Filomat 29:6 (2015), 1189–1194
DOI 10.2298/FIL1506189K
Published by Faculty of Sciences and Mathematics,
University of Ni
ˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
A New Approach to the Study of Fixed Point Theory
for Simulation Functions
Farshid Khojasteh
a
, Satish Shukla
b
, Stojan Radenovi´c
c
a
Department of Mathematics, Arak-Branch, Islamic Azad University, Arak, Iran.
b
Department of Applied Mathematics, Shri Vaishnav Institute of Technology and Science, Gram Baroli Sanwer Road, Indore 453331, India.
c
Faculty of Mechanical Engineering, University of Belgrade, Kraljice Marije 16, 11 120 Beograd, Serbia
Abstract. Let (X, d) be a metric space and T: X → X be a mapping. In this work, we introduce the mapping
ζ: [0, ∞) × [0, ∞) → R, called the simulation function and the notion of Z-contraction with respect to ζ
which generalize the Banach contraction principle and unify several known types of contractions involving
the combination of d(Tx, Ty) and d(x, y). The related fixed point theorems are also proved.
1. Introduction and Preliminaries
Let (X, d) be a metric space and T: X → X be a mapping, then T is called a contraction (Banach
contraction) on X if
d(Tx, Ty) ≤ λd(x, y) for all x, y ∈ X,
where λ is a real such that λ ∈ [0, 1). A point x ∈ X is called a fixed point of T if Tx = x.
The well known Banach contraction principle [1] ensures the existence and uniqueness of fixed point of
a contraction on a complete metric space. After this interesting principle, several authors generalized this
principle by introducing the various contractions on metric spaces (see, e.g., [2, 4–9]). Rhoades [8], in his
work compare several contractions defined on metric spaces.
In this work, we introduce a mapping namely simulation function and the notion of Z-contraction
with respect to ζ. The Z-contraction generalize the Banach contraction and unify several known type of
contractions involving the combination of d(Tx, Ty) and d(x, y) and satisfies some particular conditions in
complete metric spaces.
2. Main Results
In this section, we define the simulation function, give some examples and prove a related fixed point
result.
2010 Mathematics Subject Classification. Primary 54H25 ; Secondary 47H10, 54C30
Keywords. Contraction mapping; Simulation function; Z-contraction; Fixed point.
Received: 06 October 2013; Accepted: 15 April 2015
Communicated by Vladimir Rako
ˇ
cevi
´
c
Research supported by Islamic Azad University of Arak
Email addresses: f-khojaste@iau-arak.ac.ir (Farshid Khojasteh), satishmathematics@yahoo.co.in (Satish Shukla),
radens@beotel.net (Stojan Radenovi
´
c)
F. Khojasteh et al. / Filomat 29:6 (2015), 1189–1194 1190
Definition 2.1. Let ζ: [0, ∞) × [0, ∞) → R be a mapping, then ζ is called a simulation function if it satisfies the
following conditions:
(ζ1) ζ(0, 0) = 0;
(ζ2) ζ(t, s) < s − t for all t, s > 0;
(ζ3) if {t
n
}, {s
n
} are sequences in (0, ∞) such that lim
n→ ∞
t
n
= lim
n→ ∞
s
n
> 0 then
lim sup
n→ ∞
ζ(t
n
, s
n
) < 0.
We denote the set of all simulation functions by Z.
Next, we give some examples of the simulation function.
Example 2.2. Let ζ
i
: [0, ∞) × [0, ∞) → R, i = 1, 2, 3 be defined by
(i) ζ
1
(t, s) = ψ(s) − φ(t) for all t, s ∈ [0, ∞), where φ, ψ: [0, ∞) → [0, ∞) are two continuous functions such that
ψ(t) = φ(t) = 0 if and only if t = 0 and ψ(t) < t ≤ φ(t) for all t > 0.
(ii) ζ
2
(t, s) = s −
f (t, s)
1(t, s)
t for all t, s ∈ [0, ∞), where f, 1 : [0, ∞) → (0, ∞) are two continuous functions with
respect to each variable such that f (t, s) > 1(t, s) for all t, s > 0.
