CUBO, A Mathematical Journal

Vol.22, N◦02, (273–288). August 2020

http://dx.doi.org/10.4067/S0719-06462020000200273

Received: 11 September, 2019 | Accepted: 27 July, 2020

Fixed point theorems on cone S-metric spaces using implicit
relation

G. S. Saluja

Department of Mathematics,

Govt. Kaktiya P. G., College Jagdalpur,

Jagdalpur - 494001 (C.G.), India.

saluja1963@gmail.com

ABSTRACT

In this paper, we establish some fixed point theorems in the framework of cone S-metric

spaces using implicit relation. Our results extend, unify and generalize several results

from the current existing literature. Especially, they extend the corresponding results

of Sedghi and Dung [24] to the setting of complete cone S-metric spaces.

RESUMEN

En este art́ıculo, establecemos algunos teoremas de punto fijo en el marco de espa-

cios S-métricos del cono usando una relación impĺıcita. Nuestros resultados extienden,

unifican y generalizan diversos resultados de la literatura actual existente. Especial-

mente, extienden los resultados correspondientes de Sedghi y Dung [24] en el contexto

de espacios S-métricos de cono completo.

Keywords and Phrases: Fixed point, implicit relation, cone S-metric space, cone.

2020 AMS Mathematics Subject Classification: 47H10, 54H25.

c©2020 by the author. This open access article is licensed under a Creative
Commons Attribution-NonCommercial 4.0 International License.

http://dx.doi.org/10.4067/S0719-06462020000200273


274 G. S. Saluja CUBO
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1 Introduction and Preliminaries

In 2007, Huang and Zhang [8] introduced the concept of cone metric spaces as a generalization of

metric spaces by replacing the set of real numbers by a general Banach space E which is partially

ordered with respect to a cone P ⊂ E and establish some fixed point theorems for contractive

mappings in normal cone metric spaces.

In 2012, Sedghi et al. [23] introduced the concept of S-metric space which is different from

other space and proved fixed point theorems in S-metric space. They also give some examples of

S-metric space which shows that S-metric space is different from other spaces.

In 2016, Rahman and Sarwar [20] have discussed the fixed point results of Altman integral

type mappings in S-metric spaces and in the same year Ozgur and Tas [14] have studied new

contractive conditions of integral type in complete S-spaces.

Recently, Dhamodharan and Krishnakumar [6] introduced the concept of cone S-metric space

and proved some fixed point theorems using various contractive conditions in the above said space.

Due to great importance of the fixed point theory, it is immensely interesting to study fixed

point theorems on different concepts. Many authors studied the fixed points for mappings satisfying

contractive conditions in complete S-metric spaces (see, e.g., [6, 11, 13, 14, 20, 23, 25, 26]) and

others).

Popa [15] and [16], on the other hand, considered an implicit contraction type condition

instead of the usual explicit condition. This direction of research produced a consistent literature

on fixed point and common fixed point theorems in various ambient spaces. For more details see

[1, 2, 3, 9, 17, 18, 19, 24].

Motivated and inspired by Popa [15, 16], Sedghi and Dung [24] and others, this paper is aimed

to study and establish some fixed point theorems in the setting of complete cone S-metric spaces

under implicit contractive condition which is used in [24]. Following the current literature there

is ample vicinity to explore and improve this new avenue of research area. Here, we prove an

important result of cone S-metric space and then obtain some classical fixed point theorems as

corollaries, for example, Banach’s contraction mapping principle, Kannan’s fixed point theorem,

Chatterjae’s fixed point theorem, Reich fixed point theorem and Ćirić’s fixed point theorem in this

setting. Our results extend and generalize several results from the existing literature, especially,

the results of Sedghi and Dung [24] from complete S-metric spaces to the setting of complete cone

S-metric spaces.

The present work is to encouraged by its possible application, especially in discrete models

for numerical analysis, where iterative schemes are extensively used due to their versatility for

computer simulation. These models play an important role in applied mathematics.



