Applied Mathematics E-Notes, 14(2014), 1-12 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Implicit Relation And Eldestein-Suzuki Type Fixed
Point Results In Cone Metric Spaces
Fridoun Moradlouy, Peyman Salimiz, Stojan Radenovićx
Received 15 January 2013
Abstract
In this paper, we prove Eldestein-Suzuki type …xed point results in cone metric
spaces by using implicit relation. Our results generalize, extend, unify, enrich and
complement many existing results in the literature. Examples are given showing
the validity of our results.
1
Introduction
In 1962, M. Edelstein [6] proved another version of Banach Contraction Principle.
He assumed a compact metric space (X; d) and a self-mapping T on X such that
d(T x; T y) < d(x; y) for all x; y 2 X with x 6= y, and he proved T has a unique …xed
point. In 2009, T. Suzuki [20] improved the results of Banach and Edelstein. Suzuki
replaced the condition d(T x; T y) < d(x; y) by
1
d(x; T x) < d(x; y) ) d(T x; T y) < d(x; y)
2
for all x; y 2 X. By this assumption he established T has a unique …xed point. Recently
D. Dorić et al. in [5] proved the following theorem and extended the results of Edelstein
and Suzuki on compact cone metric spaces.
THEOREM 1. Let (X; d) be a compact cone metric space over a normal and solid
cone P and let T : X ! X be given. Assume that
d(T x; T y)
Ad(x; y) + Bd(x; T x) + Cd(y; T y) + Dd(x; T y) + Ed(y; T x);
for all x; y 2 X; x 6= y and
1
d(x; T x)
2
d(x; y) 2
= intP .
Mathematics Subject Classi…cations: 47H10, 54H25, 55M20.
of Mathematics, Sahand University of Technology, Tabriz, Iran.
z Department of Mathematics, Sahand University of Technology, Tabriz, Iran.
x University of Belgrade, Faculty of Mechanical Engineering, Kraljice Marije 16, 11120, Beograd,
Serbia.
y Department
1
2
Implicit Relation and Eldestein-Suzuki Type Fixed Point Results
where A; B; C; D; E 0, A + B + C + 2D = 1 and C 6= 1. Then T has a …xed point
in X. If E B + C + D; then the …xed point of T is unique.
In 2007, Huang and Zhang [8] introduced cone metric spaces and de…ned some
properties of convergence of sequences and completeness in cone metric spaces, also
they proved a …xed point theorem of cone metric spaces. A number of authors were
attracted to these results of Huang and Zhang and stimulated to investigate the …xed
point theorems in cone metric spaces. During the recent years, cone metric spaces
and properties of these spaces have been studied by a number of authors. Also many
mathematicians have been extensively investigated …xed point theorems in cone metric
spaces (see [16–21]).
Furthermore, many authors considered implicit relation technique to investigation
of …xed point theorems in metric spaces (see [2–4, 9, 11–14, 18]).
In this paper, we introduce an implicit relation. This helps us to extend result of
D. Dorić et. al. (Theorem 3.8 of [5]).
2
Preliminaries
We begin with the following:
DEFINITION 1. Let E be a real Banach space with norm k:k and P be a subset
of E. P is called a cone if and only if the following conditions are satis…ed:
(i) P is closed, nonempty and P 6= f g,
(ii) a; b
0 and x; y 2 P implies ax + by 2 P ,
(iii) x 2 P and
x 2 P implies x = .
Let P
E be a cone, we de…ne a partial ordering
on E with respect to P by
x y if and only if y x 2 P . We write x y whenever x y and x 6= y, while x
y
will stand for y x 2 intP (interior of P ). The cone P
E is called normal if there
exists a positive real number K such that for all x; y 2 E,
x y ) kxk Kkyk.
The least positive number satisfying the above inequality is called the normal constant
of P . If K = 1, then the cone P is called monotone.
DEFINITION 2. A cone metric space is an ordered pair (X; d), where X is any set
and d : X X ! E is a mapping satisfying:
(d1)
d(x; y) for all x; y 2 X, and d(x; y) =
(d2) d(x; y) = d(y; x) for all x; y 2 X,
(d3) d(x; y)
d(x; z) + d(z; y) for all x; y; z 2 X.
if and only if x = y,
Moradlou et al.