(iii) ζ
3
(t, s) = s − ϕ(s) − t for all t, s ∈ [0, ∞), where ϕ: [0, ∞) → [0, ∞) is a continuous function such that ϕ(t) = 0
if and only if t = 0.
Then ζ
i
for i = 1, 2, 3 are simulation functions.
Definition 2.3. Let (X, d) be a metric space, T : X → X a mapping and ζ ∈ Z. Then T is called a Z-contraction
with respect to ζ if the following condition is satisfied
ζ(d(Tx, Ty), d(x, y)) ≥ 0 for all x, y ∈ X. (1)
A simple example of Z-contraction is the Banach contraction which can be obtained by taking λ ∈ [0, 1)
and ζ(t, s) = λs − t for all s, t ∈ [0, ∞) in above definition.
We now prove some properties of Z-contractions defined on a metric space.
Remark 2.4. It is clear from the definition simulation function that ζ(t, s) < 0 for all t ≥ s > 0. Therefore, if T is a
Z-contraction with respect to ζ ∈ Z then
d(Tx, Ty) < d(x, y) for all distinct x, y ∈ X.
This shows that every Z−contraction mapping is contractive, therefore it is continuous.
In the following lemma the uniqueness of fixed point of a Z-contraction is proved.
Lemma 2.5. Let (X, d) be a metric space and T : X → X be a Z-contraction with respect to ζ ∈ Z. Then the fixed
point of T in X is unique, provided it exists.
Proof. Suppose u ∈ X be a fixed point of T. If possible, let v ∈ X be another fixed point of T and it is distinct
from u, that is, Tv = v and u , v. Now it follows from (1) that
0 ≤ ζ(d(Tu, Tv), d(u, v)) = ζ(d(u, v), d(u, v)).
In view of Remark 2.4, above inequality yields a contradiction and proves result.
A self map T of a metric space (X, d) is said to be asymptotically regular at point x ∈ X if lim
n→ ∞
d(T
n
x, T
n+1
x) = 0
(see [3]).
The next lemma shows that a Z-contraction is asymptotically regular at every point of X.
F. Khojasteh et al. / Filomat 29:6 (2015), 1189–1194 1191
Lemma 2.6. Let (X, d) be a metric space and T : X → X be a Z-contraction with respect to ζ ∈ Z. Then T is
asymptotically regular at every x ∈ X.
Proof. Let x ∈ X be arbitrary. If for some p ∈ N we have T
p
x = T
p−1
x, that is, Ty = y, where y = T
p−1
x, then
T
n
y = T
n−1
Ty = T
n−1
y = . . . = Ty = y for all n ∈ N. Now for sufficient large n ∈ N we have
d(T
n
x, T
n+1
x) = d(T
n−p+1
T
p−1
x, T
n−p+2
T
p−1
x) = d(T
n−p+1
y, T
n−p+2
y)
= d(y, y) = 0,
Therefore, lim
n→ ∞
d(T
n
x, T
n+1
x) = 0.
Suppose T
n
x , T
n−1
x, for all n ∈ N, then it follows from (1) that
0 ≤ ζ(d(T
n+1
x, T
n
x), d(T
n
x, T
n−1
x))
= ζ(d(TT
n
x, TT
n−1
x), d(T
n
x, T
n−1
x))
≤ d(T
n
x, T
n−1
x) − d(T
n+1
x, T
n
x).
The above inequality shows that {d(T
n
x, T
n−1
x)} is a monotonically decreasing sequence of nonnegative
reals and so it must be convergent. Let lim
n→ ∞
d(T
n
x, T
n+1
x) = r ≥ 0. If r > 0 then since T is Z-contraction with
respect to ζ ∈ Z therefore by (ζ3), we have
0 ≤ lim sup
n→ ∞
ζ(d(T
n+1
x, T
n
x), d(T
n
x, T
n−1
x)) < 0
This contradiction shows that r = 0, that is, lim
n→ ∞
d(T
n
x, T
n+1
x) = 0. Thus T is an asymptotically regular
mapping at x.
The next lemma shows that the Picard sequence {x
n
} generated by a Z-contraction is always bounded.