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Fixed point theorems on cone S-metric spaces using implicit relation 275

We need the following definitions and lemmas in the sequel.

Definition 1. ([8]) Let E be a real Banach space. A subset P of E is called a cone whenever the

following conditions hold:

(c1) P is closed, nonempty and P 6= {0};

(c2) a,b ∈ R, a,b ≥ 0 and x,y ∈ P imply ax + by ∈ P;

(c3) P ∩ (−P) = {0}.

Given a cone P ⊂ E, we define a partial ordering ≤ in E with respect to P by x ≤ y if and

only if y − x ∈ P. We shall write x < y to indicate that x ≤ y but x 6= y, while x ≪ y will stand

for y − x ∈ P0, where P0 stands for the interior of P. If P0 6= ∅ then P is called a solid cone (see

[28]).

There exist two kinds of cones- normal (with the normal constant K) and non-normal ones

([7]).

Let E be a real Banach space, P ⊂ E a cone and ≤ partial ordering defined by P . Then P is

called normal if there is a number K > 0 such that for all x,y ∈ P ,

0 ≤ x ≤ y imply ‖x‖ ≤ K‖y‖, (1.1)

or equivalently, if (∀n) xn ≤ yn ≤ zn and

lim
n→∞

xn = lim
n→∞

zn = x imply lim
n→∞

yn = x. (1.2)

The least positive number K satisfying (1.1) is called the normal constant of P .

The cone P is called regular if every increasing sequence which is bounded from above is

convergent, that is, if {xn} is a sequence such that x1 ≤ x2 ≤ · · · ≤ xn ≤ · · · ≤ y for some y ∈ E,

then there is x ∈ E such that ‖xn − x‖ → 0 as n → ∞. Equivalently, the cone P is regular if and

only if every decreasing sequence which is bounded from below is convergent. It is well known that

a regular cone is a normal cone. Suppose E is a Banach space, P is a cone in E with int(P) 6= ∅

and ≤ is partial ordering in E with respect to P .

Example 1. ([12]) Let K > 1 be given. Consider the real vector space

E =
{

ax + b : a,b ∈ R;x ∈
[

1 −
1

K
,1
]}

with supremum norm and the cone

P =
{

ax + b ∈ E : a ≥ 0,b ≥ 0
}

in E. The cone P is regular and so normal.



276 G. S. Saluja CUBO
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Definition 2. ([8, 29]) Let X be a nonempty set. Suppose that the mapping d: X × X → E

satisfies:

(CM1) 0 ≤ d(x,y) for all x,y ∈ X with x 6= y and d(x,y) = 0 ⇔ x = y;

(CM2) d(x,y) = d(y,x) for all x,y ∈ X;

(CM3) d(x,y) ≤ d(x,z) + d(z,y) for all x,y,z ∈ X.

Then d is called a cone metric [8] on X and (X,d) is called a cone metric space [8] or simply

CMS.

The concept of a cone metric space is more general than that of a metric space, because each

metric space is a cone metric space where E = R and P = [0,+∞).

Lemma 1. ([22]) Every regular cone is normal.

Example 2. ([8]) Let E = R2, P = {(x,y) ∈ R2 : x ≥ 0,y ≥ 0}, X = R and d: X × X → E

defined by d(x,y) = (|x − y|,α|x − y|), where α ≥ 0 is a constant. Then (X,d) is a cone metric

space with normal cone P where K = 1.

Clearly, the above example shows that the class of cone metric spaces contains the class of

metric spaces.

Definition 3. ([23, 14]) Let X be a nonempty set and S : X3 → [0,∞) be a function satisfying

the following conditions for all x, y, z, t ∈ X:

(SM1) S(x,y,z) ≥ 0;

(SM2) S(x,y,z) = 0 if and only if x = y = z;

(SM3) S(x,y,z) ≤ S(x,x,t) + S(y,y,t) + S(z,z,t).

Then the function S is called an S-metric on X and the pair (X,S) is called an S-metric

space or simply SMS.