3
Let (X; d) be a cone metric space, P be a normal cone in X with normal constant
K, x 2 X and fxn g a sequence in X. The sequence fxn g is converges to x if and only
if d(xn ; x) ! . Limit point of every sequence is unique.
It is well known that, there exists a norm k:k1 on E, equivalent with the given k:k,
such that the cone P is monotone w.r.t. k:k1 (see [1], [22]). By using this fact, from
now on, we assume that the cone P is solid and monotone. In this case, we can de…ne
a metric on X by D(x; y) = kd(x; y)k. Furthermore, it is proved that D and d give
same topology on X (see [15]).
We will use the following Lemma in the proof of the next theorem.
LEMMA 1. Let (X; d) be a cone metric space. Then
x
y ) kxk < kyk:
PROOF. According ([22], Proposition (2.2), page 20) [ (y x); y x] is neighbory x.
hood of . Hence, for su¢ ciently large n, we have n1 y 2 [ (y x); y x], i.e., ny
From this it follows that x
1 n1 y, that is kxk
1 n1 kyk < kyk.
3
Implicit Relation
In this section, we prove a theorem in the context of compact cone metric spaces over a
monotone and solid cone by using implicit relation technique. Our result extends [20,
Theorem 8] and [5, Theorem 3.8].
Let F : P 6 ! R be a function satis…es the following conditions:
(M1) u
v implies F (:; :; :; v; :; :)
(M2) F (u; v; v; u + v; u; )
F (:; :; :; u; :; :),
0 implies kuk
kvk,
(M3) F (u; v; v; u + v; u; ) < 0 implies kuk < kvk, where u
(M4) F (u; v; ; v; ; v) < 0 implies kuk < kvk, where u
and v
and v
,
.
EXAMPLE 1.
(A) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = kp1 k
kp2 k;
(B) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = 2kp1 k
kp4 k;
(C) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = 2kp1 k
(kp2 k + kp5 k);
(D) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = 5kp1 k
(kp2 k + kp3 k + kp4 k + kp5 k + kp6 k);
(E) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = 2kp1 k
maxfkp2 k; kp3 k; kp4 k; kp5 k; kp6 kg;
(F) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = 2kp1 k2 (kp2 k2 + kp5 k2 ).
It is easy to see that, (M 1) (M 4) are satis…ed for F in (A), (B), (C), (D), (E)
and (F ).
4
Implicit Relation and Eldestein-Suzuki Type Fixed Point Results
(G) F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = kp1 k (akp2 k + bkp3 k + ckp4 k + dkp5 k + ekp6 k), where
a; b; c; d and e are nonnegative numbers, a+b+2c+d = 1, d 6= 1 and e b+c+d.
Clearly (M 1) holds. Now, let
(1
d)kuk
(a + b)kvk + cku + vk = F (u; v; v; u + v; u; )
0:
Then we see that
(1
c
d)kuk
(a + b + c)kvk
F (u; v; v; u + v; u; )
0:
So by the assumption 1 c d = a + b + c, we observe that 1 c d
0 implies
a = b = c = 0. Therefore, d = 1, which is a contradiction. Hence 1 c d > 0.
Thus kuk kvk. So (M 2) is satis…ed. Similar argument shows that (M 3) is satis…ed.
Moreover, if
F (u; v; ; v; ; v) = kuk (a + c + e)kvk < 0;
then kuk < (a + c + e)kvk. So by the hypothesis, we can write
kuk < (a + c + e)kvk
(a + b + 2c + d)kvk = kvk:
Therefore, (M 4) is satis…ed.
(H)
F (p1 ; p2 ; p3 ; p4 ; p5 ; p6 )
= kp1 k
a minfkp2 k; kp3 kg + b minfkp3 k; kp4 kg
+c minfkp4 k; kp5 kg + kp6 k ;
where a; b and c are nonnegative numbers, a + b + c = 1 and c 6= 1.
Clearly (M 1) holds. If F (u; v; v; u + v; u; )
kuk
0, then
akvk + b minfkvk; ku + vkg + c minfku + vk; kukg
akvk + b minfkvk; kuk + kvkg + c minfkuk + kvk; kukg ;
which implies
(1
c)kuk
(a + b)kvk:
So, by using a + b + c = 1 and c 6= 1, we conclude that kuk kvk. This means (M 2)
is satis…ed. Similarly, we can show that (M 3) is satis…ed. Also, it is easy to see that
(M 4) is satis…ed.
