Unique Fixed Point Theorems In Metric Space
D.P. Shukla Department of Mathematics
Govt. Model Science College, Rewa-486001,(M.P.),India
S.K. Tiwari Department of Mathematics
Govt. Model Science College, Rewa-486001,(M.P.), India
S.P. Pandey Department of Mathematics
Govt. Model Science College, Rewa-486001,(M.P.) India
Abstract - In this paper, we have established unique fixed point theorems in complete metric space and generalized in n-dimensional space.
AMS Subject Classification (2010):
47
H
10
,54
H
25
,54
E
50
Key words: Complete metric space, contraction mapping, Banach fixed point theorem, continuous mapping
I. INTRODUCTION
Fixed point theory plays a basic role in applications of many branches of mathematics and other many areas. The problem of finding fixed point has been studied in several directions. The study of metric fixed point theory has been researched extensively in past decades. Recently, some generalizations of the notion of a metric space have been proposed by many authors and finding a fixed point of contractive mapping in complete metric space.
There are many works concerning the fixed point of contractive maps
(
see
,
for
example
,
[10,13])
. In[10]
, Polish mathematician Banach(1922)
proved a very important result regarding a contraction mapping, known as the Banach contraction principle . It is also known as Banach fixed point theorem.The aim of this paper, is to prove some fixed point theorems, which based on contraction mappings, Banach
contraction principle, and focused on the some results of
[1
−
16]
.1.1 Continuous mapping : A self map
T
on a Banach spaceX
is said to be continuous at a pointx
∈
X
if and only if).
(
)
(
x
T
x
T
x
x
n→
n→
1.2 Complete metric space : A metric space
(
X
,
d
)
is said to be complete if every Cauchy sequence inX
, converges inX
.1.3 Contraction mapping : A self map
T
defined on a metric space(
X
,
d
)
is said to be a contraction , if satisfying)
,
(
)
,
(
Tx
Ty
d
x
y
d
≤
α
for all
x
,
y
∈
X
,x
≠
y
and0
<
α
<
1.
1.4 Banach fixed point theorem : Let
(
X
,
d
)
be a complete metric space andT
:
X
→
X
be a contraction onX
. ThenT
has a unique fixed point inX
.II. MAINRESULTS
)
,
(
)
,
(
)
,
(
)
,
(
Tx
Ty
d
x
Tx
d
y
Ty
d
x
y
d
≤
α
+
β
+
γ
(1)for all
x
,
y
∈
X
,x
≠
y
and for someα
,
β
,
γ
∈
[0,1)
withα
+
β
+
γ
<
1
, thenT
has a unique fixed point inX
.Proof : Let
x
0 be an arbitrary point in X and∞
0 =
}
{
x
n n be the sequence of iterations ofT
atx
0, i.e.)
(
=
1 n
n
T
x
x
+∀
n
∈
N
. If
x
n=
x
n+1 for somen
then the result follows trivially . So, letx
n≠
x
n+1 for alln.
Obviously, we have
)).
(
),
(
(
=
)
,
(
x
n+1x
nd
T
x
nT
x
n−1d
Using
(1)
in above, we get)
,
(
))
(
,
(
))
(
,
(
)
,
(
x
n+1x
n≤
d
x
nT
x
n+
d
x
n−1T
x
n−1+
d
x
nx
n−1d
α
β
γ
)
,
(
)
,
(
)
,
(
)
,
(
+1≤
+1+
−1+
−1
d
x
nx
nα
d
x
nx
nβ
d
x
nx
nγ
d
x
nx
n).
,
(
)
1
(
)
,
(
+1 −1−
+
≤
n n n
n
x
d
x
x
x
d
α
γ
β
Similarly, proceeding this work, we get
).
,
(
)
1
(
)
,
(
x
1x
d
x
1x
0d
n n nα
γ
β
−
+
≤
+By the triangle inequality, we have for
m
≥
n
)
,
(
...
