International Journal of Pure and Applied Mathematics
Volume 102 No. 3 2015, 547-561
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)
url: http://www.ijpam.eu
doi: http://dx.doi.org/10.12732/ijpam.v102i3.12
AP
ijpam.eu
COMMON FIXED POINTS FOR WEAKLY COMPATIBLE
MAPPINGS SATISFYING IMPLICIT FUNCTIONS
IN MULTIPLICATIVE METRIC SPACES
Chahn Yong Jung1 , Parveen Kumar2 , Sanjay Kumar3 , Shin Min Kang4 §
1 Department
of Business Administration
Gyeongsang National University
Jinju 660-701, KOREA
2,3 Department of Mathematics
Deenbandhu Chhotu Ram University of Science and Technology
Murthal, Sonipat 131039, Haryana, INDIA
4 Department of Mathematics and RINS
Gyeongsang National University
Jinju 660-701, KOREA
Abstract: In this paper we prove common fixed point theorems for weakly
compatible mapping in multiplicative metric spaces. Next, we prove common
fixed point theorems for weakly compatible mappings along with E.A and common limit range properties.
AMS Subject Classification: 47H10, 54H25
Key Words: multiplicative metric space, weakly compatible mapping, implicit function, E.A and common limit range properties
1. Introduction and Preliminaries
It is well know that the set of positive real numbers R+ is not complete according
Received:
May 21, 2015
§ Correspondence
author
c 2015 Academic Publications, Ltd.
url: www.acadpubl.eu
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
to the usual metric. To overcome this problem, in 2008, Bashirov et al. [4]
introduced the concept of multiplicative metric spaces as follows:
Definition 1.1. Let X be a nonempty set. A multiplicative metric is a
mapping d : X × X → R+ satisfying the following conditions:
(i) d(x, y) ≥ 1 for all x, y ∈ X and d(x, y) = 1 if and only if x = y;
(ii) d(x, y) = d(y, x) for all x, y ∈ X;
(iii) d(x, y) ≤ d(x, z) · d(z, y) for all x, y, z ∈ X (multiplicative triangle
inequality).
Example 1.2. ([11]) Let Rn+ be the collection of all n-tuples of positive
real numbers. Let d : Rn+ × Rn+ → R be defind as follows:
x1 x2 xn d(x, y) = · · · · ,
y1
y2
yn
where x = (x1 , . . . , xn ), y = (y1 , . . . , yn ) ∈ Rn+ and | · | : R+ → R+ is defined by
(
a if a ≥ 1;
|a| = 1
if a < 1.
a
Then it is obvious that all conditions of a multiplicative metric are satisfied.
Example 1.3. ([15]) Let d : R × R → [1, ∞) be defined as d(x, y) = a|x−y| ,
where x, y ∈ R and a > 1. Then d is a multiplicative metric.
Remark 1.4. We note that the Example 1.2 is valid for positive real
numbers and Example 1.3 is valid for all real numbers.
One can refer to [6, 11] for detailed multiplicative metric topology.
Definition 1.5. Let (X, d) be a multiplicative metric space. Then a
sequence {xn } in X said to be
(1) a multiplicative convergent to x if for every multiplicative open ball
Bǫ (x) = {y | d(x, y) < ǫ}, ǫ > 1, there exists a natural number N such that
n ≥ N, then xn ∈ Bǫ (x), that is, d(xn , x) → 1 as n → ∞.
(2) a multiplicative Cauchy sequence if for all ǫ > 1, there exists N ∈ N such
that d(xn , xm ) < ǫ for all m, n > N , that is, d(xn , xm ) → 1 as n → ∞.
(3) We call a multiplicative metric space complete if every multiplicative
Cauchy sequence in it is multiplicative convergent to x ∈ X.
COMMON FIXED POINTS FOR WEAKLY COMPATIBLE...
549
Remark 1.6. The set of positive real numbers R+ is not complete according to the usual metric. Let X = R+ and the sequence {xn } = { n1 }. It is
obvious {xn } is a Cauchy sequence in X with respect to usual metric and X is
not a complete metric space, since 0 ∈
/ R+ . In case of a multiplicative metric
1
n
space, we take a sequence {xn } = {a }, where a > 1. Then {xn } is a Cauchy
sequence since for n ≥ m,
1 xn a n 1 − 1 =
= a n m d(xn , xm ) = xm a m1 1
1
1
≤ a m − n < a m < ǫ if m >
log a
,
log ǫ
(
a
if a ≥ 1,
Also, {xn } → 1 as n → ∞ and 1 ∈ R+ . Hence (X, d)
if a < 1.
is a complete multiplicative metric space.
where |a| =
1
a
In 2012, Özavsar and Çevikel [11] gave the concept of multiplicative contraction mappings and proved some fixed point theorem of such mappings in a
multiplicative metric space.
