A Suzuki-Type Common Fixed Point Theorem
for generalized (
ψ
,
φ
)
- weak contractions
Ombir Dahiya1, Ravinder Kumar2 and Raj Kamal3
1 Department of Mathematics, B.P.S.I.H.L, BPSMV Khanpur-kalan, Sonipat, 131305
Email Id- [email protected]
2Department of Mathematics, A.I.J.H.M. College, Rohtak, 124001
3Department of Mathematics, S. K. Govt. Postgraduate College Kanwali, Rewari, 123411
Abstract
Common fixed point results are presented for generalized (
ψ
,
φ) - weak contractive mappings with constants in complete metric spaces. Our results extend previous results of Chugh (2014), Đorić (2009), Zhang and Song (2009), as well as of Kikkawa and Suzuki (2008), Rhoades (2001), Nadler (1969) and others.
MSC: 54H25 and 47H10
Keywords: Banach contraction principle; φ - weak contractive;
generalized φ - weak contractive; generalized (
ψ
,
φ) - weak contractive; Lower semi-continuous; Metric completeness; Common fixed point.1.
Introduction
Let (X, d) be a complete metric space and T be a self map
on X. Then T is called a contraction if there exists r∈ [0, 1) such that
d(Tx, Ty) ≤ r d(x, y) for all x, y ∈ X.
We know that if X is complete, then every contraction has
a fixed point (Banach contraction mapping principle).
The following general common fixed point theorem is due to Sastry and Naidu [18].
Theorem 1.1. Let X be a complete metric space and S, T: X → X. Assume there exists r ∈ [0, 1) such that for every x, y ∈ X,
(
)
( ,
)
( ,
)
d
, max
( , ), ( ,
), ( ,
),
.
2
d x Ty
d y Sx
Sx Ty
≤
r
d x y d x Sx d y Ty
+
(1.1.) Then S and T have a unique common fixed point.
Very recently, Suzuki [22] introduced a weaker
notation of contraction and obtained the following
theorem, which is a new type of generalizations of the
classical Banach contraction principle.
Theorem 1.2. Define a non-increasing function θ from
[0, 1) onto
1
, 1
2
byθ(r) = 2
1
1
if 0
( 5 1)
2
1
1
1
if
( 5 1)
2
2
1
1
if
1.
1
2
r
r
r
r
r
r
≤ ≤
−
−
− ≤ ≤
≤ <
+
Let X be a complete metric space and let T: X → X.
Assume there exists r ∈[0, 1) such that for each x, y ∈ X, θ(r) d(x, Tx) ≤ d(x, y) implies d(Tx, Ty) ≤ r d(x, y).
This type of generalization of contraction
mapping has been a very active field of research during
last five years. Suzuki contractive condition has been dealt
with in a number of papers [4], [5], [7], [9-15] and [19-23].
A mapping T: X → X is said to be φ – weak contractive if
there exists a map φ: [0, +∞) → [0, +∞) with φ(0) = 0 and
φ(t) > 0 for all t > 0 such that
d(Tx, Ty) ≤ d(x, y) − φ(d(x, y)) for all x, y ∈ X .
The concept of φ−weak contractive mappings was defined
by Daffer and Kaneko [3] in 1995.
Rhoades [16] proved the following fixed point
theorem for φ – weak contractive single-valued map
generalizing the Banach contraction principle.
Theorem 1.3. Let (X, d) be a complete metric space and let T: X → X be a map such that
d(Tx, Ty) ≤ d(x, y) − φ(d(x, y)), for all x, y ∈ X ,
where φ: [0, +∞) → [0, +∞) is a continuous and
non-decreasing function with φ(0) = 0 and φ(t) > 0 for all t > 0.
Then T has a unique fixed point.
Also, two maps S, T: X → X are called generalized φ –
weak contraction if there exists a map
φ: [0, +∞) → [0, +∞) with φ(0) = 0 and φ(t) > 0 for all t > 0 such that
d(Sx, Ty) ≤ M(x, y) − φ(M(x, y)),
for all x, y ∈ X, where
M(x,y) =max
( , ), ( ,
), ( ,
),
( ,
)
( ,
)
.
2
d x Ty
d y Sx
d x y d x Sx d y Ty
+
In 2014, Chugh [2] proved the following result.
Theorem 1.4. Let (X, d) be a complete metric space and let S, T : X → X. Assume that there exists r ∈ [0, 1) such
that for every x, y ∈ X,
min{d(x, Sx), d(y, Ty)} ≤ (1+ r) d(x, y)
implies
d(Sx, Ty) ≤ d(x, y) − φ d(x, y),
Then there exists z ∈ X such that z ∈ Sz ∩Tz.