)
,
(
)
,
(
)
,
(
x
nx
md
x
nx
n 1d
x
n 1x
n 2d
x
m 1x
md
≤
++
+ ++
+
−
)
,
(
)
...
(
0 11 1
x
x
d
K
K
K
n+
n++
+
m−≤
where
)
1
(
=
α
γ
β
−
+
K
∞
→
−
K
d
x
x
as
m
K
n)
,
(
)
1
(
=
0 1
.
0
→
∞
→
as
n
Since
X
is complete, there exist a pointp
∈
X
such thatx
n→
p
. Further, the continuity ofT
inX
,implies that
)
lim
(
=
)
(
nn
x
T
p
T
∞ →
p
x
T
nn
=
)
(
lim
=
∞ → Therefore
p
is a fixed point ofT
inX
.Now, we prove that
p
is a unique fixed point ofT
. If there exist another fixed pointq
(
≠
p
)
∈
X
, then we have))
(
),
(
(
=
)
,
(
p
q
d
T
p
T
q
d
)
,
(
))
(
,
(
))
(
,
(
p
T
p
d
q
T
q
d
p
q
d
β
γ
α
+
+
≤
)
,
(
)
,
(
)
,
(
p
p
d
q
q
d
p
q
d
β
γ
α
+
+
≤
1),
<
(
)
,
(
<
)
,
(
p
q
d
p
q
since
γ
d
which is a contradiction. Hence
p
is a unique fixed point ofT
inX
.Theorem 2.2: Let
T
be a self map defined on a complete metric space(
X
,
d
)
, which satisfying the condition (1). If for some positive integerr
,T
r is continuous, thenT
has a unique fixed point.Proof : Consider a sequence
{
x
n}
as in theorem (2.1). It converges to some pointp
∈
X
. Therefore, itssubsequence
{
x
nk}
also converges top
. Also,)
lim
(
=
)
(
k n k r r
x
T
p
T
∞ →
p
x
r k n k
=
lim
=
+ ∞ →
Therefore
p
is a fixed point ofT
r.
Now, we have to show thatp
is a fixed point ofT
, i.e.,T
(
p
)
=
p
. Letm
be the smallest positive integer such thatT
m(
u
)
=
u
butT
l(
p
)
≠
p
forl
=
1,2,3,
⋅
⋅
⋅⋅
,
m
−
1
. If1
>
m
, then we have)))
(
(
),
(
(
=
))
(
),
(
(
=
)
),
(
(
T
p
p
d
T
p
T
p
d
T
p
T
T
1p
d
m m−
))
(
,
(
))
(
),
(
(
))
(
,
(
1 1p
T
p
d
p
T
p
T
d
p
T
p
d
+
m− m+
m−≤
α
β
γ
))
(
,
(
))
(
),
(
(
))
(
,
(
=
α
d
p
T
p
+
β
d
T
m−1p
T
p
+
γ
d
p
T
m−1p
).
),
(
(
)
1
(
=
)
),
(
(
T
p
p
d
T
1p
p
d
m−−
+
α
γ
β
(2) Again, we have
))
(
),
(
(
=
))
(
,
(
p
T
1p
d
T
p
T
1p
d
m− m m−
))
(
),
(
(
))
(
),
(
(
))
(
),
(
(
T
1p
T
p
d
T
2p
T
1p
d
T
1p
T
2p
d
m− m+
m− m−+
m− m−≤
α
β
γ
implying that
)).
(
),
(
(
)
1
(
))
(
,
(
p
T
1p
d
T
2p
T
1p
d
m− m− m−−
+
≤
α
γ
β
Similarly, proceeding this work, we obtain
).