THEOREM 2. Let (X; d) be a compact cone metric space and T be a self-mapping
on X. Suppose that F : P 6 ! R is a continuous function such that (M 1)-(M 3) are
satis…ed. Assume that
F d(T x; T y); d(x; y); d(x; T x); d(x; T y); d(y; T y); d(y; T x) < 0;
(1)
Moradlou et al.
5
for x; y 2 X and
1
D(x; T x) < D(x; y):
2
Then T has at least one …xed point. Moreover, if F satis…es (M 4), then T has a unique
…xed point.
PROOF. Let = inffD(x; T x) : x 2 Xg. There exists a sequence fxn g in X such
that limn!1 D(xn ; T xn ) = . By compactness of X, there exist w1 ; w2 2 X such that
limn!1 xn = w1 and limn!1 T xn = w2 . Hence
lim D(xn ; w2 ) = lim D(xn ; T xn ) = D(w1 ; w2 ) = :
n!1
n!1
Now, we show that must be equal to 0. If > 0, then there exists N 2 N such
that for all n N , 32 < D(xn ; w2 ) and D(xn ; T xn ) < 43 . Therefore for all n N ,
2
1
2 D(xn ; T xn ) < 3 < D(xn ; w2 ). Now, by (1), we have that
F d(T xn ; T w2 ); d(xn ; w2 ); d(xn ; T xn ); d(xn ; T w2 ); d(w2 ; T w2 ); d(w2 ; T xn ) < 0: (2)
By taking the limit as n ! 1 in (2), we get
F d(w2 ; T w2 ); d(w1 ; w2 ); d(w1 ; w2 ); d(w1 ; T w2 ); d(w2 ; T w2 );
0:
By triangle inequality and (M 1), we get
F d(w2 ; T w2 ); d(w1 ; w2 ); d(w1 ; w2 ); d(w1 ; w2 ) + d(w2 ; T w2 ); d(w2 ; T w2 );
so by (M 2), we have that D(w2 ; T w2 )
. Therefore, D(w2 ; T w2 ) =
that 21 D(w2 ; T w2 ) < D(w2 ; T w2 ). Now by (1), we can obtain that
0;
> 0. It follows
F d(T w2 ; T 2 w2 ); d(w2 ; T w2 ); d(w2 ; T w2 ); d(w2 ; T 2 w2 ); d(T w2 ; T 2 w2 );
< 0;
which implies that
F d(T w2 ; T 2 w2 ); d(w2 ; T w2 ); d(w2 ; T w2 );
d(T w2 ; T 2 w2 ) + d(w2 ; T w2 ); d(T w2 ; T 2 w2 );
< 0:
By (M 3), we get D(T w2 ; T 2 w2 ) < D(w2 ; T w2 ) = , which is a contradiction of the
de…nition of . So = 0, that is, w1 = w2 .
Now, we must show that T has at least one …xed point. Assume towards a contradiction that T does not have a …xed point. Hence,
0<
1
D(xn ; T xn ) < D(xn ; T xn ):
2
Then by (1), we have that
F d(T xn ; T 2 xn ); d(xn ; T xn ); d(xn ; T xn ); d(xn ; T 2 xn ); d(T xn ; T 2 xn ); d(T xn ; T xn ) < 0:
6
Implicit Relation and Eldestein-Suzuki Type Fixed Point Results
By taking the limit as n ! 1 in above inequality, we get
F
lim d(w1 ; T 2 xn ); ; ; lim d(w1 ; T 2 xn ); lim d(w1 ; T 2 xn );
n!1
n!1
n!1
It follows from (M 2) that limn!1 D(w1 ; T 2 xn )
more, by using (1) and (M 1), we obtain that
0:
0, so limn!1 T 2 xn = w1 . Further-
F d(T xn ; T 2 xn ); d(xn ; T xn ); d(xn ; T xn );
d(T xn ; T 2 xn ) + d(xn ; T xn ); d(T xn ; T 2 xn );
< 0:
Then by (M 3) we have that D(T xn ; T 2 xn ) < D(xn ; T xn ).