Definition 1.7. Let f be a mapping of a multiplicative metric space (X, d)
into itself. Then f is said to be a multiplicative contraction if there exists a real
constant λ ∈ [0, 1) such that
d(f x, f y) ≤ dλ (x, y)
for all x, y ∈ X.
Gu et. al. [5] introduced the notion of commutative and weak commutative
mappings in a multiplicative metric space and proved some fixed point theorems
for these mappings.
Definition 1.8. Let f and g be two mappings of a multiplicative metric
space (X, d) into itself. Then f and g are said to be
(1) commutative mappings if f gx = gf x for all x ∈ X.
(2) weak commutative mappings if d(f gx, gf x) ≤ d(f x, gx) for all x ∈ X.
Notice that commuting mappings are obviously weakly commuting. However, the converse need not be true.
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
Example 1.9. Let X = [0, 1] be a multiplicative metric d on X defined
by d(x, y) = a|x−y| , where for all x, y ∈ X and a > 1. Define mappings f and
x
and gx = x3 for all x ∈ X. For any x ∈ X,
g : X → X by f x = 3−x
d(f gx, gf x) = a
2x2
(9−x)(9−3x)
≤a
x2
9−3x
= d(f x, gx).
Then f and g are weakly commuting but f and g are not commuting since
f gx =
x
x
<
= gf x
9−x
9 − 3x
for any non-zero x ∈ X.
In metric spaces, they introduced the notions of weak compatibility [9, 10],
E.A. property [1] and common limit range property [8, 17].
Now, we introduce the notions in multiplicative metric spaces
Definition 1.10. Let f and g be two mappings of a multiplicative metric
space (X, d) into itself. Then f and g are said to be weakly compatible if they
commute at coincidence points, that is, if f t = gt for some t ∈ X implies that
f gt = gf t.
Notice that weakly commuting mappings are obviously weakly compatible.
However, the converse need not be true.
Example 1.11. Let X = [0, ∞) be a multiplicative metric d on X defined
by d(x, y) = a|x−y| , where for all x, y ∈ X and a > 1. Define mappings f and
g : X → X by f x = x2 and gx = 2x2 for all x ∈ X. So we have f gx = 4x4 and
gf x = 2x4 for all x ∈ X. For any x ∈ X,
4
2
d(f gx, gf x) = a2x ax = d(f x, gx).
Then f and g are not weakly commuting but f and g are weakly compatible
since f 0 = g0 for some 0 ∈ X implies f g0 = gf 0.
Definition 1.12. Let f and g be two mappings of a multiplicative metric
space (X, d) into itself. Then f and g are said to satisfy E.A property if there
exists a sequence {xn } in X such that lim f xn = lim gxn = t for some t ∈ X.
n→∞
n→∞
COMMON FIXED POINTS FOR WEAKLY COMPATIBLE...
551
Example 1.13. Let X = [1, ∞). Define d : X 2 → R+ by d(x, y) = a|x−y| ,
a > 1. Then (X, d) be a multiplicative metric space. Define f, g : X → X as
f x = x1 and gx = x12 . Consider the sequence {xn } = {a1/n }, a > 1 in X. Now
lim f xn = lim gxn = 1 ∈ X. Hence f and g satisfies E.A. property.
n→∞
n→∞
Definition 1.14. Let f and g be two mappings of a multiplicative metric
space (X, d) into itself. Then f and g are said to satisfy CLRg property (common limit range of g property) if there exists a sequence {xn } in X such that
lim f xn = lim gxn = gt for some t ∈ X.
n→∞
n→∞
Example 1.15. Let X = [1, ∞). Define d : X 2 → R+ by d(x, y) = a| x−y|,
a > 1. Then (X, d) be a multiplicative metric space. Define f, g : X → X as
f x = x2 and gx = x3 . Consider the sequence {xn } = {a1/n }, a > 1 in X.
Now lim f xn = lim gxn = 1 = g1 and 1 ∈ X. Hence f and g satisfies CLRg
n→∞
n→∞
property.
