
n
+
2
2
+
G
w
,
S
w
,
T
x
3
n
+
1
G
w
,
x
3
n
+
1
,
x
3
n
+
2
G
w
,
S
w
,
T
x
3
n
+
1
+
G
w
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
+
1
,
T
x
3
n
+
1
,
R
x
3
n
+
2

1
+
B
·
G
x
3
n
+
1
,
T
x
3
n
+
1
,
R
x
3
n
+
2
1
+
G
w
,
S
w
,
T
x
3
n
+
1
1
+
G
w
,
x
3
n
+
1
,
x
3
n
+
2

1
+
C
·
G
w
,
x
3
n
+
1
,
x
3
n
+
2
=
A
·
G
w
,
S
w
,
x
3
n
+
2
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
+
G
w
,
x
3
n
+
1
,
x
3
n
+
2
2
+
G
w
,
S
w
,
x
3
n
+
2
G
w
,
x
3
n
+
1
,
x
3
n
+
2
G
w
,
s
w
,
x
3
n
+
2
+
G
w
,
x
3
n
+
1
,
x
3
n
+
2
+
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3

1
+
B
·
G
x
3
n
+
1
,
x
3
n
+
2
,
x
3
n
+
3
1
+
G
w
,
S
w
,
x
3
n
+
2
1
+
G
w
,
x
3
n
+
1
,
x
3
n
+
2

1
+
C
·
G
w
,
x
3
n
+
1
,
x
3
n
+
2
.
As
x
n
is
G
Cauchy sequence and converges to
w
, therefore, by taking limit
n
→
∞
, we get
G
(
S
w
,
w
,
w
)
≤
0
which is held only if
G
(
S
w
,
w
,
w
)
=
0
implies that
S
w
=
w
. Similarly, it can be shown that
T
w
=
w
and
R
w
=
w
. Hence,
w
is a common fixed point of
S
,
T
and
R
.
Uniqueness. Suppose that
S
,
T
, and
R
have two common fixed points
z
and
w
such that
z
≠
w
. Since condition
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
=
0
, we have that
G
(
z
,
S
z
,
T
w
)
+
G
(
z
,
w
,
w
)
+
G
(
z
,
T
w
,
R
w
)
=
0
implies
G
(
S
z
,
T
w
,
R
w
)
=
0
. Therefore, one can get the following:
(14)
G
S
z
,
T
w
,
R
w
=
G
z
,
w
,
w
=
0
i
m
p
l
i
e
s
t
h
a
t
z
=
w
,
which is a contradiction. Therefore, the common fixed point is unique.
Corollary 9.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
S
,
T
,
R
:
X
→
X
be three selfmappings satisfying the condition
(15)
G
S
x
,
T
y
,
R
z
≤
A
·
G
x
,
S
x
,
T
y
G
x
,
T
y
,
R
z
+
G
x
,
y
,
z
2
+
G
x
,
S
x
,
T
y
G
x
,
y
,
z
G
x
,
S
x
,
T
y
+
G
x
,
y
,
z
+
G
x
,
T
y
,
R
z

1
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
A
≥
0
with
0
≤
A
<
1
,
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
≠
0
. Then,
S
,
T
, and
R
have a common fixed point. Further, if
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
=
0
, then
S
,
T
, and
R
have a unique common fixed point in
X
.
Proof.
The proof follows by taking
B
=
C
=
0
in Theorem 8.
Corollary 10.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
S
,
T
,
R
:
X
→
X
be three selfmappings satisfying the condition
(16)
G
S
x
,
T
y
,
R
z
≤
B
·
G
y
,
T
y
,
R
z
1
+
G
x
,
S
x
,
T
y
1
+
G
x
,
y
,
z
+
C
·
G
x
,
y
,
z
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
B
,
C
≥
0
with
0
≤
B
+
C
<
1
,
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
≠
0
. Then
S
,
T
, and
R
have a common fixed point. Further, if
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
R
z
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
=
0
, then
S
,
T
, and
R
have a unique common fixed point in
X
.
Proof.
The proof follows by taking
A
=
0
in Theorem 8.
Corollary 11.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
S
,
T
:
X
→
X
be two selfmappings satisfying the condition
(17)
G
S
x
,
T
y
,
T
z
≤
A
·
G
x
,
S
x
,
T
y
G
x
,
T
y
,
T
z
+
G
x
,
y
,
z
2
+
G
x
,
S
x
,
T
y
G
x
,
y
,
z
G
x
,
S
x
,
T
y
+
G
x
,
y
,
z
+
G
x
,
T
y
,
T
z