Example 3. ([27]) Let X be a nonempty set and d be the ordinary metric on X. Then S(x,y,z) =

d(x,z) + d(y,z) is an S-metric on X.

Example 4. ([23]) Let X = Rn and ‖.‖ a norm on X, then S(x,y,z) = ‖y + z − 2x‖ + ‖y − z‖

is an S-metric on X.

Example 5. ([23]) Let X = Rn and ‖.‖ a norm on X, then S(x,y,z) = ‖x − z‖ + ‖y − z‖ is an

S-metric on X.

Example 6. ([24]) Let X = R be the real line. Then S(x,y,z) = ‖x − z‖ + ‖y − z‖ for all

x,y,z ∈ R is an S-metric on X. This S-metric on X is called the usual S-metric on X.



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Fixed point theorems on cone S-metric spaces using implicit relation 277

Definition 4. ([6]) Suppose that E is a real Banach space, P is a cone in E with intP 6= ∅ and

≤ is partial ordering with respect to P. Let X be a nonempty set and let the function S : X3 → E

satisfy the following conditions:

(CSM1) S(x,y,z) ≥ 0;

(CSM2) S(x,y,z) = 0 if and only if x = y = z;

(CSM3) S(x,y,z) ≤ S(x,x,a) + S(y,y,a) + S(z,z,a), ∀x,y,z,a ∈ X.

Then the function S is called a cone S-metric on X and the pair (X,S) is called a cone

S-metric space or simply CSMS.

Example 7. ([6]) Let E = R2, P = {(x,y) ∈ R2 : x ≥ 0,y ≥ 0}, X = R and d be the ordinary

metric on X. Then the function S : X3 → E defined by S(x,y,z) =
(

d(x,z) + d(y,z),α(d(x,z) +

d(y,z))
)

, where α > 0 is a cone S-metric on X.

Lemma 2. ([6]) Let (X,S) be a cone S-metric space. Then we have S(x,x,y) = S(y,y,x).

Definition 5. ([6]) Let (X,S) be a cone S-metric space.

(i) A sequence {un} in X converges to u if and only if S(un,un,u) → 0 as n → ∞, that is,

there exists n0 ∈ N such that for all n ≥ n0, S(un,un,u) ≪ c for each c ∈ E, 0 ≪ c. We denote

this by limn→∞ un = u or limn→∞ S(un,un,u) = 0.

(ii) A sequence {un} in X is called a Cauchy sequence if S(un,un,um) → 0 as n,m → ∞,

that is, there exists n0 ∈ N such that for all n,m ≥ n0, S(un,un,um) ≪ c for each c ∈ E, 0 ≪ c.

(iii) The cone S-metric space (X,S) is called complete if every Cauchy sequence is convergent.

In the following lemma, we see the relationship between a cone metric and a cone S-metric.

Lemma 3. ([6]) Let (X,d) be a cone metric space. Then, the following properties are satisfied:

(1) S(u,v,z) = d(u,z) + d(v,z) for all u,v,z ∈ X, is a cone S-metric on X.

(2) un → u in (X,d) if and only if un → u in (X,Sd).

(3) {un} is Cauchy in (X,d) if and only if {un} is Cauchy in (X,Sd).

(4) (X,d) is complete if and only if (X,Sd) is complete.

Lemma 4. ([24]) Let f : X → Y be a map from an S-metric space X to an S-metric space Y .

Then f is continuous at x ∈ X if and only if f(xn) → f(x) whenever xn → x.

Now, we introduce an implicit relation to investigate some fixed point theorems on cone S-

metric spaces. Let ψ be the family of all continuous functions of five variables φ: R5
+
→ R+. For

some k ∈ [0,1), we consider the following conditions.



278 G. S. Saluja CUBO
22, 2 (2020)

(A1) For all x,y,z ∈ R+, if y ≤ φ(x,x,y,z,0) with z ≤ 2x + y, then y ≤ kx.

(A2) For all y ∈ R+, if y ≤ φ(y,0,0,y,y), then y = 0.