In 2009, Zhang and Song [24] proved the
following theorem for generalized φ- weak contraction
(see also [17]) which is defined for two mappings and gave
conditions for the existence of a common fixed point.
Theorem 1.5. Let (X, d) be a complete metric space and let S, T: X → X be two mappings such that for all x, y ∈ X
d(Sx, Ty) ≤ M(x, y) − φ(M(x, y)),
where
φ: [0, +∞) → [0, +∞) is lower semi-continuous function with φ(t) = 0 if and only if t = 0.
Then there exists the unique point z∈ X such that z = Tz =
Sz.
Dutta and Choudhary [8] gave the following
theorem by introducing a new generalization of contraction
principle.
Theorem 1.6. Let (X, d) be a complete metric space and let T: X → X be a map satisfying the inequality
ψ
d(Tx, Ty) ≤ψ
d(x, y) − φ(d(x, y)),where
ψ
, φ: [0, +∞) → [0, +∞) are both continuous andmonotone non-decreasing functions with φ(t) =
ψ
(t) = 0 ifand only if t = 0. Then T has a unique fixed point.
Đorić [6] used generalized (
ψ
,
φ) weak contraction whichis defined for two maps and gave conditions for the
existence of a common fixed point as follows.
Theorem 1.7. Let (X, d) be a complete metric space and Let S, T: X → X be two selfmaps such that for all x, y ∈ X
ψ
d(Sx, Ty) ≤ψ
(M(x, y)) − φ(M(x, y)),where
(i)
ψ
: [0, +∞) → [0, +∞) is continuous and monotone(ii) φ: [0, +∞) → [0, +∞) is lower semi-continuous function with φ(t) = 0 if and only if t = 0.
Then there exists the unique point z∈ X such that
z = Tz = Sz.
In this paper, we established a common fixed point
theorem which is generalization of Theorem 1.7. The idea
is in line with Theorem 1.3 where a generalization of
Theorem 1.1 has been established by use of Suzuki
contractive condition.
2. Main Results
In this paper, the following theorem is our main result.
Theorem 2.1. Let (X, d) be a complete metric space and let S and T be maps on X. Assume that for each x, y ∈ X,
1
2
min{d(x, Sx), d(y, Ty)} ≤ d(x, y) implies
ψ
(d(Sx, Ty))≤ψ
(M(x, y)) − φ(M(x, y)),(2.1)
where
ψ
and φ are defined as above. Then S and T have acommon fixed point.
Proof: Take
x
0 ∈ X. Puttingx
1=
Tx
0 andx
2=
Sx
1 ,then let
x
3=
Tx
2 andx
4=
Sx
3.
Inductively, Choose a sequence {
x
n} in X such that2n 1 2n
x
+=
Tx
andx
2n+2=
Sx
2n+1for all n ≥ 0.
As
1
2
d x
(
n−1,
x
n)
≤
d x
(
n−1,
x
n).
(2.2)Now if n is odd and suppose
1
(
n,
n)
d x
−x
≤
d x
(
n,
x
n+1).
(2.3)
Then by (2.2) and (2.3)
{
1 1}
11
min
(
,
), ( ,
)
(
,
).
2
d x
n−x
nd x x
n n+≤
d x
n−x
nAnd this implies (2.1), that is, we have
1 1 1
( (
d Sx Tx
n, ))
n−(
M x x
(
n,
n−))
M x x
(
n,
n−).
ψ
≤ ψ
− ϕ
Suppose if
d x
(
n,
x
n+1)
≤
d x
(
n−1,
x
n).
(2.4)
And by (2.2),
1
2
d x
(
n−1,
x
n)
≤
d x
(
n−1,
x
n).
So1
2
d x
(
n,
x
n+1)
≤
d x
(
n−1,
x
n).
(2.5)Then by (2.4) and (2.5)
{
1 1}
11
min
(
,
), ( ,
)
(
,
).
2
d x
n−x
nd x x
n n+≤
d x
n−x
n Andthis implies (2.1), that is, we have
1 1 1
( (
d Sx Tx
n, ))
n−(
M x x
(
n,
n−))
M x x
(
n,
n−).