1
+
B
·
G
y
,
T
y
,
T
z
1
+
G
x
,
S
x
,
T
y
1
+
G
x
,
y
,
z

1
+
C
·
G
x
,
y
,
z
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
A
,
B
,
C
≥
0
with
0
≤
A
+
B
+
C
<
1
,
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
T
z
)
≠
0
. Then,
S
and
T
have a common fixed point. Further, if
G
(
x
,
S
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
T
z
)
=
0
implies
G
(
S
x
,
T
y
,
T
z
)
=
0
, then
S
and
T
have a unique common fixed point in
X
.
Proof.
The proof follows by taking
R
=
T
in Theorem 8.
By setting
R
=
T
=
S
in Theorem 8, we have the following corollary.
Corollary 12.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
T
:
X
→
X
be a selfmapping satisfying the condition
(18)
G
T
x
,
T
y
,
T
z
≤
A
·
G
x
,
T
x
,
T
y
G
x
,
T
y
,
T
z
+
G
x
,
y
,
z
2
+
G
x
,
T
x
,
T
y
G
x
,
y
,
z
G
x
,
T
x
,
T
y
+
G
x
,
y
,
z
+
G
x
,
T
y
,
T
z

1
+
B
·
G
y
,
T
y
,
T
z
1
+
G
x
,
T
x
,
T
y
1
+
G
x
,
y
,
z

1
+
C
·
G
x
,
y
,
z
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
A
,
B
,
C
≥
0
with
0
≤
A
+
B
+
C
<
1
,
G
(
x
,
T
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
T
z
)
≠
0
. Then,
T
has a unique fixed point. Further, if
G
(
x
,
T
x
,
T
y
)
+
G
(
x
,
y
,
z
)
+
G
(
x
,
T
y
,
T
z
)
=
0
implies
G
(
T
x
,
T
y
,
T
z
)
=
0
, then
T
has a unique common fixed point in
X
.
The second main result in this section is the following.
Theorem 13.
Let
(
X
,
G
)
be a
G
complete
G
metric space. Let
R
,
S
,
T
,
I
,
J
,
Q
:
X
→
X
be six continuous selfmaps and let
{
S
,
I
}
,
{
T
,
J
}
, and
{
R
,
Q
}
be weakly commuting pairs of selfmapping such that
T
(
X
)
⊂
I
(
X
)
,
S
(
X
)
⊂
J
(
X
)
, and
R
(
X
)
⊂
Q
(
X
)
, satisfying the condition
(19)
G
R
x
,
S
y
,
T
z
≤
A
·
G
Q
x
,
S
x
,
I
z
G
R
x
,
S
x
,
I
x
+
G
Q
x
,
J
y
,
I
z
2
+
G
R
x
,
S
x
,
I
x
G
Q
x
,
J
y
,
I
z
G
R
x
,
S
x
,
I
x
+
G
Q
x
,
J
y
,
I
z
+
G
R
x
,
S
x
,
I
x