(A3) If xi ≤ yi + zi for all xi,yi,zi ∈ R+, i ≤ 5, then

φ(x1, . . . ,x5) ≤ φ(y1, . . . ,y5) + φ(z1, . . . ,z5).

Moreover, for all y ∈ X, φ(0,0,2y,y,0) ≤ ky.

Remark 1. Note that the coefficient k in conditions (A1) and (A3) may be different, for example,

k1 and k3 respectively. But we may assume that they are equal by taking k = max{k1,k3}.

2 Main Results

In this section, we shall prove some fixed point theorems using implicit relation in the setting of

cone S-metric spaces.

Theorem 1. Let T be a self-map on a complete cone S-metric space (X,S), P be a normal cone

with normal constant K and

S(Tx,Tx,Ty) ≤ φ
(

S(x,x,y),S(x,x,Tx),S(y,y,Ty),

S(x,x,Ty),S(y,y,Tx)
)

(2.1)

for all x,y ∈ X and some φ ∈ ψ. Then we have

(1) If φ satisfies the condition (A1), then T has a fixed point. Moreover, for any x0 ∈ X and

the fixed point x, we have

S(Txn,Txn,x) ≤
( 2kn

1 − k

)

S(x0,x0,Tx0).

(2) If φ satisfies the condition (A2) and T has a fixed point, then the fixed point is unique.

(3) If φ satisfies the condition (A3) and T has a fixed point x, then T is continuous at x.



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Fixed point theorems on cone S-metric spaces using implicit relation 279

Proof. (1) For each x0 ∈ X and n ∈ N, put xn+1 = Txn. It follows from (2.1) and Lemma 2 that

S(xn+1,xn+1,xn+2) = S(Txn,Txn,Txn+1)

≤ φ
(

S(xn,xn,xn+1),S(xn,xn,Txn),S(xn+1,xn+1,Txn+1),

S(xn,xn,Txn+1),S(xn+1,xn+1,Txn)
)

= φ
(

S(xn,xn,xn+1),S(xn,xn,xn+1),S(xn+1,xn+1,xn+2),

S(xn,xn,xn+2),S(xn+1,xn+1,xn+1)
)

= φ
(

S(xn,xn,xn+1),S(xn,xn,xn+1),S(xn+1,xn+1,xn+2),

S(xn,xn,xn+2),0
)

. (2.2)

By condition (CSM3) and Lemma 2, we have

S(xn,xn,xn+2) ≤ 2S(xn,xn,xn+1) + S(xn+2,xn+2,xn+1)

= 2S(xn,xn,xn+1) + S(xn+1,xn+1,xn+2). (2.3)

Since φ satisfies the condition (A1), there exists k ∈ [0,1) such that

S(xn+1,xn+1,xn+2) ≤ kS(xn,xn,xn+1) ≤ k
n+1S(x0,x0,x1). (2.4)

Thus for all n < m, by using (CSM3), Lemma 2 and equation (2.4), we have

S(xn,xn,xm) ≤ 2S(xn,xn,xn+1) + S(xm,xm,xn+1)

= 2S(xn,xn,xn+1) + S(xn+1,xn+1,xm)

. . .

≤ 2[kn + · · · + km−1]S(x0,x0,x1)

≤
( 2kn

1 − k

)

S(x0,x0,x1).

This implies that

‖S(xn,xn,xm)‖ ≤
(2knK

1 − k

)

‖S(x0,x0,x1)‖.

Taking the limit as n,m → ∞, we get

‖S(xn,xn,xm)‖ → 0,

since 0 < k < 1. Thus, we have S(xn,xn,xm) → 0 as n,m → ∞.

This shows that the sequence {xn} is a Cauchy sequence in the complete cone S-metric space

(X,S). By the completeness of the space, we have limn→∞ xn = x ∈ X. Moreover, taking the

limit as m → ∞ we get

S(xn,xn,x) ≤
(2kn+1

1 − k

)

S(x0,x0,x1).



