Gen. Math. Notes, Vol. 14, No. 1, January 2013, pp. 43-50
ISSN 2219-7184; Copyright © ICSRS Publication, 2013
www.i-csrs.org
Available free online at http://www.geman.in
Some Common Fixed Point Theorems of
Compatible Mappings of Type (A) in Metric
Space
M. Bina Devi
CMJ University Shillong, Meghalaya
E-mail: ymnehor2008@yahoo.com
(Received: 9-10-12 / Accepted: 4-12-12)
Abstract
In this note we prove two common fixed point theorems for compatible
mappings of Type (A) in complete metric space. Our results generalize some of
the results in the literature.
Keywords: common fixed point, compatible mappings, compatible mappings
of type (A).
1
Introduction
The concept of compatible mappings was introduced by Jungck [2] as a
generalization of commuting mappings. Jungck, Murthy and Cho [3] introduced a
new type of compatible mappings and named as compatible mapping of type (A).
It can be mentioned that compatible mappings and compatible mappings of type
(A) are equivalent under certain conditions. Fixed points of compatible mappings
of type (A) in different spaces can be seen in [1, 3, 4, 5].
44
2
M. Bina Devi
Preliminaries
Following are definition of types of compatible mappings.
Definition 2.1 [2]: Let S and T be mappings from a complete metric space X into
itself. The mappings S and T are said to be compatible mappings if
lim d ( STxn , TSxn ) = 0 whenever {xn} is a sequence in X such that
n →∞
lim Sxn = lim Txn = z for some z ∈ X.
n →∞
n →∞
Definition 2.2 [3]: Let S and T be mappings from a complete metric space X into
itself. The mappings S and T are said to be compatible mappings of type (A) if
lim d ( STx n , TTx n ) = 0 and lim d (TSxn , SSxn ) = 0 and whenever {xn} is a sequence
n →∞
n→∞
in X such that lim Sxn = lim Txn = z for some z∈ X.
n →∞
n →∞
We need following propositions for our main results.
Proposition 2.3 [4]: If S and T be continuous mappings from a metric space X
into itself then S and T are compatible if and only if they are compatible mappings
of type (A).
Proposition 2.4 [4]: Let S and T be mappings from a complete metric space (X, d)
into itself. If a pair {S, T} is compatible of type (A) on X and lim Sxn = lim Txn =
n→∞
n→∞
z for some z ∈ X, then we have
(i) d(TSxn, Sz) →0 as n → ∞ if S is continuous.
(ii) d(STxn, Tz) →0 as n → ∞ if T is continuous.
(iii) STz=TSz and Sz=Tz if S and T are continuous at z.
3
Main Results
Theorem 3.1 Suppose S, T and A are three self mappings of a complete metric
space (X, d) into itself satisfying following conditions
(i)
(ii)
Sx)]
S(X) ∪ T (X) ⊂ A(X).
d(Sx, Ty) ≤ a1d(Ax, Ay)+ a2d(Sx, Ax)+ a3d(Ty, Ay)+ a4d(Ax, Ty)+ a5d(Ay,
for all x, y ∈ X and a1, a2, a3, a4, a5 are non-negative real numbers such that a1
+a2 +a3 +2a4 +a5< 1.
(iii)
(iv)
one of S, T and A is continuous.
(S, A) and (T, A) are compatible mappings of type (A).
Then S, T and A have a unique common fixed point.
Some Common Fixed Point Theorems of…
Proof. Let x0∈ X arbitrary. Construct a sequence {xn}, as follows.
Ax2n+1 = Sx2n, Ax2n+2 = Tx2n+1, n = 0, 1, 2, ….
45
(1)
d(Ax2n+1, Ax2n+2) = d(Sx2n, Tx2n+1)
≤ a1d(Ax2n, Ax2n+1) + a2d(Sx2n, Ax2n) + a3d(Tx2n+1, Ax2n+1)
+ a4d(Ax2n, T x2n+1) + a5d(Ax2n+1, Sx2n)
= a1d(Ax2n, Ax2n+1) + a2d(Ax2n+1, Ax2n) + a3d(Ax2n+2, Ax2n+1)
+ a4d(Ax2n, Ax2n+2) + a5d(Ax2n+1, Ax2n+1)
≤ (a1+ a2 + a4)d(Ax2n, Ax2n+1) + (a3 + a4)d(Ax2n+1, Ax2n+2)
a1 + a2 + a4
d(Ax2n, Ax2n+1)
1 − a3 − a4
d(Ax2n+1, Ax2n+2) ≤ hd(Ax2n, Ax2n+1)
d(Ax2n+1, Ax2n+2) ≤
where h =
a1 + a2 + a4
< 1.