1
+
B
·
G
Q
x
,
J
y
,
I
z
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
A
,
B
≥
0
with
0
≤
A
+
B
<
1
,
G
(
R
x
,
S
x
,
I
x
)
+
G
(
Q
x
,
J
y
,
I
z
)
+
G
(
R
x
,
S
x
,
I
x
)
≠
0
. Then
R
,
S
,
T
,
I
,
J
,
Q
have a common fixed point. Further, if
G
(
R
x
,
S
x
,
I
x
)
+
G
(
Q
x
,
J
y
,
I
z
)
+
G
(
R
x
,
S
x
,
I
x
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
+
G
(
Q
x
,
J
y
,
I
z
)
=
0
, then
R
,
S
,
T
,
I
,
J
,
Q
have a unique common fixed point in
X
.
Proof.
Take
x
0
as arbitrary point of
X
. Since
R
(
X
)
⊂
Q
(
X
)
, we can find a point
x
1
in
X
such that
R
x
0
=
Q
x
1
. For
S
(
X
)
⊂
J
(
X
)
, we can find a point
x
2
in
X
such that
R
x
1
=
Q
x
2
and for
T
(
X
)
⊂
I
(
X
)
we can find a point
x
3
in
X
such that
T
x
2
=
I
x
3
. Generally, for a point
x
3
n
, choose
x
3
n
+
1
such that
R
x
3
n
=
Q
x
3
n
+
1
; for a point
x
3
n
+
1
, choose
x
3
n
+
2
such that
S
x
3
n
+
1
=
J
x
3
n
+
2
; and for a point
x
3
n
+
2
, choose
x
3
n
+
3
such that
T
x
3
n
+
2
=
I
x
3
n
+
3
for
n
=
0,1
,
2,3
,
…
.
Suppose
G
3
n
=
G
(
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
)
≠
0
and
G
3
n
+
1
=
G
(
R
x
3
n
+
1
,
S
x
3
n
+
2
,
T
x
3
n
+
3
)
≠
0
. Then, from condition (19), we have
(20)
G
3
n
+
1
=
G
R
x
3
n
+
1
,
S
x
3
n
+
2
,
T
x
3
n
+
3
≤
A
·
G
Q
x
3
n
+
1
,
S
x
3
n
+
1
,
I
x
3
n
+
3
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
I
x
3
n
+
1
+
G
Q
x
3
n
+
1
,
J
x
3
n
+
2
,
I
x
3
n
+
3
2
+
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
I
x
3
n
+
1
G
Q
x
3
n
+
1
,
J
x
3
n
+
2
,
I
x
3
n
+
3
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
I
x
3
n
+
1
+
G
Q
x
3
n
+
1
,
J
x
3
n
+
2
,
I
x
3
n
+
3
+
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
I
x
3
n
+
1

1
+
B
·
G
Q
x
3
n
+
1
,
J
x
3
n
+
2
,
I
x
3
n
+
3
=
A
·
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
T
x
3
n
+
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
2
+
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
T
x
3
n
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
T
x
3
n
+
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
+
G
R
x
3
n
+
1
,
S
x
3
n
+
1
,
T
x
3
n

1
+
B
·
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
=
A
·
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
+
B
·
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
=
A
+
B
·
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
.
Hence,
(21)
G
R
x
3
n
+
1
,
S
x
3
n
+
2
,
T
x
3
n
+
3
≤
A
+
B
·
G
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
,
G
3
n
+
1
≤
h
·
G
3
n
,
where
h
=
A
+
B
. Continuing this procedure, in the end we get
(22)
G
3
n
+
1
≤
h
·
G
3
n
≤
h
2
·
G
3
n

1
≤
h
3
·
G
3
n

2
≤
h
4
·
G
3
n

3
≤
⋯
≤
h
3
n
+
1
·
G
0
.
Clearly,
G
3
n
+
1
→
0
as
n
→
∞
. So,
G
(
R
x
3
n
,
S
x
3
n
+
1
,
T
x
3
n
+
2
)
→
0
; we get the following sequence:
(23)
R
x
0
,
S
x
1
,
T
x
2
,
R
x
3
,
S
x
4
,
T
x
5
,
R
x
6
,
S
x
7
,
T
x
8
,
…
,
R
x
3
n
+
1
,
S
x
3
n
+
2
,
T
x
3
n
+
3
,
…
,
which is a Cauchy sequence in
G
complete
G
metric space and therefore converges to a limit point
w
. But all subsequences of a convergent sequence converge; so, we have
(24)
l
i
m
n
→
∞
R
x
3
n
=
l
i
m
n
→
∞
Q
x
3
n
+
1
=
w
,
l
i
m
n
→
∞
S
x
3
n
=
l
i
m
n
→
∞
J
x
3
n
+
1
=
w
,
l
i
m
n
→
∞
T
x
3
n