),
(
(
)
1
(
))
(
,
(
p
T
1p
1d
T
p
p
d
m− m−−
+
≤
α
γ
β
(3)
From
(2)
and(3)
, we have)
),
(
(
)
1
(
)
),
(
(
T
p
p
d
T
p
p
d
mα
γ
β
−
+
≤
since
α
+
β
+
γ
<
1
),
),
(
(
<
d
T
p
p
which is a contradiction. Hence
T
(
p
)
=
p
, i.e.p
is a fixed point ofT
. We have to show that, the fixed point ofT
is unique. supposeq
(
≠
p
)
∈
X
be another fixed point ofT
, then we have))
(
),
(
(
=
)
,
(
q
p
d
T
q
T
p
d
)
,
(
))
(
,
(
))
(
,
(
q
T
q
d
p
T
p
d
p
q
d
β
γ
α
+
+
)
,
(
p
q
d
γ
≤
1),
<
(
)
,
(
<
d
p
q
since
γ
which is a contradiction. Hence fixed point ofT
is unique.In the next theorem, we generalized the theorem
(2.1)
and(2.2)
.Theorem 2.3: Let
T
be a self map defined on a complete metric space(
X
,
d
)
and for some positive integert
′
'
, satisfying the condition)
,
(
))
(
,
(
))
(
,
(
))
(
),
(
(
T
x
T
y
d
x
T
x
d
y
T
y
d
x
y
d
t t≤
α
t+
β
t+
γ
(4)for all
x
,
y
∈
X
,
x
≠
y
and for someα
,
β
,
γ
∈
[0,1)
withα
+
β
+
γ
<
1
. IfT
t is continuous, thenT
has a unique fixed point.Proof : The proof of theorem
(2.3)
is similar to the theorem(2.1)
and(2.2)
.In the next theorem, we will study the existence of a unique common fixed point of two mappings which are not necessarily continuous or commuting.
Theorem 2.4: Let
T
1 andT
2 be two self mappings defined on a complete metric space(
X
,
d
)
, satisfying thefollowing conditions
(i) For some
α
,
β
,
γ
∈
[0,1)
withα
+
β
+
γ
<
1
such that)
,
(
))
(
,
(
))
(
,
(
))
(
),
(
(
T
1x
T
2y
d
x
T
1x
d
y
T
2y
d
x
y
d
≤
α
+
β
+
γ
(5)
for all
x
,
y
∈
X
,
x
≠
y
.
(ii)
T
1T
2 is continuous onX
.(iii) There exist an
x
0∈
X
and the sequence{
x
n}
,°
¯
°
®
− −
n
whenniseve
x
T
whennisodd
x
T
x
where
nn
n
(
)
:
:
)
(
=
2 11 1
such that
x
n≠
x
n+1 for alln
. ThenT
1andT
2 having a unique common fixed point.Proof : Consider,
)
,
(
=
)
,
(
x
2n+1x
2nd
T
1x
2nT
2x
2n−1d
using
(5)
in above equation, we have)
,
(
)
,
(
)
,
(
2 1 2+
2 −1 2 2 −1+
2 2 −1≤
α
d
x
nT
x
nβ
d
x
nT
x
nγ
d
x
nx
n)
,
(
)
1
(
)
,
(
2 +1 2 2 2 −1−
+
≤
n n n
n
x
d
x
x
x
d
α
γ
β
continuing this work, we have
).
,
(
)
1
(
2nd
x
1x
0α
γ
β
−
+
≤
Similarly,
).
,
(
)
1
(
)
,
(
1 01 2 1
2 2
2
x
d
x
x
x
d
n+ n+ n+−
+
≤
α
γ
β
p
x
k
n
}
→
{
. Then, we have
.
=
=
(
=
)
(
1 )
2 1 2
1
T
p
T
T
lim
x
lim
x
p
T
k n k k n k
+ ∞ → ∞
→
i.e.
p
is a fixed point ofT
1T
2. Next, we have to show thatT
2(
p
)
=
p
. SupposeT
2(
p
)
≠
p
, then))
(
),
(
(
=
)
),
(
(
T
2p
p
d
T
2p
T
1T
2p
d
))
(
,
(
))
(
),
(
(
))
(
,
(
p
T
2p
d
T
2p
T
1T
2p
d
p
T
2p
d
β
γ
α
+
+
≤
))
(
,
(
)
(
=
α
+
β
+
γ
d
p
T
2p
since
α
+
β
+
γ
<
1
)),
(
,
(
<
d
p
T
2p
which is a contradiction. Hence
T
2(
p
)
=
p
, i.e.p
is a fixed point ofT
2.Also,
0.