ψ
≤ ψ
− ϕ
It follows from property of the function φ that if n is an
odd,
1 1
1 1 1
1
1 1
1 1
1
1 1
( (
,
))
( (
,
))
( (
, ))
(
(
,
))
(
,
)
(
,
), (
,
),
max
(
,
)
(
,
)
(
,
),
2
(
,
), (
,
),
max
(
,
),
n n n n
n n n n n n
n n n n
n n n n
n n
n n n n n n
d x
x
d Sx Tx
d Sx Tx
M x x
M x x
d x x
d x Sx
d x Tx
d x
Sx
d x
Tx
d x x
d x Sx
d
d x
Tx
+ −
− − −
−
− −
− −
−
− −
ψ
= ψ
ψ
≤ ψ
− ϕ
= ψ
+
−ϕ
1 1
1 1
1 1
1
1 1
1 1
1
(
,
)
(
,
)
2
(
,
), (
,
),
max
(
,
)
(
,
)
(
,
),
2
(
,
), (
,
),
max
(
,
)
(
,
)
(
,
),
2
n n n n
n n n n
n n n n n n
n n n n
n n n n n n
x Tx
d x
Sx
d x x
d x x
d x x
d x
x
d x
x
d x x
d x x
d x x
d x
x
d x
x
− −
− +
− +
−
− +
− +
−
+
= ψ
+
−ϕ
+
{
}
(
)
{
}
(
)
1 1
1
1 1
max
(
,
), (
,
)
( (
,
))
max
(
,
), (
,
)
n n n n n n
n n n n
d x x
d x x
d x
x
d x x
d x x
− +
+
− +
ψ
ψ
≤
−ϕ
(
)
1 1 1
1
( (
,
))
( ( ,
))
( ,
)
( ( ,
))
n n n n n n
n n
d x
x
d x x
d x x
d x x
+ − −
−
ψ
≤ ψ
− ϕ
≤ ψ
i.e.
ψ
( (
d x
n+1,
x
n))
≤ψ
( (
d x
n,
x
n−1)).
So by the property of
ψ
, we have1
(
n,
n)
d x
+x
≤d x
(
n,
x
n−1).
Similarly if n is even, we obtain
d x
(
n+1,
x
n)
≤d x
(
n,
x
n−1).
Therefore, for all n ≥ 0,
d x
(
n+1,
x
n)
≤d x
(
n,
x
n−1)
andso
{
d x
(
n+1,
x
n)
}
is monotonic non-increasing and bounded below, so their exists r ≥ 0 such that1 1
lim (
n,
n)
lim (
n,
n)
n→∞
d x
+x
= =
r
n→∞d x
x
− (2.6)Then (by lower semi-continuity of φ)
Φ(r) ≤
lim inf
n→∞ φ(
d x
(
n,
x
n−1)
).We claim that r = 0. In fact taking upper limits as n → ∞
on either side of the following inequality:
1
( (
d x
n+,
x
n))
ψ
≤
ψ
( (
d x
n,
x
n−1))
−φ(d x
(
n,
x
n−1)
).and using (2.6),We have
ψ
(r) ≤ψ
(r) - φ(r),i.e. φ(r) ≤ 0.
Then φ(r) = 0 by the property of function φ, and furthermore by property of function φ
φ(r) = 0 implies r = 0.
So
lim (
n 1,
n)
0.
n→∞
d x
+x
=
r
=
(2.7)
Next we claim that {xn} is Cauchy. Let
{
}
sup
(
,
) : ,
.
n j k
c
=
d x
x
j k
≥
n
Then {Cn} is decreasing.
If
lim
n0,
n→∞
C
=
Then we are done.
Assume that lim
nn→∞
C
=
C > 0.Choose ∈
<
8
C
small enough and select
N such thatfor all n ≥ N,
d x
(
n+1,
x
n)
<ε
andC
n< C +ε
.
By the definition of
C
N+1, there exists m, n ≥ N + 1 such that
d x
(
m,
x
n)
>C
n−ε
≥
C
−ε
.
Replacing
x by x
m m+1 if necessary, we have1
(
n,
m)
.
d x
x
+> − ε
C
(2.8)i.e.
d x
(
n,
x
m+1)
−d x
(
m+1,
x
m)
> C−ε
−d x
(
m+1,
x
m)
i.e.
1 1
(
n,
m)
(
n,
m)
(
m,
m)
d x
x
≥
d x
x
+−
d x
+x
> C−
ε
−d x
(
m+1,
x
m)
i.e.
d x
(
m,
x
n)
> C−ε
−d x
(
m+1,
x
m)
i.e.
d x
(
m,
x
n)
> C−ε
−ε
i.e.d x
(
m,
x
n)
> − ε
C
2
(2.9)
We may assume that m is even, n is odd.