1
=
l
i
m
n
→
∞
I
x
3
n
=
w
.
Since
{
S
,
I
}
are weakly commuting mappings, thus we have
(25)
G
S
I
x
3
n
,
I
S
x
3
n
,
I
S
x
3
n
≤
G
I
x
3
n
,
S
x
3
n
,
S
x
3
n
.
Taking limit
n
→
∞
and noting that
S
and
I
are continuous mappings, we have
(26)
G
S
w
,
I
w
,
I
w
≤
G
w
,
w
,
w
,
which gives the notion that
S
w
=
I
w
. Analogously, we can get
T
w
=
J
w
and
R
w
=
Q
w
. We claim that
R
w
≠
S
w
and
S
w
≠
T
w
and then from condition (3)
(27)
G
R
w
,
S
w
,
T
w
≤
A
·
G
R
w
,
S
w
,
T
w
G
R
w
,
S
w
,
S
w
+
G
R
w
,
T
w
,
S
w
2
+
G
R
w
,
S
w
,
S
w
G
R
w
,
T
w
,
S
w
G
R
w
,
S
w
,
S
w
+
G
R
w
,
T
w
,
S
w
+
G
R
w
,
S
w
,
S
w

1
+
B
·
G
R
w
,
T
w
,
S
w
,
G
R
w
,
S
w
,
T
w
≤
A
+
B
G
R
w
,
T
w
,
S
w
,
which is a contraction:
(28)
G
R
w
,
S
w
,
T
w
=
0
i
m
p
l
i
e
s
R
w
=
S
w
=
T
w
.
Similarly, using similar arguments to those given above, we obtain a contradiction for
R
w
≠
S
w
and
S
w
=
T
w
or for
R
w
=
S
w
and
S
w
≠
T
w
. Hence, in all the cases, we conclude that
R
w
=
S
w
=
T
w
. We prove that any fixed point of
R
is a fixed point of
S
,
T
,
Q
,
I
, and
J
. Assume that
w
∈
X
is such that
R
w
=
w
. Now, we prove that
w
=
T
w
=
S
w
. If it is not the case, then, for
w
≠
S
w
and
w
≠
T
w
, we get
(29)
G
w
,
S
w
,
T
w
=
G
R
w
,
S
w
,
T
w
≤
A
·
G
R
w
,
S
w
,
T
w
G
R
w
,
S
w
,
S
w
+
G
R
w
,
T
w
,
S
w
2
+
G
R
w
,
S
w
,
S
w
G
R
w
,
T
w
,
S
w
G
R
w
,
S
w
,
S
w
+
G
R
w
,
T
w
,
S
w
+
G
R
w
,
S
w
,
S
w

1
+
B
·
G
R
w
,
T
w
,
S
w
,
G
w
,
S
w
,
T
w
≤
A
+
B
G
w
,
T
w
,
S
w
,
where
G
(
w
,
S
w
,
T
w
)
=
0
which implies that
w
=
S
w
=
T
w
; in a similar argument, we can prove the other cases.
Uniqueness. Suppose that
S
,
T
,
R
,
I
,
J
, and
Q
have two common fixed points
z
and
w
such that
z
≠
w
. Since condition
G
(
R
x
,
S
x
,
I
x
)
+
G
(
Q
x
,
J
y
,
I
z
)
+
G
(
R
x
,
S
x
,
I
x
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
+
G
(
Q
x
,
J
y
,
I
z
)
=
0
, we have that
G
(
R
z
,
S
z
,
I
z
)
+
G
(
Q
z
,
J
z
,
I
w
)
+
G
(
R
z
,
S
z
,
I
z
)
=
0
implies
G
(
S
z
,
T
z
,
R
w
)
+
G
(
Q
z
,
J
z
,
I
w
)
=
0
, which can be written as
G
(
S
z
,
T
z
,
R
w
)
=
0
or
G
(
Q
z
,
J
z
,
I
w
)
=
0
.
Therefore, one can get the following:
(30)
G
z
,
z
,
w
=
0
o
r
G
z
,
z
,
w
=
0
i
m
p
l
i
e
s
t
h
a
t
z
=
w
.
Theorem 13 produces the following corollaries.
Corollary 14.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
R
,
S
,
T
,
I
,
J
,
Q
:
X
→
X
be three selfmaps and let
{
S
,
I
}
,
{
T
,
J
}
, and
{
R
,
Q
}
be weakly commuting pairs of selfmapping such that
T
(
X
)
⊂
I
(
X
)
,
S
(
X
)
⊂
J
(
X
)
, and
R
(
X
)
⊂
Q
(
X
)
, satisfying
(31)
G
R
x
,
S
y
,
T
z
≤
B
·
G
Q
x
,
J
y
,
I
z
for all
x
,
y
,
z
in
X
with
x
≠
y
≠
z
≠
x
with
0
≤
B
<
1
. Then,
R
,
S
,
T
,
I
,
J
, and
Q
have a unique common fixed point in
X
.
Proof.
It follows by taking
A
=
0
in Theorem 13.
Corollary 15.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
R
,
S
,
T
,
I
,
J
,
Q
:
X
→
X
be three selfmaps and let
{
S
,
I
}
,
{
T
,
J
}
, and
{
R
,
Q
}
be weakly commuting pairs of selfmapping such that
T
(
X
)
⊂
I
(
X
)
,
S
(
X
)
⊂
J
(
X
)
, and
R
(
X
)
⊂
Q
(
X
)
, satisfying
(32)
G
R
x
,
S
y
,
T
z
≤
A
·
G
Q
x
,
S
x
,
I
z
G
R
x
,
S
x
,
I
x
+
G
Q
x
,
J
y
,
I
z
2
+
G
R
x
,
S
x
,
I
x
G
Q
x
,
J
y
,
I
z
G
R
x
,
S
x
,
I
x
+
G
Q
x
,
J
y
,
I
z
+
G
R
x
,
S
x
,
I
x

