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Article

On the Fixed-Circle Problem and Khan Type Contractions

1
Department of Mathematics and General Sciences, Prince Sultan University, 11586 Riyadh, Saudi Arabia
2
Department of Mathematics, Balıkesir University, 10145 Balıkesir, Turkey
*
Author to whom correspondence should be addressed.
Axioms 2018, 7(4), 80; https://doi.org/10.3390/axioms7040080
Submission received: 18 October 2018 / Revised: 31 October 2018 / Accepted: 5 November 2018 / Published: 8 November 2018
(This article belongs to the Special Issue Mathematical Analysis and Applications II)

Abstract

:
In this paper, we consider the fixed-circle problem on metric spaces and give new results on this problem. To do this, we present three types of F C -Khan type contractions. Furthermore, we obtain some solutions to an open problem related to the common fixed-circle problem.
MSC:
Primary: 47H10; Secondary: 54H25, 55M20, 37E10

1. Introduction

Recently, the fixed-circle problem has been considered for metric and some generalized metric spaces (see [1,2,3,4,5,6] for more details). For example, in [1], some fixed-circle results were obtained using the Caristi type contraction on a metric space. Using Wardowski’s technique and some classical contractive conditions, new fixed-circle theorems were proved in [5,6]. In [2,3], the fixed-circle problem was studied on an S-metric space. In [7], a new fixed-circle theorem was proved using the modified Khan type contractive condition on an S-metric space. Some generalized fixed-circle results with geometric viewpoint were obtained on S b -metric spaces and parametric N b -metric spaces (see [8,9] for more details, respectively). Also, it was proposed to investigate some fixed-circle theorems on extended M b -metric spaces [10]. On the other hand, an application of the obtained fixed-circle results was given to discontinuous activation functions on metric spaces (see [1,4,11]). Hence it is important to study new fixed-circle results using different techniques.
Let ( X , d ) be a metric space and C x 0 , r = x X : d ( x , x 0 ) = r be any circle on X. In [5], it was given the following open problem.
Open Problem C C : What is (are) the condition(s) to make any circle C x 0 , r as the common fixed circle for two (or more than two) self-mappings?
In this paper, we give new results to the fixed-circle problem using Khan type contractions and to the above open problem using both of Khan and Ćirić type contractions on a metric space. In Section 2, we introduce three types of F C -Khan type contractions and obtain new fixed-circle results. In Section 3, we investigate some solutions to the above Open Problem C C . In addition, we construct some examples to support our theoretical results.