Definition 1.16. Let f, g and h, k be mappings of a multiplicative metric
space (X, d) into itself. Then the pairs f, g and h, k are said to share common
limit in the range of g property if there exist sequences {xn } and {yn } in X
such that lim f xn = lim gxn = lim hyn = lim kyn = gt for some t ∈ X.
n→∞
n→∞
n→∞
n→∞
2. Main Results
Recently, Popa [12] used the implicit function rather than contraction conditions
to prove fixed point theorems in metric spaces. The strength of implicit relation
unifies several contraction conditions at the same time. This fact is seen from
examples furnished in Popa [12]. Implicit relations on metric spaces have been
used by many authors (for details see [2, 3, 7, 13, 14, 16, 18] and their references
therein).
In this section, we define a suitable class of the implicit function involving
five real non-negative arguments as follows:
Let Φ denote the family of functions such that φ : R5+ → R+ is continuous
and increasing in each coordinate variable and φ(t, 1, 1, t, t) ≤ t, φ(1, t, 1, t, 1) ≤
t, φ(1, 1, t, 1, t) ≤ t, φ(t1 , t1 , t, 1, t1 t) ≤ t1 t, φ(t1 , t, t1 , t1 t, 1) ≤ t1 t for every
t, t1 ∈ R+ (t, t1 ≥ 1).
Obviously φ(1, 1, 1, 1, 1) = 1.
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
There exists many functions φ which belongs to Φ:
Example 2.1. Let φ : R5+ → R+ be defined by
φ(t1 , t2 , t3 , t4 , t5 ) = t1 + t2 + t3 − t4 − t5 ,
then φ ∈ Φ.
Example 2.2. Let φ : R5+ → R+ be defined by
φ(t1 , t2 , t3 , t4 , t5 ) = t1 + 1 − max{t2 , t3 , t4 , t5 },
then φ ∈ Φ.
Example 2.3. Let φ : R5+ → R+ be defined by
φ(t1 , t2 , t3 , t4 , t5 ) = max{t1 , t2 , t3 , t4 , t5 },
then φ ∈ Φ.
Example 2.4. Let φ : R5+ → R+ be defined by
φ(t1 , t2 , t3 , t4 , t5 ) = [max{t1 , t2 , t3 , t4 , t5 }]1/2 ,
then φ ∈ Φ.
Now we prove the following theorems for weakly compatible mappings satisfying the implicit function in a multiplicative metric space as follow:
Theorem 2.5. Let A, B, S and T be mappings of a multiplicative metric
space (X, d) into itself satisfying
(C1)
(C2)
SX ⊂ BX,
T X ⊂ AX;
d(Sx, T y) ≤ [φ d(Ax, By), d(Ax, Sx),
d(By, T y), d(Sx, By), d(Ax, T y)}]λ
for all x, y ∈ X, where λ ∈ (0, 12 ) and φ ∈ Φ;
(C3) Assume that the pairs A, S and B, T are weakly compatible;
(C4) One of the subspace AX or BX or SX or T X is complete.
Then A, B, S and T have a unique common fixed point.
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553
Proof. Let x0 ∈ X be an arbitrary point. Since SX ⊂ BX, there exists
x1 ∈ X such that Sx0 = Bx1 = y0 . Now for this x1 there exists x2 ∈ X such
that T x1 = Ax2 = y1 . Similarly, we can inductively define a sequence {yn }
such that
Sx2n = Bx2n+1 = y2n ,
T x2n+1 = Ax2n+2 = y2n+1 .
From (C2), we have
d(y2n , y2n+1 )
= d(Sx2n , T x2n+1 )
≤ [φ d(Ax2n , Bx2n+1 ), d(Ax2n , Sx2n ),
d(Bx2n+1 , T x2n+1 ), d(Sx2n , Bx2n+1 ), d(Ax2n , T x2n+1 )}]λ
= [φ d(y2n−1 , y2n ), d(y2n−1 , y2n ),
d(y2n , y2n+1 ), d(y2n , y2n ), d(y2n−1 , y2n+1 )}]λ
≤ [φ d(y2n−1 , y2n ), d(y2n−1 , y2n ),
d(y2n , y2n+1 ), 1, d(y2n−1 , y2n ) · d(y2n , y2n+1 )}]λ
≤ dλ (y2n−1 , y2n ) · dλ (y2n , y2n+1 ).