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It implies that

S(Txn,Txn,x) ≤
( 2kn

1 − k

)

S(x0,x0,Tx0).

Now we prove that x is a fixed point of T . By using inequality (2.1) again we obtain

S(xn+1,xn+1,Tx) = S(Txn,Txn,Tx)

≤ φ
(

S(xn,xn,x),S(xn,xn,Txn),S(x,x,Tx),

S(xn,xn,Tx),S(x,x,Txn)
)

= φ
(

S(xn,xn,x),S(xn,xn,xn+1),S(x,x,Tx),

S(xn,xn,Tx),S(x,x,xn+1)
)

.

Note that φ ∈ ψ, then using Lemma 3 and taking the limit as n → ∞, we get

S(x,x,Tx) ≤ φ
(

0,0,S(x,x,Tx),S(x,x,Tx),0
)

.

Since φ satisfies the condition (A1), then S(x,x,Tx) ≤ k.0 = 0. This shows that x = Tx. Thus x

is a fixed point of T .

(2) Let x1,x2 be fixed points of T . We shall prove that x1 = x2. It follows from equation

(2.1) and Lemma 2 that

S(x1,x1,x2) = S(Tx1,Tx1,Tx2)

≤ φ
(

S(x1,x1,x2),S(x1,x1,Tx1),S(x2,x2,Tx2),

S(x1,x1,Tx2),S(x2,x2,Tx1)
)

= φ
(

S(x1,x1,x2),S(x1,x1,x1),S(x2,x2,x2),

S(x1,x1,x2),S(x2,x2,x1)
)

= φ
(

S(x1,x1,x2),0,0,S(x1,x1,x2),S(x2,x2,x1)
)

= φ
(

S(x1,x1,x2),0,0,S(x1,x1,x2),S(x1,x1,x2)
)

.

Since φ satisfies the condition (A2), then S(x1,x1,x2) = 0. This shows that x1 = x2. Thus the

fixed point of T is unique.

(3) Let x be the fixed point of T and yn → x ∈ X. By Lemma 4, we need to prove that



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Fixed point theorems on cone S-metric spaces using implicit relation 281

Tyn → Tx. It follows from inequality (2.1) and Lemma 2 that

S(x,x,Tyn) = S(Tx,Tx,Tyn)

≤ φ
(

S(x,x,yn),S(x,x,Tx),S(yn,yn,Tyn),

S(x,x,Tyn),S(yn,yn,Tx)
)

= φ
(

S(x,x,yn),S(x,x,x),S(yn,yn,Tyn),

S(x,x,Tyn),S(yn,yn,x)
)

= φ
(

S(x,x,yn),0,S(Tyn,Tyn,yn),

S(Tyn,Tyn,x),S(x,x,yn)
)

.

Since φ satisfies the condition (A3), by Lemma 2 and (CSM3), we have

S(Tyn,Tyn,yn) ≤ 2S(Tyn,Tyn,x) + S(yn,yn,x)

= 2S(Tyn,Tyn,x) + S(x,x,yn)

then we have

S(x,x,Tyn) ≤ φ
(

S(x,x,yn),0,0,0,S(x,x,yn)
)

+φ
(

0,0,2S(Tyn,Tyn,x),S(Tyn,Tyn,x),0
)

≤ φ
(

S(x,x,yn),0,0,0,S(x,x,yn)
)

+kS(Tyn,Tyn,x)

= φ
(

S(x,x,yn),0,0,0,S(x,x,yn)
)

+kS(x,x,Tyn). (by Lemma 2)

Therefore

S(x,x,Tyn) ≤
( 1

1 − k

)

φ
(

S(x,x,yn),0,0,0,S(x,x,yn)
)

.

Note that φ ∈ ψ, hence taking the limit as n → ∞, we get S(x,x,Tyn) → 0. This shows that

Tyn → x = Tx. This completes the proof.