Lemma 2.7. Let (X, d) be a metric space and T : X → X be a Z-contraction with respect to ζ. Then the Picard
sequence {x
n
} generated by T with initial value x
0
∈ X is a bounded sequence, where x
n
= Tx
n−1
for all n ∈ N.
Proof. Let x
0
∈ X be arbitrary and {x
n
} be the Picard sequence, that is, x
n
= Tx
n−1
for all n ∈ N. On the
contrary, assume that {x
n
} is not bounded. Without loss of generality we can assume that x
n+p
, x
n
for all
n, p ∈ N. Since {x
n
} is not bounded, there exists a subsequence {x
n
k
} such that n
1
= 1 and for each k ∈ N, n
k+1
is the minimum integer such that
d(x
n
k+1
, x
n
k
) > 1
and
d(x
m
, x
n
k
) ≤ 1 for n
k
≤ m ≤ n
k+1
− 1.
Therefore by the triangular inequality we have
1 < d(x
n
k+1
, x
n
k
) ≤ d(x
n
k+1
, x
n
k+1
−1
) + d(x
n
k+1
−1
, x
n
k
)
≤ d(x
n
k+1
, x
n
k+1
−1
) + 1.
Letting k → ∞ and using Lemma 2.6 we obtain
lim
k→ ∞
d(x
n
k+1
, x
n
k
) = 1.
By (1) we have d(x
n
k+1
, x
n
k
) ≤ d(x
n
k+1
−1
, x
n
k
−1
), therefore using the triangular inequality we obtain
1 < d(x
n
k+1
, x
n
k
) ≤ d(x
n
k+1
−1
, x
n
k
−1
)
≤ d(x
n
k+1
−1
, x
n
k
) + d(x
n
k
, x
n
k
−1
)
≤ 1 + d(x
n
k
, x
n
k
−1
).
F. Khojasteh et al. / Filomat 29:6 (2015), 1189–1194 1192
Letting k → ∞ and using Lemma 2.6 we obtain
lim
k→ ∞
d(x
n
k+1
−1
, x
n
k
−1
) = 1.
Now since T is a Z-contraction with respect to ζ ∈ Z therefore by (ζ3), we have
0 ≤ lim sup
k→ ∞
ζ(d(Tx
n
k+1
−1
, Tx
n
k
−1
), d(x
n
k+1
−1
, x
n
k
−1
))
= lim sup
k→ ∞
ζ(d(x
n
k+1
, x
n
k
), d(x
n
k+1
−1
, x
n
k
−1
)) < 0
This contradiction proves result.
In the next theorem we prove the existence of fixed point of a Z-contraction.
Theorem 2.8. Let (X, d) be a complete metric space and T: X → X be a Z-contraction with respect to ζ. Then T has
a unique fixed point u in X and for every x
0
∈ X the Picard sequence {x
n
}, where x
n
= Tx
n−1
for all n ∈ N converges
to the fixed point of T.
Proof. Let x
0
∈ X be arbitrary and {x
n
} be the Picard sequence, that is, x
n
= Tx
n−1
for all n ∈ N. We shall
show that this sequence is a Cauchy sequence. For this, let
C
n
= sup{d(x
i
, x
j
): i, j ≥ n}.
Note that the sequence {C
n
} is a monotonically decreasing sequence of positive reals and by Lemma 2.7
the sequence {x
n
} is bounded, therefore C
n
< ∞ for all n ∈ N. Thus {C
n
} is monotonic bounded sequence,
therefore convergent, that is, there exists C ≥ 0 such that lim
n→ ∞
C
n
= C. We shall show that C = 0. If C > 0
then by the definition of C
n
, for every k ∈ N there exists n
k
, m
k
such that m
k
> n
k
≥ k and
C
k
−
1
k
< d(x
m
k
, x
n
k
) ≤ C
k
.
Hence
lim
k→ ∞
d(x
m
k
, x
n
k
) = C. (2)
Using (1) and the triangular inequality we have
d(x
m
k
, x
n
k
) ≤ d(x
m
k
−1
, x
n
k
−1
)
≤ d(x
m
k
−1
, x
m
k
) + d(x
m
k
, x
n
k
) + d(x
n
k
, x
n
k
−1
).