2. New Fixed-Circle Theorems

In this section, using Khan type contractions, we give new fixed-circle theorems (see [12,13,14,15] for some Khan type contractions used to obtain fixed-point theorems). At first, we recall the following definitions.
Definition 1
([16]). Let F be the family of all functions F : ( 0 , ) R such that
( F 1 ) F is strictly increasing,
( F 2 ) For each sequence α n n = 1 of positive numbers, lim n α n = 0 if and only if lim n F ( α n ) = ,
( F 3 ) There exists k ( 0 , 1 ) such that lim α 0 + α k F ( α ) = 0 .
Definition 2
([16]). Let ( X , d ) be a metric space. A mapping T : X X is said to be an F-contraction on ( X , d ) , if there exist F F and τ ( 0 , ) such that
d ( T x , T y ) > 0 τ + F ( d ( T x , T y ) ) F ( d ( x , y ) ) ,
for all x , y X .
Definition 3
([15]) .Let F k be the family of all increasing functions F : ( 0 , ) R , that is, for all x , y ( 0 , ) , if x < y then F ( x ) F ( y ) .
Definition 4
([15]). Let ( X , d ) be a metric space and T : X X be a self-mapping. T is said to be an F-Khan-contraction if there exist F F k and t > 0 such that for all x , y X if max d ( T y , x ) , d ( T x , y ) 0 then T x T y and
t + F ( d ( T x , T y ) ) F d ( T x , x ) d ( T y , x ) + d ( T y , y ) d ( T x , y ) max d ( T y , x ) , d ( T x , y ) ,
and if max d ( T y , x ) , d ( T x , y ) = 0 then T x = T y .
Now we modify the definition of an F-Khan-contractive condition, which is used to obtain a fixed point theorem in [15], to get new fixed-circle results. Hence, we define the notion of an F C -Khan type I contractive condition as follows.
Definition 5.
Let ( X , d ) be a metric space and T : X X be a self-mapping. T is said to be an F C -Khan type I contraction if there exist F F k , t > 0 and x 0 X such that for all x X if the following condition holds
max d ( T x 0 , x 0 ) , d ( T x , x ) 0 ,
then
t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) max d ( T x 0 , x 0 ) , d ( T x , x ) ,
where h 0 , 1 2 and if max d ( T x 0 , x 0 ) , d ( T x , x ) = 0 then T x = x .
One of the consequences of this definition is the following proposition.
Proposition 1.
Let ( X , d ) be a metric space. If a self-mapping T on X is an F C -Khan type I contraction with x 0 X then we get T x 0 = x 0 .
Proof. 
Let T x 0 x 0 . Then using the hypothesis, we find
max d ( T x 0 , x 0 ) , d ( T x , x ) 0
and
t + F ( d ( T x 0 , x 0 ) ) F h d ( T x 0 , x 0 ) d ( T x 0 , x 0 ) + d ( T x 0 , x 0 ) d ( T x 0 , x 0 ) d ( T x 0 , x 0 ) = F ( 2 h d ( T x 0 , x 0 ) ) < F ( d ( T x 0 , x 0 ) ) .
This is a contradiction since t > 0 and so it should be T x 0 = x 0 . □
Consequently, the condition (1) can be replaced with d ( T x , x ) 0 and so T x x . Considering this, now we give a new fixed-circle theorem.
Theorem 1.
Let ( X , d ) be a metric space, T : X X be a self-mapping and
r = inf d ( T x , x ) : T x x .
If T is an F C -Khan type I contraction with x 0 X then C x 0 , r is a fixed circle of T.
Proof. 
Let x C x 0 , r . Assume that T x x . Then we have d ( T x , x ) 0 and by the F C -Khan type I contractive condition, we obtain
t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) max d ( T x 0 , x 0 ) , d ( T x , x ) = F ( h r ) F ( h d ( T x , x ) ) < F ( d ( T x , x ) ) ,
a contradiction since t > 0 . Therefore, we have T x = x and so T fixes the circle C x 0 , r . □
Corollary 1.