λ
This implies that d(y2n , y2n+1 ) ≤ d 1−λ (y2n−1 , y2n ). On putting
λ
1−λ
= h.
d(y2n , y2n+1 ) ≤ dh (y2n−1 , y2n ).
Similarly we obtain
d(y2n+1 , y2n+2 ) ≤ dh (y2n , y2n+1 ).
Hence
2
n
d(yn , yn+1 ) ≤ dh (yn−1 , yn ) ≤ dh (yn−2 , yn−1 ) ≤ · · · ≤ dh (y0 , y1 )
for all n ≥ 2. Let m, n ∈ N such that m ≥ n. Then we get
d(ym , yn ) ≤ d(ym , ym−1 ) · d(ym−1 , ym−2 ) · · · d(yn+1 , yn )
≤ dh
m−1
(y1 , y0 ) · dh
m−2
n
(y1 , y0 ) · · · dh (y1 , y0 )
hn
≤ d 1−h (y1 , y0 ).
Letting limit as m, n → ∞, we have d(ym , yn ) → 1. Therefore {yn } is a multiplicative Cauchy sequence.
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
Now, suppose that AX is complete there exists u ∈ AX such that
y2n+1 = T x2n+1 = Ax2n+2 → u
as n → ∞.
Consequently, we can find v ∈ X such that Av = u. Further a multiplicative
Cauchy sequence {yn } has a convergent subsequence {y2n+1 }, therefore, the
sequence {yn } converges and hence a subsequence {y2n } also converges. Thus
we have
y2n = Sx2n = Bx2n+1 → u as n → ∞.
We claim Sv = u. Putting x = v and y = x2n+1 in (C2), we get
d(Sv, y2n+1 ) = d(Sv, T x2n+1 )
≤ [φ d(Av, Bx2n+1 ), d(Av, Sv),
d(Bx2n+1 , T x2n+1 ), d(Sv, Bx2n+1 ), d(Av, T x2n+1 ) ]λ .
Letting n → ∞, we have
d(Sv, u) ≤ [φ 1, d(u, Sv), 1, d(Sv, u), 1 ]λ ≤ dλ (Sv, u),
this implies that d(Sv, u) = 1 and hence u = Sv. Since u = Sv ∈ SX ⊂ BX
there exists w ∈ X such that u = Bw.
We claim T w = u. Putting x = v and y = w in (C2), we have
d(u, T w) = d(Sv, T w)
≤ [φ d(Av, Bw), d(Av, Sv),
d(Bw, T w), d(Sv, Bw), d(Av, T w) ]λ
= [φ 1, 1, d(u, T w), 1, d(u, T w) ]λ
≤ dλ (u, T w),
this implies that d(u, T w) = 1 and hence u = T w. Hence we get u = Av = Sv,
that is, v is a coincidence point of A and S and u = Bw = T w, that is w is a
coincidence point of B and T. Therefore Av = Sv = Bw = T w = u.
Since the pairs A, S and B, T are weakly compatible, we have
Su = S(Av) = A(Sv) = Au = w1 (say)
and
T u = T (Bw) = B(T w) = Bu = w2 (say).
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555
From (C2), we have
d(w1 , w2 ) = d(Su, T u)
≤ [φ d(Au, Bu), d(Au, Su),
d(Bu, T u), d(Su, Bu), d(Au, T u) ]λ
= [φ d(w1 , w2 ), 1, 1, d(w1 , w2 ), d(w1 , w2 ) ]λ
≤ dλ (w1 , w2 ),
this implies that w1 = w2 . and hence we have Su = Au = T u = Bu.
Again using (C2), we have
d(Sv, T u) ≤ [φ d(Av, Bu), d(Av, Sv),
d(Bu, T u), d(Sv, Bu), d(Av, T u) ]λ
= [φ d(Sv, T u), 1, 1, d(Sv, T u), d(Sv, T u) ]λ
≤ dλ (Sv, T u),
this implies that Sv = T u (u = T u) and hence we have u = Su = Au = T u =
Bu. Therefore u is a common fixed point of A, B, S and T.
Similarly, we can complete the proof for cases in which BX or T X or SX
is complete.
The uniqueness can be easily follows from (C2). This completes the proof.
In Theorem 2.3. if we put S = T , then we obtains the following corollary.