Next, we give some analogues of fixed point theorems in metric spaces for cone S-metric

spaces by combining Theorem 1 with φ ∈ ψ and φ satisfies the conditions (A1), (A2) and (A3).

The following corollary is an analogue of Banach’s contraction principle.

Corollary 1. Let (X,S) be a complete cone S-metric space and P be a normal cone with normal

constant K. Suppose that the mapping T : X → X satisfies the following condition:

S(Tx,Tx,Ty) ≤ hS(x,x,y)



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for all x,y ∈ X, where h ∈ [0,1) is a constant. Then T has a unique fixed point in X. Moreover,

T is continuous at the fixed point.

Proof. The assertion follows using Theorem 1 with φ(x,y,z,s,t) = hx for some h ∈ [0,1) and all

x,y,z,s,t ∈ R+.

The following corollary is an analogue of R. Kannan’s result [10].

Corollary 2. Let (X,S) be a complete cone S-metric space and P be a normal cone with normal

constant K. Suppose that the mapping T : X → X satisfies the following condition:

S(Tx,Tx,Ty) ≤ q [S(x,x,Tx) + S(y,y,Ty)]

for all x,y ∈ X, where q ∈ [0, 1
2
) is a constant. Then T has a unique fixed point in X. Moreover,

T is continuous at the fixed point.

Proof. The assertion follows using Theorem 1 with φ(x,y,z,s,t) = q(y + z) for some q ∈ [0, 1
2
)

and all x,y,z,s,t ∈ R+. Indeed, φ is continuous. First, we have φ(x,x,y,z,0) = q(x + y). So,

if y ≤ φ(x,x,y,z,0) with z ≤ 2x + y, then y ≤
(

q

1−q

)

x with
(

q

1−q

)

< 1. Thus, T satisfies the

condition (A1).

Next, if y ≤ φ(y,0,0,y,y), then y = 0. Thus, T satisfies the condition (A2).

Finally, if xi ≤ yi + zi for i ≤ 5, then

φ(x1, . . . ,x5) = q(x2 + x3)

≤ q[(y2 + z2) + (y3 + z3)]

= q(y2 + y3) + q(z2 + z3)

= φ(y1, . . . ,y5) + φ(z1, . . . ,z5).

Moreover

φ(0,0,2y,y,0) = q(0 + 2y) = 2qy

where 2q < 1. Thus, T satisfies the condition (A3).

The following corollary is an analogue of S. K. Chatterjae’s result [4].

Corollary 3. Let (X,S) be a complete cone S-metric space and P be a normal cone with normal

constant K. Suppose that the mapping T : X → X satisfies the following condition:

S(Tx,Tx,Ty) ≤ p [S(x,x,Ty) + S(y,y,Tx)]

for all x,y ∈ X, where p ∈ [0, 1
2
) is a constant. Then T has a unique fixed point in X. Moreover,

T is continuous at the fixed point.



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Fixed point theorems on cone S-metric spaces using implicit relation 283

Proof. The assertion follows using Theorem 1 with φ(x,y,z,s,t) = p(s + t) for some p ∈ [0, 1
2
)

and all x,y,z,s,t ∈ R+. Indeed, φ is continuous. First, we have φ(x,x,y,z,0) = p(z + 0). So,

if y ≤ φ(x,x,y,z,0) with z ≤ 2x + y, then y ≤
(

2p

1−p

)

x with
(

2p

1−p

)

< 1. Thus, T satisfies the

condition (A1).

Next, if y ≤ φ(y,0,0,y,y) = 2py, then y = 0 since p < 1
2
. Thus, T satisfies the condition (A2).

Finally, if xi ≤ yi + zi for i ≤ 5, then

φ(x1, . . . ,x5) = p(x4 + x5)

≤ p[(y4 + z4) + (y5 + z5)]

= p(y4 + y5) + p(z4 + z5)

= φ(y1, . . . ,y5) + φ(z1, . . . ,z5).

Moreover

φ(0,0,2y,y,0) = p(y + 0) = py

where p < 1. Thus, T satisfies the condition (A3).