Then
d x
(
m−1,
x
n−1)
> − ε
C
4
and since1 1 1
(
m,
m)
(
m,
n)
d x
−x
≤
d x
−x
− and1 1 1
(
n,
n)
(
m,
n).
d x
−x
≤
d x
−x
−So
1
2
min{
d x
(
m−1,
x
m), (
d x
n−1,
x
n)
}
≤
d x
(
m−1,
x
n−1)
i.e.1
2
min{
d x
(
m−1,
Sx
m−1), (
d x
n−1,
Tx
n−1)
}
≤
1 1
(
m,
n).
d x
−x
−1 1 1 1
1 1
( (
,
))
(
(
,
))
(
(
,
))
m n m n
m n
d Sx
Tx
M x
x
M x
x
− − − −
− −
ψ
≤ ψ
− ϕ
i.e.
ψ
( (
d x
m,
x
n))
= ψ
( (
d Sx
m−1,
Tx
n−1))
1 1 1 1
1 1
1 1 1 1
1 1 1 1
1 1
1 1 1 1
(
,
), (
,
),
max
(
,
),
(
,
)
(
,
)
2
(
,
), (
,
),
max
(
,
),
(
,
)
(
,
)
2
m n m m
n n
m n n m
m n m m
n n
m n n m
d x
x
d x
Sx
d x
Tx
d x
Tx
d x
Sx
d x
x
d x
Sx
d x
Tx
d x
Tx
d x
Sx
− − − − − − − − − − − − − − − − − − − −
≤ ψ
+
−ϕ
+
.1 1 1
1 1
1
1 1 1
1 1
1
(
,
), (
,
),
max
(
,
)
(
,
)
(
,
),
2
(
,
), (
,
),
max
(
,
)
(
,
)
(
,
),
2
m n m m
m n n m
n n
m n m m
m n n m
n n
d x
x
d x
x
d x
x
d x
x
d x
x
d x
x
d x
x
d x
x
d x
x
d x
x
− − − − − − − − − − − −
≤ ψ
+
−ϕ
+
. i.e.1 1 1 1
( (
d x
m,
x
n))
( (
d x
m−,
x
n−))
( (
d x
m−,
x
n−))
ψ
≤ ψ
− ϕ
We have proved that
ψ
(
C
N+1)
<ψ
( )
C
N−
φ2
C
(ifε is small enough).
This is impossible. Thus we must have C = 0.
That is, the sequence
{ }
x
n is Cauchy sequence. Since X is complete, so the sequence{ }
x
n is convergent. That is, there exists z ∈ X such thatx
n→
z as n
→ ∞
.
Moreover
x
2n→
z
and
x
2n+1→
z n
→ ∞
.
Now we prove that z is fixed point of S and T.
Since
x
n→
z
, there existsn
0∈ N such that( ,
n)
d z x
≤
1
3
d z y
( , )
for all
y
≠
z
with
n
≥
n
0.
Then we have
2 1 2 1 2 1 2 1 2 1 2
2 1 2
2 1
2 1 2 1 2 1 2 1
1
(
,
)
(
,
)
(
,
)
2
(
, )
( ,
)
2
1
( , )
( , )
( , )
( , )
(
, )
3
3
1
(
, )
(
,
)
(
, ).
2
n n n n n n
n n
n
n n n n
d x
Sx
d x
Sx
d x
x
d x
z
d z x
d y z
d y z
d y z
d y z
d x
z
d x
y
d x
Sx
d x
y
− − − − − − − − − − −
≤
≤
≤
+
≤
=
−
≤
−
≤
≤
≤
(2.10)Now suppose if d(y, Ty) ≤
d x
(
2n−1,
Sx
2n−1).
Then
1
2
min{ d(y, Ty),d x
(
2n−1,
Sx
2n−1)
}≤d x
(
2n−1, ).
y
And ifd x
(
2n−1,
Sx
2n−1)
≤ d(y, Ty),then
1
2
min{ d(y, Ty),d x
(
2n−1,
Sx
2n−1)
}≤d x
(
2n−1, ).
y
This implies (2.1), that is, we have2 1 2 1 2 1
( (
d Sx
n−,
Ty
))
(
M x
(
n−, ))
y
(
M x
(
n−, )).
y
ψ
≤ ψ
− ϕ
2 1
( (
d Sx
n−,
Ty
))
ψ
≤
2 1 2 1 2 1
2 1 2 1
2 1 2 1 2 1
2 1 2 1
(
, ), (
,
),
max
( ,
)
(
,
)
( ,
),
2
(
, ), (
,
),
max
( ,
)
(
,
)
( ,
),
2
n n m
n n
n n m
n n
d x
y d x
Sx
d y Sx
d x
Ty
d y Ty
d x
y d x
Sx
d y Sx
d x
Ty
d y Ty
− − − − − − − − − −
+
−ϕ
+
ψ
(d(z, Ty)) ≤ψ
(max{d(z, y), d(y,Ty)})
−ϕ
(max{d(z, y), d(y, Ty)})That is,
ψ
(d(z, Ty)) ≤ψ
(max{d(z, y), d(y, Ty)}).That is, d(z, Ty) ≤ max{d(z, y), d(y, Ty)}.