Let ( X , d ) be a metric space, T : X X be a self-mapping and r be defined as in (2). If T is an F C -Khan type I contraction with x 0 X then T fixes the disc D x 0 , r = x X : d ( x , x 0 ) r .
We recall the following theorem.
Theorem 2
([12]). Let ( X , d ) be a metric space and T : X X satisfy
d ( T x , T y ) k d ( x , T x ) d ( x , T y ) + d ( y , T y ) d ( y , T x ) d ( x , T y ) + d ( y , T x ) i f d ( x , T y ) + d ( y , T x ) 0 0 i f d ( x , T y ) + d ( y , T x ) = 0 ,
where k [ 0 , 1 ) and x , y X . Then T has a unique fixed point x X . Moreover, for all x X , the sequence { T n x } n N converges to x .
We modify the inequality (3) using Wardowski’s technique to obtain a new fixed-point theorem. We give the following definition.
Definition 6.
Let ( X , d ) be a metric space and T : X X be a self-mapping. T is said to be an F C -Khan type II contraction if there exist F F k , t > 0 and x 0 X such that for all x X if d ( T x 0 , x 0 ) + d ( T x , x ) 0 then T x x and
t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x 0 ) + d ( T x , x ) ,
where h 0 , 1 2 and if d ( T x 0 , x 0 ) + d ( T x , x ) = 0 then T x = x .
An immediate consequence of this definition is the following result.
Proposition 2.
Let ( X , d ) be a metric space. If a self-mapping T on X is an F C -Khan type II contraction then we get T x 0 = x 0 .
Proof. 
Let T x 0 x 0 . Then using the hypothesis, we find
d ( T x 0 , x 0 ) + d ( T x , x ) 0
and
t + F ( d ( T x 0 , x 0 ) ) F h d ( T x 0 , x 0 ) d ( T x 0 , x 0 ) + d ( T x 0 , x 0 ) d ( T x 0 , x 0 ) 2 d ( T x 0 , x 0 ) = F ( h d ( T x 0 , x 0 ) ) < F ( d ( T x 0 , x 0 ) ) ,
which is a contradiction since t > 0 . Hence it should be T x 0 = x 0 . □
Theorem 3.
Let ( X , d ) be a metric space, T : X X be a self-mapping and r be defined as in (2). If T is an F C -Khan type II contraction with x 0 X then C x 0 , r is a fixed circle of T.
Proof. 
Let x C x 0 , r . Assume that T x x . Then using Proposition 2, we get
d ( T x 0 , x 0 ) + d ( T x , x ) = d ( T x , x ) 0 .
By the F C -Khan type II contractive condition, we obtain
t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x 0 ) + d ( T x , x ) = F ( h r ) F ( h d ( T x , x ) ) < F ( d ( T x , x ) ) ,
a contradiction since t > 0 . Therefore, we have T x = x and T fixes the circle C x 0 , r . □
Corollary 2.
Let ( X , d ) be a metric space, T : X X be a self-mapping and r be defined as in (2). If T is an F C -Khan type II contraction with x 0 X then T fixes the disc D x 0 , r .
In the following theorem, we see that the F C -Khan type I and F C -Khan type II contractive conditions are equivalent.
Theorem 4.
Let ( X , d ) be a metric space and T : X X be a self-mapping. T satisfies the F C -Khan type I contractive condition if and only if T satisfies the F C -Khan type II contractive condition.
Proof. 
Let the F C -Khan type I contractive condition be satisfied by T. Using Proposition 1 and Proposition 2, we get
t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) max d ( T x 0 , x 0 ) , d ( T x , x ) = F h d ( T x , x ) d ( T x 0 , x ) d ( T x , x ) = F ( h d ( T x 0 , x ) ) = F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x 0 ) + d ( T x , x ) .
Using the similar arguments, the converse statement is clear. Consequently, the F C -Khan type I contractive and the F C -Khan type II contractive conditions are equivalent. □