1 − a3 − a4
Similarly we can show that
d(Ax2n+1, Ax2n+2) ≤ h2n+1d(Ax0, Ax1)
For k > n, we have
d(Axn, Axn+k ) ≤ d(Axn+i−1, Axn+i)
≤ hn+i−1d(Ax0, Ax1)
≤ d(Ax0, Ax1)
→0 as n→∞.
Hence {Axn} is a Cauchy sequence. Since X is complete metric space, there exist a
z ∈ X such that Axn → z. Then the subsequences of {Axn}, {Sx2n} and {Tx2n+1}
also converges to z. i.e. {Sx2n} → z and {Tx2n+1} → z. Suppose that A is
continuous and the pair (S, A) is compatible mappings of type of (A).
Then from condition (ii) we have
d(SAx2n, T x2n+1) ≤ a1d(AAx2n, Ax2n+1) + a2d(SAx2n, AAx2n) + a3d(Tx2n+1,
Ax2n+1)
+ a4d(AAx2n, Tx2n+1) + a5d(Ax2n+1, SAx2n)
Since A is continuous, AAx2n→Az as n→∞. The pair (S, A) is compatible
mappings of type (A), then SAx2n → Az as n → ∞.
Letting n → ∞, we have
46
M. Bina Devi
d(Az, z) ≤ a1d(Az, z) + a2d(Az, Az) + a3d(z, z) + a4d(Az, z) + a5d(z, Az)
= (a1 + a4+ a5)d(Az, z)
Hence d(Az, z) = 0 and Az = z, since a1 + a4+ a5< 1.
Again we have
d(Sz, Tx2n+1) ≤ a1d(Az, Ax2n+1) + a2d(Sz, Az) + a3d(Tx2n+1, Ax2n+1)
+ a4d(Az, Tx2n+1) + a5d(Ax2n+1, Sz)
Letting n → ∞ and using Az = z, we have
d(Sz, z) ≤ a1d(z, z) + a2d(Sz, z) + a3d(z, z) + a4d(z, z) + a5d(z, Sz)
= (a2 + a5)d(Sz, z)
Hence d(Sz, z) = 0 and Sz = z, since a2+ a5 < 1. So we have Az = Sz = z.
By condition (ii), we have
d(z, Tz) = d(Sz, Tz) ≤ a1d(Az, Az) +a2d(Sz, Az)+a3d(Tz, Az)+a4d(Az, Tz)
+a5d(Az, Sz)]
= (a3 + a4)d(Tz, z)
⇒ d(z, Tz)=0 i.e. z = Tz, since a3+a4 < 1. Showing that z is common fixed
point of S, T and A. Similarly we can show that z is a common fixed point of S, T
and A when the pair (T, A) is compatible mappings of type (A).
In order to prove the uniqueness of fixed point, let z and z′ be two common fixed
point of S, T and A. So we have z = Sz =Tz = Az and z′ = S z′ = T z′ = A z′ .
d(z, z′ )=d(Sz, T z′ ) ≤ a1d(Az, T z′ )+a2d(Sz, Az)+a3d(T z′ , A z′ )
+a4d(Az, T z′ )+a5d(A z′ , Sz)
= a1d(z, z′ ) + a2d(z, z) + a3d( z′ , z′ )+ a4d(z, z′ ) + a5d( z′ , z)
= (a1 + a4 + a5)d(z, z′ )
Hence d(z, z′ ) = 0 and z = z′ , since a1 + a4 + a5 < 1. Thus the common fixed
point is unique.
Theorem 3.2 Suppose S, A, T and B are four self mapping of complete metric
space (X, d) into itself satisfying following conditions
(i) S(X) ⊂ B(X), T(X) ⊂ A(X).
(ii) d(Sx, Ty) ≤ a1d(Ax, By)+ a2d(Ax, Sx)+ a3d(By, Ty)+ a4d(Ax, Ty)+ a5d(By, Sx)
for all x, y ∈ X and a1, a2, a3, a4, a5 are non-negative real such that a1 +a2 +a3
+2a4 +a5< 1
Some Common Fixed Point Theorems of…
47
(iii) one of S, A, T and B is continuous
(iv) (S, A) and (T, B) are compatible mappings of type (A)
then S, A, T and B have unique common fixed point.