Now, suppose that both of the following inequalities hold for some n 2 N,
1
D(xn ; T xn )
2
D(xn ; w1 ) and
1
D(T xn ; T 2 xn )
2
D(T xn ; w1 );
then, we have that
D(xn ; T xn )
D(xn ; w1 ) + d(w1 ; T xn )
1
1
D(xn ; T xn ) + D(T xn ; T 2 xn )
2
2
1
1
<
D(xn ; T xn ) + D(xn ; T xn ) = D(xn ; T xn );
2
2
which is a contradiction. Thus, for each n 2 N, either
1
D(xn ; T xn ) < D(xn ; w1 );
2
or
1
D(T xn ; T 2 xn ) < D(T xn ; w1 );
2
holds. So by hypotheses, we conclude that one of the following inequalities holds for
all n in an in…nite subset of N:
F d(T xn ; T w1 ); d(xn ; w1 ); d(xn ; T xn ); d(xn ; T w1 ); d(w1 ; T w1 ); d(w1 ; T xn ) < 0;
or
F d(T 2 xn ; T w1 ); d(T xn ; w1 ); d(T xn ; T 2 xn ); d(T xn ; T w1 ); d(w1 ; T w1 ); d(w1 ; T 2 xn ) < 0:
If we take the limit as n ! 1 in each of these inequalities, then we have that
F d(w1 ; T w1 ); ; ; d(w1 ; T w1 ); d(w1 ; T w1 );
So (M 2) implies that D(w1 ; T w1 )
a …xed point of T .
0:
0, i.e., w1 = T w1 . Hence, we conclude that w1 is
Moradlou et al.
7
To prove the uniqueness of w1 , suppose that w0 is another …xed point of T such
that w1 6= w0 . Hence, 0 = 21 D(w1 ; T w1 ) < D(w1 ; w0 ). By (1), we have that
F d(T w1 ; T w0 ); d(w1 ; w0 ); d(w1 ; T w1 ); d(w1 ; T w0 ); d(w0 ; T w0 ); d(w0 ; T w1 ) < 0:
So
F d(w1 ; w0 ); d(w1 ; w0 ); ; d(w1 ; w0 ); ; d(w0 ; w1 ) < 0:
Considering (M 4), we have that D(w1 ; w0 ) < D(w1 ; w0 ), which is a contradiction.
Therefore w1 = w0 . Then w1 is the unique …xed point of T .
THEOREM 3. Let (X; d) be a cone metric space and let G and T be two selfmappings on X such that T X GX and GX is compact. Suppose that F : P 6 ! R
is a continuous function such that (M 1) (M 3) are satis…ed. Assume that
F d(T x; T y); d(Gx; Gy); d(Gx; T x); d(Gx; T y); d(Gy; T y); d(Gy; T x) < 0;
(3)
for all x; y 2 X and
1
D(Gx; T x) < D(Gx; Gy):
2
Then G and T have at least one point of coincidence. Moreover, if F satis…es (M 4)
and G and T are weakly compatible, then G and T have a unique common …xed point.
PROOF. De…ne H : GX ! GX by H(G(w)) = T w. Replacing T x and T y by
H(Gx) and H(Gy), respectively, in (3), we have that, for Gx; Gy 2 GX,
1
d(Gx; H(Gx)) < d(Gx; Gy);
2
which implies that, for Gx; Gy 2 GX,
F d(H(Gx); H(Gy)); d(Gx; Gy); d(Gx; H(Gx));
d(Gx; H(Gy)); d(Gy; H(Gy)); d(Gy; H(Gx)) < 0:
Since GX is compact, by Theorem 2, H has a …xed point, i.e., there exists z 2 X
such that Gz = H(Gz) = T z := u. Moreover, if F satis…es (M 4), then H has a
unique …xed point. So we conclude that z is a unique point of coincidence of G and
T . Furthermore, if G and T are weakly compatible mappings, we get GT z = T Gz, so
Gu = T u. Therefore z = u and Gz = T z = z. This yields z as the unique common
…xed point of G and T .