Remark 1.
By Theorem 4, we see that Theorem 1 and Theorem 3 are equivalent.
Now we give an example.
Example 1.
Let X = R be the metric space with the usual metric d ( x , y ) = x y . Let us define the self-mapping T : R R as
T x = x i f x < 6 x + 1 i f x 6 ,
for all x R . The self-mapping T is both of an F C -Khan type I and an F C -Khan type II contraction with F = ln x , t = ln 2 , x 0 = 0 and h = 1 3 . Indeed, we get
d ( T x , x ) = 1 0 ,
for all x R such that x 6 . Then we have
ln 2 ln 1 3 x ln 2 + ln 1 ln h d ( x , 0 ) = ln h d ( x , x 0 ) t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) max d ( T x 0 , x 0 ) , d ( T x , x )
and
ln 2 ln 1 3 x ln 2 + ln 1 ln h d ( x , 0 ) = ln h d ( x , x 0 ) t + F ( d ( T x , x ) ) F h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x 0 ) + d ( T x , x ) .
Also we obtain
r = min d ( T x , x ) : T x x = 1 .
Consequently, T fixes the circle C 0 , 1 = { 1 , 1 } and the disc D 0 , 1 = x X : x 1 . Notice that the self-mapping T has other fixed circles. The above results give us only one of these circles. Also, T has infinitely many fixed circles.
Now we consider the case if T : X X is a self-mapping, then for all x , y X ,
x y d ( T y , x ) + d ( T x , y ) 0 .
Definition 7.
Let ( X , d ) be a metric space and T : X X be a self-mapping. Then T is called a C-Khan type contraction if there exists x 0 X such that
d ( T x , x ) h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x ) + d ( T x , x 0 ) ,
where h 0 , 1 for all x X x 0 .
We can give the following fixed-circle result.
Theorem 5.
Let ( X , d ) be a metric space, T : X X be a self-mapping and C x 0 , r be a circle on X. If T satisfies the C-Khan type contractive condition (4) for all x C x 0 , r with T x 0 = x 0 , then T fixes the circle C x 0 , r .
Proof. 
Let x C x 0 , r . Suppose that T x x . Using the C-Khan type contractive condition with T x 0 = x 0 , we find
d ( T x , x ) h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x ) + d ( T x , x 0 ) = h r d ( T x , x ) r + d ( T x , x 0 ) h r d ( T x , x ) r = h d ( T x , x ) ,
which is a contradiction since h < 1 . Consequently, T fixes the circle C x 0 , r . □
Theorem 6.
Let ( X , d ) be a metric space, x 0 X and T : X X be a self-mapping. If T is a C-Khan type contraction for all x X x 0 with T x 0 = x 0 , then T is the identity map I X on X.
Proof. 
Let x X x 0 be any point. If T x x then using the C-Khan type contractive condition (4) with T x 0 = x 0 , we find
d ( T x , x ) h d ( T x , x ) d ( T x 0 , x ) + d ( T x 0 , x 0 ) d ( T x , x 0 ) d ( T x 0 , x ) + d ( T x , x 0 ) = h d ( T x , x ) d ( x 0 , x ) d ( x 0 , x ) + d ( T x , x 0 ) h d ( T x , x ) d ( x 0 , x ) + d ( T x , x ) d ( T x , x 0 ) d ( x 0 , x ) + d ( T x , x 0 ) = h d ( T x , x ) d ( x 0 , x ) + d ( T x , x 0 ) d ( x 0 , x ) + d ( T x , x 0 ) = h d ( T x , x ) ,
which is a contradiction since h < 1 . Consequently, we have T x = x and hence T is the identity map I X on X. □
Example 2.
Let X = R be the usual metric space and consider the circle C 0 , 3 = { 3 , 3 } . Let us define the self-mapping T : R R as
T x = 9 x + 8 2 x 9 i f x { 3 , 3 } 0 i f x R { 3 , 3 } ,
for all x R . Then the self-mapping T satisfies the C-Khan type contractive condition for all x C 0 , 3 and T 0 = 0 . Consequently, C 0 , 3 is a fixed circle of T.