The following corollary is an analogue of S. Reich’s result [21].

Corollary 4. Let (X,S) be a complete cone S-metric space and P be a normal cone with normal

constant K. Suppose that the mapping T : X → X satisfies the following condition:

S(Tx,Tx,Ty) ≤ aS(x,x,y) + bS(x,x,Tx) + cS(y,y,Ty)

for all x,y ∈ X, where a,b,c ≥ 0 are constants with a+b+c < 1. Then T has a unique fixed point

in X. Moreover, if c < 1
2
, then T is continuous at the fixed point.

Proof. The assertion follows using Theorem 1 with φ(x,y,z,s,t) = ax+by+cz for some a,b,c ≥ 0

are constants with a + b + c < 1 and all x,y,z,s,t ∈ R+. Indeed, φ is continuous. First, we have

φ(x,x,y,z,0) = ax + bx + cy. So, if y ≤ φ(x,x,y,z,0) with z ≤ 2x + y, then y ≤
(

a+b
1−c

)

x with
(

a+b
1−c

)

< 1. Thus, T satisfies the condition (A1).

Next, if y ≤ φ(y,0,0,y,y) = ay, then y = 0 since a < 1. Thus, T satisfies the condition (A2).

Finally, if xi ≤ yi + zi for i ≤ 5, then

φ(x1, . . . ,x5) = ax1 + bx2 + cx3

≤ a(y1 + z1) + b(y2 + z2) + c(y3 + z3)

= (ay1 + by2 + cy3) + (az1 + bz2 + cz3)

= φ(y1, . . . ,y5) + φ(z1, . . . ,z5).



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Moreover

φ(0,0,2y,y,0) = a.0 + b.0 + c.2y = 2cy

where 2c < 1. Thus, T satisfies the condition (A3).

The following corollary is an analogue of L. B. Ćirić’s result [5].

Corollary 5. Let (X,S) be a complete cone S-metric space and P be a normal cone with normal

constant K. Suppose that the mapping T : X → X satisfies the following condition:

S(Tx,Tx,Ty) ≤ h max
{

S(x,x,y),S(x,x,Tx),S(y,y,Ty),

S(x,x,Ty),S(y,y,Tx)
}

for all x,y ∈ X, where h ∈ [0, 1
3
) is a constant. Then T has a unique fixed point in X. Moreover,

T is continuous at the fixed point.

Proof. The assertion follows using Theorem 1 with φ(x,y,z,s,t) = h max{x,

y,z,s,t} for some h ∈ [0, 1
3
) and all x,y,z,s,t ∈ R+. Indeed, φ is continuous. First, we have

φ(x,x,y,z,0) = h max{x,x,y,z,0}. So, if y ≤ φ(x,x,y,z,0) with z ≤ 2x + y, then y ≤ hx or

y ≤ hz ≤ h(2x + y). Then y ≤ kx with k = max
{

h, 2h
1−h

}

< 1. Thus, T satisfies the condition

(A1).

Next, if y ≤ φ(y,0,0,y,y) = h max{y,0,0,y,y} = hy, then y = 0 since h < 1
3
. Thus, T

satisfies the condition (A2).

Finally, if xi ≤ yi + zi for i ≤ 5, then

φ(x1, . . . ,x5) = h max{x1, . . . ,x5}

≤ h max{y1 + z1, . . . ,y5 + z5}

≤ h max{y1, . . . ,y5} + h max{z1, . . . ,z5}

= φ(y1, . . . ,y5) + φ(z1, . . . ,z5).

Moreover

φ(0,0,2y,y,0) = h max{0,0,2y,y,0} = 2hy

where 2h < 1. Thus, T satisfies the condition (A3).

Example 8. Let E = R2, the Euclidean plane, P = {(x,y) ∈ R2 : x ≥ 0,y ≥ 0} a normal cone

in E and X = R. Then the function S : X3 → E defined by S(x,y,z) = |x − z| + |y − z| for all

x,y,z ∈ X. Then (X,S) is a cone S-metric space. Now, we consider the mapping T : X → X by

T(x) = x
2
and {xn} = {

1

2n
} for all n ∈ N is a sequence converging to zero.