Proof. Let x0 ∈ X be arbitrary. Choose a point x1 in X such that Sx0 = Bx1. This can
be done since S(X) ⊂ B(X). Let x2 be a point in X such that Tx1 = Ax2. This can be
done since T(X) ⊂ A(X). In general we can choose x2n, x2n+1, x2n+2, · · · , such that
Sx2n = Bx2n+1 and Tx2n+1 = Ax2n+2. So that we obtain a sequence Sx0, Tx1, Sx2, Tx3, ·
··
(2)
Using condition (ii), we have
d(Sx2n, Tx2n+1) ≤ a1d(Ax2n, Bx2n+1) + a2d(Ax2n, Sx2n) + a3d(Bx2n+1, Tx2n+1)
+ a4d(Ax2n, Tx2n+1) + a5d(Bx2n+1, Sx2n)
= a1d(Tx2n−1, Sx2n) + a2d(Tx2n−1, Sx2n) + a3d(Sx2n, Tx2n+1)
+ a4d(Tx2n−1, Tx2n+1) + a5d(Sx2n, Sx2n)
≤ a1d(Tx2n−1, Sx2n) + a2d(Tx2n−1, Sx2n) + a3d(Sx2n, Tx2n+1)
+ a4d(Tx2n−1, Sx2n) + a4d(Sx2n, Tx2n+1)
= (a1+ a2 + a4)d(Tx2n−1, Sx2n) + (a3 + a4)d(Sx2n, Tx2n+1)
Hence
d(Sx2n, Tx2n+1) ≤ kd(Sx2n, Tx2n−1)
a1 + a2 + a4
< 1. Similarly, we can show
1 − a3 − a4
d(Sx2n, Tx2n−1) ≤ kd(Sx2n−2, Tx2n−1)
where k =
Therefore
d(Sx2n, Tx2n+1) ≤ k2d(Sx2n−2, Tx2n−1)
≤ k2nd(Sx0, Tx1)
which implies that the sequence (1) is a Cauchy sequence and since (X, d) is
complete, so the sequence (1) has a limit point z in X. Hence the subsequences
{Sx2n} = {Bx2n−1} and {Tx2n−1} = {Ax2n} also converges to the point z in X.
Suppose that the mapping A is continuous then AAx2n→Az and ASx2n→Az as
n→∞. Since the pair (S, A) is compatible mappings of type (P). We get SAx2n →
Az as n → ∞. Now by (ii)
d(SAx2n, Tx2n+1) ≤ a1d(AAx2n, Bx2n+1) + a2d(AAx2n, SAx2n) + a3d(Bx2n+1,
Tx2n+1)
+ a4d(AAx2n, Tx2n+1) + a5d(Bx2n+1,
SAx2n)
letting n → ∞, we get
d(Az, z) ≤ a1d(Az, z) + a2d(Az, Az) + a3d(z, z)+ a4d(Az, z) + a5d(z, Az)
48
M. Bina Devi
= (a1+ a4 +a5)d(Az, z).
This gives d(Az, z) = 0, since 0 ≤ a1 + a4 +a5 < 1. Hence Az = z. Further
d(Sz, Tx2n+1) ≤ a1d(Az, Bx2n+1) + a2d(Az, Sz) + a3d(Bx2n+1, Tx2n+1)
+ a4d(Az, Tx2n+1) + a5d(Bx2n+1, Sz)
letting Bx2n+1, Tx2n+1 → z as n → ∞ and Az = z we get
d(Sz, z) ≤ a1d(z, z) + a2d(z, Sz) + a3d(z, z)+ a4d(z, z) + a5d(z, Sz)
= (a2+a5)d(Sz, z).
Hence d(Sz, z) = 0 i.e. Sz = z, since 0 ≤ a2+a5 < 1. Thus Sz = Az = z. Since S(X) ⊂
B(X), there is a point u ∈ X such that z = Sz = Bu.
Now by (ii)
d(z, Tu) = d(Sz, Tu)
≤ a1d(Az, Bu) + a2d(Az, Sz) + a3d(Bu, Tu) + a4d(Az, Tu) + a5d(Bu,
Sz)
= a1d(z, z) + a2d(z, z) + a3d(z, Tu)+ a4d(z, Tu) + a5d(z, z)
= (a3 + a4)d(z, Tz)
hence d(z, Tu) = 0 i.e. Tu = z = Bu, since 0 ≤ a3+a4< 1. Take yn = u for n ≥ 1. Then
Tyn→Tu = z and Byn → Bu = z as n → ∞. Since the pair (T, B) is compatible
mappings of type (A), we get lim( BTyn , TTyn ) = 0 implies d(Tz, Bz) = 0, since
n →∞
Tyn→ z and Byn = z for all n ≥ 1. Hence Tz = Bz.
Now
d(z, Tz) = d(Sz, Tz) ≤ a1d(Az, Bz) +a2d(Az, Sz) +a3d(Bz, Tz) +a4d(Az,
Tz)+a5d(Bz, Sz)
= a1d(z, Tz) + a2d(z, z) + a3d(Tz, Tz) + a4d(z, Tz) + a5d(Tz, z)
= (a1 + a4+ a5)d(z, Tz)
Since a1 + a4 +a5 < 1, we get Tz = z. Hence z = Tz = Bz, therefore z is common
fixed point of S, A, T and B when the continuity of A is assumed.