Corollary 2.6. Let A, B and S be mappings of a multiplicative metric
space (X, d) into itself satisfying
(c1)
(c2)
SX ⊂ BX,
SX ⊂ AX;
d(Sx, Sy) ≤ [φ d(Ax, By), d(Ax, Sx),
d(By, Sy), d(Sx, By), d(Ax, Sy)}]λ
for all x, y ∈ X, where λ ∈ (0, 21 ) and φ ∈ Φ;
(c3) the pairs A, S and B, S are weakly compatible;
(c4) one of the subspace AX or BX or SX is complete.
Then A, B and S have a unique common fixed point.
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
In Theorem 2.3, if we put A = B = I, then we obtains the corollary.
Corollary 2.7. Let S and T be mappings of a multiplicative metric space
(X, d) into itself satisfying
(c5)
d(Sx, T y) ≤ [φ d(x, y), d(x, Sx),
d(y, T y), d(Sx, y), d(x, T y) ]λ
for all x, y ∈ X, where λ ∈ (0, 12 ) and φ ∈ Φ;
(c6) one of the subspace SX or T X is complete.
Then S and T have a unique common fixed point.
Next we prove the following theorems for weakly compatible mappings with
E.A. property satisfying the implicit function in a multiplicative metric space
as follow:
Theorem 2.8. Let A, B, S and T be mappings of a multiplicative metric space (X, d) into itself satisfying the conditions (C1), (C2), (C3) and the
following conditions:
(C5) One of the subspace AX or BX or SX or T X is closed subset of X;
(C6) the pairs A, S and B, T satisfies the E.A. property.
Then A, B, S and T have a unique common fixed point.
Proof. Suppose that the pair A, S satisfies the E.A property. Then there
exists a sequence {xn } in X such that lim Axn = lim Sxn = z for some z ∈ X.
n→∞
n→∞
Since SX ⊂ BX, there exists a sequence {yn } in X such that Sxn = Byn . Hence
lim Byn = z.
n→∞
Now, suppose that BX is a closed subset of X, there exists a point u ∈ X
such that Bu = z.
We will show that lim T yn = z. From inequality (C2), we have
n→∞
d(Sxn , T yn ) ≤ [φ d(Axn , Byn ), d(Axn , Sxn ),
Letting n → ∞, we have
d(Byn , T yn ), d(Sxn , Byn ), d(Axn , T yn ) ]λ .
d(z, T yn ) ≤ [φ 1, 1, d(z, T yn ), 1, d(z, T yn ) ]λ
≤ dλ (z, T yn ),
COMMON FIXED POINTS FOR WEAKLY COMPATIBLE...
557
which implies that lim d(z, T yn ) = 1. Thus we have
n→∞
lim Axn = lim Byn = lim Sxn = lim T yn = z = Bu
n→∞
n→∞
n→∞
n→∞
for some u ∈ X.
Putting x = xn and y = u in (C2), we have
d(Sxn , T u) ≤ [φ d(Axn , Bu), d(Axn , Sxn ),
Letting n → ∞, we have
d(Bu, T u), d(Sxn , Bu), d(Axn , T u) ]λ .
d(Bu, T u) ≤ [φ 1, 1, d(Bu, T u), 1, d(Bu, T u) ]λ
≤ dλ (Bu, T u),
which implies that Bu = T u. Since the pair B, T is weakly compatible, we have
BT u = T Bu and then BBu = BT u = T Bu = T T u.
On the other way, since T X ⊂ AX, there exists v ∈ X such that T u = Av.
Next we claim that Av = Sv. Putting x = v and y = u, we have
d(Sv, T u) ≤ [φ d(Av, Bu), d(Av, Sv),
Letting n → ∞, we have
d(Bu, T u), d(Sv, Bu), d(Av, T u) ]λ .
d(Sv, Av) ≤ [φ 1, d(Av, Sv), 1, d(Av, Sv), 1 ]λ
≤ dλ (Sv, Av),
which implies that Sv = Av and hence Bu = T u = Av = Sv. Since the pair
A, S is weakly compatible, we have ASv = SAv and then SSv = SAv = ASv =
AAv.
Next we claim that SAv = Av. Putting x = Av and y = u, we have
d(SA, T u) = d(SAv, T u)
≤ [φ d(AAv, Bu), d(AAv, SAv),
d(Bu, T u), d(SAv, Bu), d(AAvv, T u) ]λ
≤ dλ (SAv, Av),
which implies that SAv = Av and hence SAv = Av = AAv. Hence Av is
common fixed point of A and S.
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
Also, one can easily prove that BBu = Bu = T Bu, that is, Bu is common
fixed point of B and T. As Av = Bu, Av is a common fixed point of A, B, S
and T.