COROLLARY 1. Let (X; d) be a cone metric space and let G and T be two selfmappings on X such that T X GX and GX is compact. Assume that
D(T x; T y) < aD(Gx; Gy) + bD(Gx; T x) + cD(Gx; T y) + dD(Gy; T y) + eD(Gy; T x);
for all x; y 2 X and
1
D(Gx; T x) < D(Gx; Gy);
2
8
Implicit Relation and Eldestein-Suzuki Type Fixed Point Results
where a; b; c; d; e 0, a + b + 2c + d = 1 and d 6= 1. Then G and T have at least one
point of coincidence. Moreover, if e b + c + d and G and T are weakly compatible,
then G and T have a unique common …xed point.
PROOF. The proof follows from Theorem 3 and part (G) of Example 1.
COROLLARY 2. Let (X; d) be a cone metric space and let G and T be two weakly
compatible self-mappings on X such that T X
GX and GX is compact. Assume
that
D(T x; T y) <
a min fD(Gx; Gy); D(Gx; T x)g + b minfD(Gx; T x); D(Gx; T y)g
+c minfD(Gx; T y); D(Gy; T y)g + D(Gy; T x);
for all x; y 2 X and
1
D(Gx; T x) < D(Gx; Gy)
2
where a + b + c = 1 and c 6= 1. Then G and T have a unique common …xed point.
PROOF. The proof follows from Theorem 3 and part (H) of Example 1.
COROLLARY 3. Let (X; d) be a cone metric space and let G and T be two weakly
compatible self-mappings on X such that T X
GX and GX is compact. Assume
that D(T x; T y) < D(Gx; Gy) for all x; y 2 X; x 6= y, and
1
D(Gx; T x) < D(Gx; Gy):
2
Then G and T have a unique common …xed point.
PROOF. The proof follows from Theorem 3 and part (A) of Example 1.
REMARK 1. We can obtain some new results by using Theorem 3 and other
examples of F .
In the rest of this section, we assume that
mappings satisfying the following conditions:
(P1) u
v implies 'p (:; :; v; :; :)
p
:P
! P and 'p : P 5 ! P are two
'p (:; :; u; :; :) 2 P ,
(P2) k
p (u)k
k'p (v; v; u + v; u; )k implies kuk
kvk,
(P3) k
p (u)k
< k'p (v; v; u + v; u; )k implies kuk < kvk, where v 6= ,
(P4) k
p (u)k
< k'p (v; ; v; ; v)k implies kuk < kvk, where v 6= .
We de…ne Fp : P 6 ! R by
Fp (p1 ; p2 ; p3 ; p4 ; p5 ; p6 ) = k
Clearly Fp satis…es (M 1)
p (p1 )k
k'p (p2 ; p3 ; p4 ; p5 ; p6 )k:
(M 4).
EXAMPLE 2. Suppose that p; p1 ; p2 ; p3 ; p4 ; p5 2 P . Let
Moradlou et al.
9
(A)
p (p)
= p and 'p (p1 ; p2 ; p3 ; p4 ; p5 ) = p1 ;
(B)
p (p)
= 2p and 'p (p1 ; p2 ; p3 ; p4 ; p5 ) = p3 ;
(C)
p (p)
= 2p and 'p (p1 ; p2 ; p3 ; p4 ; p5 ) = p1 + p4 ;
(D)
p (p) = 5p and 'p (p1 ; p2 ; p3 ; p4 ; p5 ) = p1 + p2 + p3 + p4 + p5 .
It is easy to show that (P 1) (P 4) are satis…ed for p and 'p in (A), (B), (C)
and (D).
(E)
p (p) = p and 'p (p1 ; p2 ; p3 ; p4 ; p5 ) = ap1 + bp2 + cp3 + dp4 + ep5 , where a; b; c; d
and e are nonnegative numbers, a + b + 2c + d = 1 and d 6= 1. So, (P 1) (P 3)
are satis…ed. Moreover, if e b + c + d, then (P 4) is satis…ed.
THEOREM 4. Let (X; d) be a compact cone metric space and T be a self-mapping
on X. Suppose that p : P ! P and 'p : P 5 ! P are two continuous mappings
such that (P 1) (P 3) are satis…ed. Assume that
p (d(T x; T y))
'p d(x; y); d(x; T x); d(x; T y); d(y; T y); (y; T x) ;
(4)
for all x; y 2 X, x 6= y, and
1
d(x; T x)
2
d(x; y) 2
= intP:
Then T has at least one …xed point. Moreover, if p and 'p satisfy (P 4), then T has
a unique …xed point.