1
+
B
·
G
R
z
,
T
z
,
S
z
for all
x
,
y
,
z
in
X
with
x
≠
y
≠
z
≠
x
A
≥
0
with
0
≤
A
<
1
,
G
(
R
x
,
S
x
,
I
x
)
+
G
(
Q
x
,
J
y
,
I
z
)
+
G
(
R
x
,
S
x
,
I
x
)
≠
0
. Then,
R
,
S
,
T
,
I
,
J
, and
Q
have a common fixed point. Further, if
G
(
R
x
,
S
x
,
I
x
)
+
G
(
Q
x
,
J
y
,
I
z
)
+
G
(
R
x
,
S
x
,
I
x
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
+
G
(
Q
x
,
J
y
,
I
z
)
=
0
, then
R
,
S
,
T
,
I
,
J
, and
Q
have a unique common fixed point in
X
.
Proof.
It follows by taking
B
=
0
in Theorem 13.
Corollary 16.
Let
(
X
,
G
)
be a
G
complete
G
metric space and let
T
,
R
,
I
,
J
:
X
→
X
be three selfmaps and let
{
T
,
I
}
,
{
T
,
J
}
, and
{
R
,
I
}
be weakly commuting pairs of selfmapping such that
T
(
X
)
⊂
I
(
X
)
,
T
(
X
)
⊂
J
(
X
)
, and
R
(
X
)
⊂
I
(
X
)
, satisfying
(33)
G
R
x
,
T
y
,
T
z
≤
A
·
G
I
x
,
T
x
,
I
z
G
R
x
,
T
x
,
I
x
+
G
I
x
,
J
y
,
I
z
2
+
G
R
x
,
T
x
,
I
x
G
I
x
,
J
y
,
I
z
G
R
x
,
T
x
,
I
x
+
G
I
x
,
J
y
,
I
z
+
G
R
x
,
T
x
,
I
x

1
+
B
·
G
I
x
,
J
y
,
I
z
for all
x
,
y
,
z
∈
X
with
x
≠
y
≠
z
≠
x
A
,
B
≥
0
with
0
≤
A
+
B
<
1
,
G
(
R
x
,
T
x
,
I
x
)
+
G
(
I
x
,
J
y
,
I
z
)
+
G
(
R
x
,
T
x
,
I
x
)
≠
0
. Then,
T
,
R
,
I
, and
J
have a common fixed point. Further, if
G
(
R
x
,
T
x
,
I
x
)
+
G
(
I
x
,
J
y
,
I
z
)
+
G
(
R
x
,
T
x
,
I
x
)
=
0
implies
G
(
S
x
,
T
y
,
R
z
)
+
G
(
I
x
,
J
y
,
I
z
)
=
0
, then
T
,
R
,
I
, and
J
have a unique common fixed point in
X
.
Proof.
The proof follows by setting
S
=
T
and
I
=
Q
in Theorem 13.