Now suppose that S is continuous. Then SSx2n→Sz, SAx2n→Sz as n→∞. By
condition (ii) we have
d(SSx2n, Tx2n+1) ≤ a1d(ASx2n, Bx2n+1) + a2d(ASx2n, SSx2n) + a3d(Bx2n+1,
Tx2n+1)
+ a4d(ASx2n, Tx2n+1) + a5d(Bx2n+1, SSx2n).
Letting n → ∞ and using the compatible mappings of type (P) of the pair (S, A),
we get ASxn→Sz
d(Sz, z) ≤ a1d(Sz, z) + a2d(Sz, Sz) + a3d(z, z)+a4d(Sz, z) + a5d(z, Sz)
Some Common Fixed Point Theorems of…
49
= (a1 + a4 +a5)d(Sz, z).
Since a1+a4+a5<1, we get Sz = z. But S(X) ⊂ B(X), there is a point v ∈ X such that
z=Sz=Bv.
Now by (ii)
d(SSx2n, Tu) ≤ a1d(ASx2n, Bu) + a2d(ASx2n, SSx2n) + a3d(Bu, Tu)
+ a4d(ASx2n, Tu) + a5d(Bu, SSx2n).
Letting n → ∞
d(z, Tv) = d(Sz, Tv)
≤ a1d(z, z) + a2d(z, z) + a3d(z, Tv)+ a4d(z, Tv) + a5d(z, z)
= (a3 + a4)d(z, Tv)
since a3+a4<1, we get Tv=z. Thus z =Bv=Tv. Let yn=v. Then Tyn→Tv=z and
Byn→Tv =z. Since (T, B) is compatible mappings of type (A), we have
lim( BTyn , TTyn ) = 0 this gives BTv = TTv or Bz = Tz. Further
n →∞
d(Sx2n, Tz)≤
Tz)+a5d(Bz, Sx2n)
a1d(Ax2n,
Bz)+a2d(Ax2n,
Sx2n)+a3d(Bz,
Tz)+a4d(Ax2n,
Letting n → ∞, we get
d(z, Tz) ≤ a1d(z, Tz) + a2d(z, z) + a3d(Tz, Tz)+ a4d(z, Tz) + a5d(Tz, z)
= (a1 + a4 +a5)d(z, Tz)
Since 0 ≤ a1 + a4+ a5 < 1, we get z = Tz. Again we have T(X) ⊂ A(X), there is a
point w ∈ X such that z = Tz = Aw. Now
d(Sw, z)=d(Sw, Tz)≤ a1d(Aw, Bz)+a2d(Aw, Sw)+a3d(Bz, Tz)+a4d(Aw,
Tz)+a5d(Bz, Sw)
= a1d(z, z) + a2d(z, Sw) + a3d(z, z)+ a4d(z, Tz) + a5d(z, Sw)
= (a2 + a5)d(z, Sw).
since 0 ≤ a2 + a5< 1, we get Sw = z. Take yn = w then Syn→Sw = z, Ayn→Aw = z.
Since (S, A) is compatible of type (A), we get lim( ASyn , SSyn ) = 0 this implies that
n →∞
AAw = SSw or Az = Sz. Thus we have z = Sz = Az = Bz = Tz. Hence z is a common
fixed point of S, A, T and B when S is continuous. Similarly we can prove the
same result when B or T is continuous.
In order to prove the uniqueness of fixed point, let z and z′ be two common fixed
point of S, A, T and B i.e. z=Sz =Az=Tz=Bz and z′ = S z′ = A z′ = T z′ = B z′ . By
condition (ii)
d(z, z′ )=d(Sz, T z′ )≤a1d(Az, B z′ )+a2d(Az, Sz)+a3d(B z′ ,T z′ )
+a4d(Az, T z′ )+a5d(B z′ , Sz)
50
M. Bina Devi
= a1d(z, z′ )+a2d(z, z)+a3d( z′ , z′ )+a4d(z, z′ )+a5d( z′ , z)
= (a1 + a4+ a5)d(z, z′ )
since a1 + a4+ a5 < 1, we have z = z′ . This completes the proof.
Remark Theorem 3.1 and 3.2 can be reduced to theorem 2.1 and 2.2 of Aage and
Salunke [1] by replacing following conditions
(ii) a1 = α , a2= a3= β, a4 = a5 = γ.
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[2]
[3]
[4]
[5]
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Math. Math. Sci., 9(1986), 771-779.
G. Jungck, P.P. Murthy and Y.J. Cho, Compatible mappings of type (A)
and common fixed points, Math. Japonica, 38(1993), 381-390.
Y. Rohen, Th. Indubala, O. Budhichandra and N. Leenthoi, Common fixed
point theorems for compatible mappings of type (A), IJMSEA, 6(II)
(2012), 323-333.
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