= intP . Therefore by
PROOF. Let 12 D(x; T x) < D(x; y). So 12 d(x; T x) d(x; y) 2
(4), we have that
p (d(T x; T y))
'p d(x; y); d(x; T x); d(x; T y); d(y; T y); d(y; T x) :
Thus, by Lemma 1, we get
k
p (d(T x; T y))k
< k'p d(x; y); d(x; T x); d(x; T y); d(y; T y); d(y; T x) k:
That is,
Fp d(T x; T y); d(x; y); d(x; T x); d(x; T y); d(y; T y); d(y; T x) < 0:
Clearly, Fp is continuous. Also, Fp satis…es the conditions (M 1) (M 3). Therefore,
Theorem 2 implies that T has at least one …xed point. Furthermore, if p and 'p
satisfy (P 4), then Fp satis…es (M 4) and so T has a unique …xed point.
THEOREM 5. Let (X; d) be a cone metric space and let G and T be two selfmappings on X such that T X GX and GX is compact. Suppose that p : P ! P
and 'p : P 5 ! P are two continuous mappings satisfying (P 1)–(P 3). Assume that
p (d(T x; T y))
'p d(Gx; Gy); d(Gx; T x); d(Gx; T y); d(Gy; T y); d(Gy; T x) ;
10
Implicit Relation and Eldestein-Suzuki Type Fixed Point Results
for all x; y 2 X; x 6= y, and
1
d(Gx; T x)
2
d(Gx; Gy) 2
= intP:
Then G and T have at least one point of coincidence. Moreover, if p and 'p satisfy
(P 4) and G and T are weakly compatible, then G and T have a unique common …xed
point.
COROLLARY 4. Let (X; d) be a cone metric space and let G and T be two weakly
compatible self-mappings on X such that T X
GX and GX is compact. Assume
that d(T x; T y)
d(Gx; Gy) for all x; y 2 X; x 6= y, and
1
d(Gx; T x)
2
d(Gx; Gy) 2
= intP:
Then G and T have a unique common …xed point.
COROLLARY 5. Let (X; d) be a cone metric space and let G and T be two weakly
compatible self-mappings on X such that T X
GX and GX is compact. Assume
1
that d(T x; T y)
2 d(Gx; Gy) for all x; y 2 X, x 6= y, and
1
d(Gx; T x)
2
d(Gx; Gy) 2
= intP:
Then G and T have a unique common …xed point.
COROLLARY 6. Let (X; d) be a cone metric space and let G and T be two weakly
compatible self-mappings on X such that T X
GX and GX is compact. Assume
that
1
d(T x; T y)
[d(Gx; Gy) + d(Gx; T x)]
2
for all x; y 2 X, x 6= y, and
1
d(Gx; T x)
2
d(Gx; Gy) 2
= intP :
Then G and T have a unique common …xed point.
COROLLARY 7. Let (X; d) be a cone metric space and let G and T be two weakly
compatible self-mappings on X such that T X
GX and GX is compact. Assume
that
d(T x; T y)
1
[d(Gx; Gy) + d(Gx; T x) + d(Gx; T y) + d(Gy; T y) + d(Gy; T x)];
5
for all x; y 2 X, x 6= y, and
1
d(Gx; T x)
2
d(Gx; Gy) 2
= intP :
Moradlou et al.
11
Then G and T have a unique common …xed point.
COROLLARY 8. Let (X; d) be a cone metric space and let G and T be two selfmappings on X such that T X GX and GX is compact. Assume that d(T x; T y)
M (x; y) for all x; y 2 X, x 6= y, and
1
d(Gx; T x)
2
d(Gx; Gy) 2
= intP;
where
M (x; y) = Ad(Gx; Gy) + Bd(Gx; T x) + Cd(Gx; T y) + Dd(Gy; T y) + Ed(Gy; T x);
and A; B; C; D; E
0, A + B + 2C + D = 1 and D 6= 1. Then G and T have at
least one point of coincidence. Moreover, if E B + C + D and G and T are weakly
compatible, then F and T have a unique common …xed point.
Acknowledgment. The Authors would like to thank the referee(s) for a number
of valuable suggestions regarding a previous version of this paper.
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