=
)
,
(
=
)
),
(
(
=
)
)),
(
(
(
=
)
),
(
(
T
1p
p
d
T
1T
2p
p
d
T
1T
2p
p
d
p
p
d
Hence
T
1(
p
)
=
p
, i.e.p
is a fixed point ofT
1.
SoT
1 andT
2 have a common fixed point. Now, we have toprove that
T
1 andT
2 have a unique common fixed point. If possible, letq
(
≠
p
)
∈
X
be two fixed points of1
T
andT
2. Then))
(
),
(
(
=
)
,
(
p
q
d
T
1p
T
2q
d
)
,
(
))
(
,
(
))
(
,
(
p
T
1p
d
q
T
2q
d
p
q
d
β
γ
α
+
+
≤
)
,
(
)
,
(
)
,
(
=
α
d
p
p
+
β
d
q
q
+
γ
d
p
q
)
,
(
=
γ
d
p
q
1),
<
(
)
,
(
<
d
p
q
since
γ
which is a contradiction. Hence
p
is a unique common fixed point ofT
1 andT
2.Remark 2.1: If
X
=
[0,1]
andT
1,
T
2 be two commuting continuous self mappings defined onX
, thenT
1and
T
2 need not have a common fixed point by Smart[6].
Theorem 2.5: Let
T
n be a self map defined on a complete metric space(
X
,
d
)
withp
n as a fixed point foreach
n
=
1,2,3,
⋅
⋅
⋅
⋅
respectively. IfT
n satisfying the condition(1)
for alln
and{
T
n}
converges point wiseto
T
, thenp
n→
p
asn
→
∞
if and only ifp
is a fixed point ofT
.Proof : If
p
=
p
n for somen
, then the assumption follows trivially. So, we assume thatp
≠
p
n for anyn
.Let
p
n→
p
then, we show thatp
is fixed point ofT
. Now,))
(
),
(
(
))
(
,
(
)
,
(
))
(
,
(
p
T
p
d
p
p
d
p
T
p
d
T
p
T
p
d
≤
n+
n n+
n
))
(
),
(
(
))
(
),
(
(
)
,
(
=
d
p
p
n+
d
T
np
nT
np
+
d
T
np
T
p
)).
(
),
(
(
)
,
(
))
(
,
(
))
(
,
(
)
,
(
p
p
d
p
T
p
d
p
T
p
d
p
p
d
T
p
T
p
d
n+
n n n+
n+
n+
n≤
α
β
γ
0
))
(
,
(
)
(1
−
β
d
p
T
p
≤
1)
<
(
0
))
(
,
(
p
T
p
since
β
d
≤
0
=
))
(
,
(
p
T
p
d
.
=
)
(
p
p
T
Conversely, we suppose that
T
(
p
)
=
p
.
Then obviously, we have))
(
),
(
(
=
)
,
(
p
p
d
T
p
T
p
d
n n n
))
(
),
(
(
))
(
),
(
(
T
p
T
p
d
T
p
T
p
d
n n n+
n≤
))
(
),
(
(
)
,
(
))
(
,
(
))
(
,
(
p
T
p
d
p
T
p
d
p
p
d
T
p
T
p
d
n n n+
n+
n+
n≤
α
β
γ
))
(
,
(
)
1
1
(
)
,
(
p
p
d
p
T
p
d
n nγ
β
−
+
≤
))
(
),
(
(
)
1
1
(
=
d
T
p
T
np
γ
β
−
+
.
,
0
as
n
→
∞
T
n→
T
→
Hence
p
n→
p
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