3. Common Fixed-Circle Results

Recently, it was obtained some coincidence and common fixed-point theorems using Wardowski’s technique and the Ćirić type contractions (see [17] for more details). In this section, we extend the notion of a Khan type F C -contraction to a pair of maps to obtain a solution to the Open Problem C C . At first, we give the following definition.
Definition 8.
Let ( X , d ) be a metric space and T , S : X X be two self-mappings. A pair of self-mappings ( T , S ) is called a Khan type F T , S -contraction if there exist F F k , t > 0 and x 0 X such that for all x X if the following condition holds
max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) 0 ,
then
t + F ( d ( T x , S x ) ) F h d ( T x , S x ) d ( T x , x 0 ) + d ( T x 0 , S x 0 ) d ( S x , x 0 ) max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) ,
where h 0 , 1 2 and if max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) = 0 then T x = S x .
An immediate consequence of this definition is the following proposition.
Proposition 3.
Let ( X , d ) be a metric space and T , S : X X be two self-mappings. If the pair of self-mappings ( T , S ) is a Khan type F T , S -contraction with x 0 X then x 0 is a coincidence point of T and S, that is, T x 0 = S x 0 .
Proof. 
We prove this proposition under the following cases:
Case 1: Let T x 0 = x 0 and S x 0 x 0 . Then using the hypothesis, we get
max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) = d ( S x 0 , x 0 ) 0
and so
t + F ( d ( T x 0 , S x 0 ) ) F h d ( T x 0 , S x 0 ) d ( T x 0 , x 0 ) + d ( T x 0 , S x 0 ) d ( S x 0 , x 0 ) d ( S x 0 , x 0 ) = F ( h d ( T x 0 , S x 0 ) ) ,
which is a contradiction since h 0 , 1 2 and t > 0 .
Case 2: Let T x 0 x 0 and S x 0 = x 0 . By the similar arguments used in the proof of Case 1, we get a contradiction.
Case 3: Let T x 0 = x 0 and S x 0 = x 0 . Then we get T x 0 = S x 0 .
Case 4: Let T x 0 x 0 , S x 0 x 0 and T x 0 S x 0 . Using the hypothesis, we obtain
max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) 0
and so
t + F ( d ( T x 0 , S x 0 ) ) F h d ( T x 0 , S x 0 ) d ( T x 0 , x 0 ) + d ( T x 0 , S x 0 ) d ( S x 0 , x 0 ) max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) .
Assume that d ( T x 0 , x 0 ) > d ( S x 0 , x 0 ) . Using the inequality (5), we get
t + F ( d ( T x 0 , S x 0 ) ) F h d ( T x 0 , S x 0 ) d ( T x 0 , x 0 ) + d ( T x 0 , S x 0 ) d ( S x 0 , x 0 ) d ( T x 0 , x 0 ) = F h d ( T x 0 , S x 0 ) + h d ( T x 0 , S x 0 ) d ( S x 0 , x 0 ) d ( T x 0 , x 0 ) < F ( 2 h d ( T x 0 , S x 0 ) ) < F ( d ( T x 0 , S x 0 ) ) ,
which is a contradiction. Suppose that d ( T x 0 , x 0 ) < d ( S x 0 , x 0 ) . Using the inequality (5), we find
t + F ( d ( T x 0 , S x 0 ) ) < F ( d ( T x 0 , S x 0 ) ) ,
which is a contradiction. Consequently, x 0 is a coincidence point of T and S, that is, T x 0 = S x 0 . □
Now we use the following number given in [17] (see Definition 3.1 on page 183):
M ( x , y ) = max d ( S x , S y ) , d ( S x , T x ) , d ( S y , T y ) , d ( S x , T y ) + d ( S y , T x ) 2 .
We give the following definition.
Definition 9.
Let ( X , d ) be a metric space and T , S : X X be two self-mappings. A pair of self-mappings ( T , S ) is called a Ćirić type F T , S -contraction if there exist F F k , t > 0 and x 0 X such that for all x X
d ( T x , x ) > 0 t + F ( d ( T x , x ) ) F ( M ( x , x 0 ) ) .
We get the following proposition.
Proposition 4.
Let ( X , d ) be a metric space and T , S : X X be two self-mappings. If the pair of self-mappings ( T , S ) is both a Khan type F T , S -contraction and a Ćirić type F T , S -contraction with x 0 X then x 0 is a common fixed point of T and S, that is, T x 0 = S x 0 = x 0 .
Proof. 
By the Khan type F T , S -contractive property and Proposition 3, we know that x 0 is a coincidence point of T and S, that is, T x 0 = S x 0 . Now we prove that x 0 is a common fixed point of T and S. Let T x 0 x 0 . Then using the Ćirić type F T , S -contractive condition, we get
t + F ( d ( T x 0 , x 0 ) ) F ( M ( x 0 , x 0 ) ) = F max d ( S x 0 , S x 0 ) , d ( S x 0 , T x 0 ) , d ( S x 0 , T x 0 ) , d ( S x 0 , T x 0 ) + d ( S x 0 , T x 0 ) 2 = F ( d ( S x 0 , T x 0 ) ) = F ( 0 ) ,
which is a contradiction because of the definition of F. Therefore it should be T x 0 = x 0 . Consequently, x 0 is a common fixed point of T and S, that is, T x 0 = S x 0 = x 0 . □