Similarly we can complete the proof for cases in which AX, or T X, or SX
is a closed subset of X.
The uniqueness follows easily from inequality (C2). This completes the
proof.
Finally, we prove the following theorems for weakly compatible mappings
with common limit range property satisfying the implicit function in a multiplicative metric space.
Lemma 2.9. Let A, B, S and T be mappings of a multiplicative metric
space (X, d) satisfying the conditions (C1), (C2) and the following condition:
(C7) the pair A, S satisfies CLRA property or the pair B, T satisfies CLRB
property.
Then the pairs A, S and B, T share the common limit in the range of A
property or B property.
Proof. Suppose that the pair A, S satisfies common limit range of A property. Then there exists a sequence {xn } in X such that lim Sxn = lim Axn =
n→∞
n→∞
Az for some z ∈ X. Since SX ⊂ BX, so for each xn there exists yn in X such
that Sxn = Byn . Then lim Byn = Az. Hence, we have lim Sxn = lim Axn =
n→∞
n→∞
n→∞
lim Byn = Az.
n→∞
Now we claim that lim T yn = Az. Putting x = xn and y = yn in (C2), we
n→∞
have
d(Sxn , T yn ) ≤ [φ d(Axn , Byn ), d(Axn , Sxn ),
d(Byn , T yn ), d(Sxn , Byn ), d(Axn , T yn ) ]λ .
Letting n → ∞, we have
d(Az, T yn ) ≤ [φ 1, 1, d(Az, T yn ), 1, d(Az, T yn ) ]λ
= dλ (Az, T yn ),
which implies that lim T yn = Az.
n→∞
Then the pairs A, S and B, T share the common limit in the range of A
property.
Similarly we can complete the proof for cases in which the pair B, T satisfies
common limit in the range of B property. This completes the proof.
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559
Theorem 2.10. Let A, B, S and T be mappings of a multiplicative metric
space (X, d) satisfying the conditions (C1), (C2) and (C7).
Then the pairs A, S and B, T have a coincidence point.
Moreover, assume that the pairs A, S and B, T are weakly compatible. Then
A, B, S and T have a unique common fixed point.
Proof. From Lemma 2.9, the pairs A, S and B, T share the common limit
in the range of A property, that is, there exist two sequences {xn } and {yn } in
X such that
lim Sxn = lim Axn = lim Byn = lim T yn = Av
n→∞
n→∞
n→∞
n→∞
for some v ∈ X.
Firstly, we claim that Av = Sv. Putting x = v and y = yn in (C2), we have
d(Sv, T yn ) ≤ [φ d(Av, Byn ), d(Av, Sv),
Letting n → ∞, we have
d(Byn , T yn ), d(Sv, Byn ), d(Av, T yn ) ]λ .
d(Av, Av) ≤ [φ 1, d(Av, Sv), 1, d(Sv, Av), 1 ]λ
≤ dλ (Av, Sv),
which implies that Sv = Av. Since SX ⊂ BX, there exists w ∈ X such that
Bw = Sv.
Now we claim that Bw = T w. Putting x = v and y = w, we have
d(Bw, T w) = d(Sv, T w)
≤ [φ d(Av, Bw), d(Av, Sv),
d(Bw, T w), d(Sv, Bw), d(Av, T w) ]λ
= [φ 1, 1, d(Bw, T w), 1, d(Bw, T w) ]λ
= dλ (Bw, T w),
this implies that Bw = T w and hence T w = Bw = Av = Sv. Since the pairs
A, S and B, T are weakly compatible and Av = Sv and T w = Bw. Hence
ASv = SAv = AAv = SSv,
T Bw = BT w = BBw = T T w.
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C.Y. Jung, P. Kumar, S. Kumar, S.M. Kang
Finally, we claim that SAv = Av. Putting x = Av and y = w, we have
d(SAv, Av) = d(SAv, T w)
≤ [φ d(AAv, Bw), d(AAv, SAv),
d(Bw, T w), d(SAv, Bw), d(AAv, T w) ]λ
≤ dλ (SAv, Av),
this implies that SAv = Av and hence SAv = Av = AAv, which implies that
Av is a common fixed point of A and S.
Also, one can easily prove that BBw = Bw = T Bw, that is, Bw is a
common fixed point of B and T. As Av = Bw, Av is common fixed point of
A, B, S and T.
The uniqueness follows easily from (C2). This completes the proof.
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