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Fixed point theorems on cone S-metric spaces using implicit relation 285

Result Analysis

(1) Taking x = xn−1 and y = xn in inequality (2.1) and using (CSM3), we have

S(xn,xn,xn+1) = S(Txn−1,Txn−1,Txn)

≤ φ
(

S(xn−1,xn−1,xn),S(xn−1,xn−1,Txn−1),S(xn,xn,Txn),

S(xn−1,xn−1,Txn),S(xn,xn,Txn−1)
)

= φ
(

S(xn−1,xn−1,xn),S(xn−1,xn−1,xn),S(xn,xn,xn+1),

S(xn−1,xn−1,xn+1),S(xn,xn,xn)
)

= φ
(

S(xn−1,xn−1,xn),S(xn−1,xn−1,xn),S(xn,xn,xn+1),

S(xn−1,xn−1,xn+1),0
)

≤ φ
(

S(xn−1,xn−1,xn),S(xn−1,xn−1,xn),S(xn,xn,xn+1),

2S(xn−1,xn−1,xn) + S(xn,xn,xn+1),0
)

.

Since φ satisfies the condition (A1), so there exists k ∈ [0,1) such that

S(xn,xn,xn+1) ≤ kS(xn−1,xn−1,xn)

or

2
(

xn − xn+1
)

) ≤ k.2
(

xn−1 − xn
)

or

( 1

2n
−

1

2n+1

)

) ≤ k
( 1

2n−1
−

1

2n

)

or

k ≥
1

2
.

If we take 0 < k < 1, then inequality (2.1) is satisfied. Thus all the conditions of Theorem 1 are

satisfied. Hence by Theorem 1, T has a unique fixed point. Here, note that ′0′ is the unique fixed

point of T.

(2) Let {yn} = {
1

3n
} be a sequence in X converging to the fixed point z = 0, then we have to

show that Tyn → z as n → ∞, that is, T is continuous at the fixed point of T, we have

lim
n→∞

Tyn = T( lim
n→∞

yn) = T(0) = 0 = z.



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That is,

Tyn → z as n → ∞.

Thus, T is continuous at the fixed point of T.

Example 9. Let E = R2, the Euclidean plane, P = {(x,y) ∈ R2 : x ≥ 0,y ≥ 0} a normal cone

in E and X = R. Then the function S : X3 → E defined by S(x,y,z) = |x − z| + |y − z| for all

x,y,z ∈ X. Then (X,S) is a cone S-metric space. Now, we consider the mapping T : X → X by

T(x) = x
3
. Then

S(Tx,Tx,Ty) = |Tx − Ty| + |Tx − Ty|

= 2|Tx − Ty| = 2
∣

∣

∣

(

x

3

)

−
(

y

3

)
∣

∣

∣

=
2

3
|x − y|

=
1

3

(

2|x − y|
)

≤
1

2

(

2|x − y|
)

= hS(x,x,y)

where h = 1
2
< 1. Thus T satisfies all the conditions of Corollary 1 and clearly 0 ∈ X is the unique

fixed point of T.

3 Conclusion

In this paper, we establish some fixed point theorems using implicit relation in the framework of

complete cone S-metric spaces. Our results extend, unify and generalize several results from the

existing literature. Especially, they extend the corresponding results of Sedghi and Dung [24] from

complete S-metric spaces to the setting of complete cone S-metric spaces. However, these results

have vast potential in solving various nonlinear problems in functional analysis, differential and

integral equations, computer science and engineering.

4 Acknowledgement

The author is grateful to the anonymous referees for their careful reading and valuable suggestions

to improve the manuscript.



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Fixed point theorems on cone S-metric spaces using implicit relation 287

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	Introduction and Preliminaries
	Main Results
	Conclusion
	Acknowledgement