(2.11)
And by lemma 2.1, d(y, Ty)
≤
d(y, z),(2.12)
Thus from (2.11) and (2.12), we conclude
d(z, Ty)
≤
d(z, y) for all y ∈ X−{z}.(2.13)
Now d(y, Ty)
≤
d(y, z) + d(z, Ty)
≤
d(y, z)+ d(y, z)i.e.
1
2
d(y, Ty)≤
d(y, z).Now either d(y, Ty)
≤
d(z, Sz) or d(z, Sz)≤
d(y, Ty).If d(y, Ty)
≤
d(z, Sz), then1
2
min{ ( ,
d y Ty d z Sz
), ( ,
)}
≤
d y z
( , ).
And if d(z, Sz)
≤
d(y, Ty), then1
2
min{ ( ,
d y Ty d z Sz
), ( ,
)}
≤
d y z
( , ).
So by (2.1),
( , ), ( ,
),
( (
,
))
max
( ,
)
( ,
)
( ,
),
2
( ,
)
( ,
)
max
( , ), ( ,
), ( ,
),
.
2
d z y d y Ty
d Sz Ty
d y Sz
d z Ty
d z Sz
d y Sz
d z Ty
d z y d y Ty d z Sz
ψ
≤ ψ
+
+
− ϕ
( ,
)
( ,
)
max
( , ), ( ,
), ( ,
),
.
2
d y Sz
d z Ty
d z y d y Ty d z Sz
+
−ϕ
Take y =
x
2n.2 2 2
2 2 2
2 2 2
2 2
( ,
), (
,
),
( (
,
))
max
(
,
)
( ,
)
( ,
),
2
( ,
), (
,
),
max
(
,
)
( ,
)
.
( ,
),
2
n n n
n n n
n n n
n n
d z x
d x
Tx
d Sz Tx
d x
Sz
d z Tx
d z Sz
d z x
d x
Tx
d x
Sz
d z Tx
d z Sz
ψ
≤ ψ
+
− ϕ
+
Letting
n
→ ∞
, we have( , ), ( , ),
( (
, ))
max
( ,
)
( , )
( ,
),
2
( ,
)
( , )
max
( , ), ( , ), ( ,
),
.
2
d z z d z z
d Sz z
d z Sz
d z z
d z Sz
d z Sz
d z z
d z z d z z d z Sz
ψ
≤ ψ
+
+
− ϕ
( (
d Sz z
, ))
( (
d Sz z
, ))
d Sz z
(
, ),
ψ
≤ ψ
− ϕ
which gives z = Sz. Analogously z = Tz.
The following examples show the generality of
our results.
Example 2.1 Let X = {(0, 0), (0, 4), (4, 0), (0, 5), (5, 0), (4, 5), (5, 4)} be endowed with the metric d defined by
1 2 1 2 1 1 2 2
[( ,
), ( ,
)]
.
d x x
y y
=
x
−
y
+
x
−
y
Let S and T be such that
1 1 2
1 2
1 2
( , 0)
if
( ,
)
and
(0, 0)
if
x
x
x
S x x
x
x
≤
=
>
2 1 2
1 2
2 1 2
(
, 0)
if
( ,
)
.
(0,
)
if
x
x
x
T x x
x
x
x
≤
=
>
Then S and T do not satisfy the condition (1.1) of Theorem
verified that all the hypotheses of Theorem 2.1 are satisfied
for the maps S and T with
ψ
( )
t
=
t
and ∅(t) =1
7
t
.Corollary 2.1. Let (X, d) be a complete metric space and let S and T be maps on X. Assume that for each x, y ∈ X,
1
2
min{d(x, Sx), d(y, Ty)} ≤ d(x, y)implies
d(Sx, Ty) ≤M(x, y) − φ (M(x, y)),
where φ is defined as above. Then S and T have a common
fixed point.
Proof: It comes from Theorem 2.1 by taking
ψ
an identitymap.
Corollary 2.2. Let (X, d) be a complete metric space and let T be a map on X. Assume that for each x, y ∈ X,
1
2
d(x, Tx)} ≤ d(x, y) implies
ψ
(d(Tx, Ty)) ≤ψ
(M(x, y)) − φ (M(x,y)),
where
ψ
and φ are defined as above. Then T has a uniquefixed point.
Proof: It comes from Theorem 2.1 by taking S = T.
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