Using Lemma 2, (2) and letting k → ∞ in the above inequality we obtain
lim
k→ ∞
d(x
m
k
−1
, x
n
k
−1
) = C. (3)
Since T is a Z-contraction with respect to ζ ∈ Z therefore using (1), (2), (3) and (ζ3), we have
0 ≤ lim sup
k→ ∞
ζ(d(x
m
k
−1
, x
n
k
−1
), d(x
m
k
, x
n
k
)) < 0
This contradiction proves that C = 0 and so {x
n
} is a Cauchy sequence. Since X is a complete space, there
exists u ∈ X such that lim
n→ ∞
x
n
= u. We shall show that the point u is a fixed point of T. Suppose Tu , u then
d(u, Tu) > 0. Again, using (1), (ζ2) and (ζ3), we have
0 ≤ lim sup
n→ ∞
ζ(d(Tx
n
, Tu), d(x
n
, u))
≤ lim sup
n→ ∞
[d(x
n
, u) − d(x
n+1
, Tu)]
= −d(u, Tu).
This contradiction shows that d(u, Tu) = 0, that is, Tu = u. Thus u is a fixed point of T. Uniqueness of the
fixed point follows from Lemma 2.5.
F. Khojasteh et al. / Filomat 29:6 (2015), 1189–1194 1193
Following example shows that the above theorem is a proper generalization of Banach contraction principle.
Example 2.9. Let X = [0, 1] and d : X × X → R be defined by d(x, y) = |x − y|. Then (X, d) is a complete metric
space. Define a mapping T : X → X as Tx =
x
x + 1
for all x ∈ X. T is a continuous function but it is not a Banach
contraction. But it is a Z−contraction with respect to ζ ∈ Z, where
ζ(t, s) =
s
s + 1
− t for all t, s ∈ [0, ∞).
Indeed, if x, y ∈ X, then
ζ(d(Tx, Ty), d(x, y)) =
d(x, y)
1 + d(x, y)
− d(Tx, Ty)
=
|x − y|
1 + |x − y|
− |
x
x + 1
−
y
y + 1
|
=
|x − y|
1 + |x − y|
− |
|x − y|
(x + 1)(y + 1)
| ≥ 0
Note that, all the conditions of Theorem 2.8 are satisfied and T has a unique fixed point u = 0 ∈ X.
In the following corollaries we obtain some known and some new results in fixed point theory via the
simulation function.
Corollary 2.10 (Banach Contraction principle [1]). Let (X, d) be a complete metric space and T : X → X be a
mapping satisfying the following condition:
d(Tx, Ty) ≤ λd(x, y) for all x, y ∈ X,
where λ ∈ [0, 1). Then T has a unique fixed point in X.
Proof. Define ζ
B
: [0, ∞) × [0, ∞) → R by
ζ
B
(t, s) = λs − t for all s, t ∈ [0, ∞).
Note that, the mapping T is a Z-contraction with respect to ζ
B
∈ Z. Therefore the result follows by taking
ζ = ζ
B
in Theorem 2.8.
Corollary 2.11 (Rhoades type). Let (X, d) be a complete metric space and T : X → X be a mapping satisfying the
following condition:
d(Tx, Ty) ≤ d(x, y) − ϕ(d(x, y)) for all x, y ∈ X,
where ϕ : [0, ∞) → [0, ∞) is lower semi continuous function and ϕ
−1
(0) = {0}. Then T has a unique fixed point in X.
Proof. Define ζ
R
: [0, ∞) × [0, ∞) → R by
ζ
R
(t, s) = s − ϕ(s) − t for all s, t ∈ [0, ∞).
Note that, the mapping T is a Z-contraction with respect to ζ
R
∈ Z. Therefore the result follows by taking
ζ = ζ
R
in Theorem 2.8.
Remark 2.12. Note that, in the [9] the function ϕ is assumed to be continuous and nondecreasing and lim
t→ ∞
ψ(t) = ∞.
In Corollary 2.11 we replace these conditions by lower semi continuity of ϕ. Therefore our result is stronger than the
original version of Rhoades [9].