Notice that we get a coincidence point result for a pair of self-mappings using the Khan type F T , S -contractive condition by Proposition 3. We obtain a common fixed-point result for a pair of self-mappings using the both of Khan type F T , S -contractive condition and the Ćirić type F T , S -contractive condition by Proposition 4.
We prove the following common fixed-circle theorem as a solution to the Open Problem C C .
Theorem 7.
Let ( X , d ) be a metric space, T , S : X X be two self-mappings and r be defined as in (2). If d ( T x , x 0 ) = d ( S x , x 0 ) = r for all x C x 0 , r and the pair of self-mappings ( T , S ) is both a Khan type F T , S -contraction and a Ćirić type F T , S -contraction with x 0 X then C x 0 , r is a common fixed circle of T and S, that is, T x = S x = x for all x C x 0 , r .
Proof. 
Let x C x 0 , r . We show that x is a coincidence point of T and S. Using Proposition 4, we get
max d ( T x 0 , x 0 ) , d ( S x 0 , x 0 ) = 0
and so by the definition of the Khan type F T , S -contraction we obtain
T x = S x .
Now we prove that C x 0 , r is a common fixed circle of T and S. Assume that T x x . Using Proposition 4 and the hypothesis Ćirić type F T , S -contractive condition, we find
t + F ( d ( T x , x ) ) F ( M ( x , x 0 ) ) = F max d ( S x , S x 0 ) , d ( S x , T x ) , d ( S x 0 , T x 0 ) , d ( S x , T x 0 ) + d ( S x 0 , T x ) 2 = F max d ( S x , x 0 ) , d ( S x , T x ) , d ( S x , x 0 ) + d ( x 0 , T x ) 2 = F ( max r , d ( S x , T x ) , r ) = F ( r ) ,
which contradicts with the definition of r. Consequently, we have T x = x and so C x 0 , r is a common fixed circle of T and S. □
Corollary 3.
Let ( X , d ) be a metric space, T , S : X X be two self-mappings and r be defined as in (2). If d ( T x , x 0 ) = d ( S x , x 0 ) = r for all x C x 0 , r and the pair of self-mappings ( T , S ) is both a Khan type F T , S -contraction and a Ćirić type F T , S -contraction with x 0 X then T and S fix the disc D x 0 , r , that is, T x = S x = x for all x D x 0 , r .
We give an illustrative example.
Example 3.
Let X = 1 , { 1 , 0 } be the metric space with the usual metric. Let us define the self-mappings T : X X and S : X X as
T x = x 2 i f x { 0 , 1 , 3 } 1 i f x = 1 x + 1 otherwise
and
S x = 1 x i f x { 1 , 1 } 3 x i f x { 0 , 3 } x + 1 otherwise ,
for all x X . The pair of the self-mappings ( T , S ) is both a Khan type F T , S -contraction and a Ćirić type F T , S -contraction with F = ln x , t = ln 3 2 and x 0 = 0 . Indeed, we get
max { d ( T 0 , 0 ) , d ( S 0 , 0 ) } = 0
and so T x = S x . Therefore, the pair ( T , S ) is a Khan type F T , S -contraction. Also we get
d ( T 3 , 3 ) = 6 0 ,
for x = 3 and
d ( T x , x ) = 1 0 ,
for all x X \ { 1 , 0 , 1 , 3 } . Then we have
ln 3 2 ln 9 ln 3 2 + ln 6 ln 9 ln 3 2 + ln ( d ( T 3 , 3 ) ) ln ( M ( 3 , 0 ) )
and
ln 3 2 ln x + 1 ln 3 2 + ln 1 ln x + 1 ln 3 2 + ln ( d ( T x , x ) ) ln ( M ( x , 0 ) ) .
Hence the pair ( T , S ) is a Ćirić type F T , S -contraction. Also we obtain
r = min { d ( T x , x ) : T x x } = min { 1 , 6 } = 1 .
Consequently, T fixes the circle C 0 , 1 = { 1 , 1 } and the disc D 0 , 1 .
In closing, we want to bring to the reader attention the following question, under what conditions we can prove the results in [18,19,20] in fixed circle?

Author Contributions

All authors contributed equally in writing this article. All authors read and approved the final manuscript.

Funding

The first author would like to thank Prince Sultan University for funding this work through research group Nonlinear Analysis Methods in Applied Mathematics (NAMAM) group number RG-DES-2017-01-17.

Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Mlaiki, N.; Taş, N.; Özgür, N.Y. On the Fixed-Circle Problem and Khan Type Contractions. Axioms 2018, 7, 80. https://doi.org/10.3390/axioms7040080

AMA Style

Mlaiki N, Taş N, Özgür NY. On the Fixed-Circle Problem and Khan Type Contractions. Axioms. 2018; 7(4):80. https://doi.org/10.3390/axioms7040080

Chicago/Turabian Style

Mlaiki, Nabil, Nihal Taş, and Nihal Yılmaz Özgür. 2018. "On the Fixed-Circle Problem and Khan Type Contractions" Axioms 7, no. 4: 80. https://doi.org/10.3390/axioms7040080

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