On the Fixed-Circle Problem and Khan Type Contractions

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.

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 $C_{x_0,r}=\left\{x\in X:d(x,x_0)=r\right\}$ be any circle on X. In [5], it was given the following open problem.

Open Problem $CC$: 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 $CC$. 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 $\mathbb{F}$ be the family of all functions $F:(0,\infty )\to \mathbb{R}$ such that

$\left(F1\right)$ F is strictly increasing,

$\left(F2\right)$ For each sequence ${\left\{{\alpha }_n\right\}}_{n=1}^{\infty }$ of positive numbers, ${\displaystyle \underset{n\to \infty }{lim}}{\alpha }_n=0$ if and only if ${\displaystyle \underset{n\to \infty }{lim}}F\left({\alpha }_n\right)=-\infty$,

$\left(F3\right)$ There exists $k\in (0,1)$ such that ${\displaystyle \underset{\alpha \to 0^{+}}{lim}}{\alpha }^{k}F\left(\alpha \right)=0$.

Definition 2

([16]). Let $(X,d)$ be a metric space. A mapping $T:X\to X$ is said to be an F-contraction on $(X,d)$, if there exist $F\in \mathbb{F}$ and $\tau \in (0,\infty )$ such that

$$d(Tx,Ty)>0⟹\tau +F\left(d\right(Tx,Ty\left)\right)\le F\left(d\right(x,y\left)\right),$$

for all $x,y\in X$.

Definition 3

([15]) .Let ${\mathbb{F}}_k$ be the family of all increasing functions $F:(0,\infty )\to \mathbb{R}$, that is, for all $x,y\in (0,\infty )$, if $x<y$ then $F\left(x\right)\le F\left(y\right)$.

Definition 4

([15]). Let $(X,d)$ be a metric space and $T:X\to X$ be a self-mapping. T is said to be an F-Khan-contraction if there exist $F\in {\mathbb{F}}_k$ and $t>0$ such that for all $x,y\in X$ if $max\left\{d(Ty,x),d(Tx,y)\right\}\ne 0$ then $Tx\ne Ty$ and

$$t+F\left(d(Tx,Ty)\right)\le F\left(\frac{d(Tx,x)d(Ty,x)+d(Ty,y)d(Tx,y)}{max\left\{d(Ty,x),d(Tx,y)\right\}}\right),$$

and if $max\left\{d(Ty,x),d(Tx,y)\right\}=0$ then $Tx=Ty$.

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 $T:X\to X$ be a self-mapping. T is said to be an $F_C$-Khan type I contraction if there exist $F\in {\mathbb{F}}_k$, $t>0$ and $x_0\in X$ such that for all $x\in X$ if the following condition holds

$$max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}\ne 0,$$

(1)

then

$$t+F\left(d(Tx,x)\right)\le F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}}\right),$$

where $h\in \left[0,\frac{1}{2}\right)$ and if $max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}=0$ then $Tx=x$.

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 $x_0\in X$ then we get $T{x}_0=x_0$.

Proof. 

Let $T{x}_0\ne x_0$. Then using the hypothesis, we find

$$max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}\ne 0$$

and

$$\begin{array}{ccc}\hfill t+F\left(d(T{x}_0,x_0)\right)& \le & F\left(h\frac{d(T{x}_0,x_0)d(T{x}_0,x_0)+d(T{x}_0,x_0)d(T{x}_0,x_0)}{d(T{x}_0,x_0)}\right)\hfill \\ & =& F\left(2hd(T{x}_0,x_0)\right)<F\left(d(T{x}_0,x_0)\right).\hfill \end{array}$$

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 $d(Tx,x)\ne 0$ and so $Tx\ne x$. Considering this, now we give a new fixed-circle theorem.

Theorem 1.

Let $(X,d)$ be a metric space, $T:X\to X$ be a self-mapping and

$$r=inf\left\{d(Tx,x):Tx\ne x\right\}.$$

(2)

If T is an $F_C$-Khan type I contraction with $x_0\in X$ then $C_{x_0,r}$ is a fixed circle of T.

Proof. 

Let $x\in C_{x_0,r}$. Assume that $Tx\ne x$. Then we have $d(Tx,x)\ne 0$ and by the $F_C$-Khan type I contractive condition, we obtain

$$\begin{array}{ccc}\hfill t+F\left(d\right(Tx,x\left)\right)& \le & F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}}\right)\hfill \\ & =& F\left(hr\right)\le F\left(hd\right(Tx,x\left)\right)<F\left(d\right(Tx,x\left)\right),\hfill \end{array}$$

a contradiction since $t>0$. Therefore, we have $Tx=x$ and so T fixes the circle $C_{x_0,r}$. □

Corollary 1.

Let $(X,d)$ be a metric space, $T:X\to X$ be a self-mapping and r be defined as in (2). If T is an $F_C$-Khan type I contraction with $x_0\in X$ then T fixes the disc $D_{x_0,r}=\left\{x\in X:d(x,x_0)\le r\right\}$.

We recall the following theorem.

Theorem 2

([12]). Let $(X,d)$ be a metric space and $T:X\to X$ satisfy

$$d(Tx,Ty)\le \left\{\begin{array}{ccc}k\frac{d(x,Tx)d(x,Ty)+d(y,Ty)d(y,Tx)}{d(x,Ty)+d(y,Tx)}& if& d(x,Ty)+d(y,Tx)\ne 0\\ 0& if& d(x,Ty)+d(y,Tx)=0\end{array}\right.,$$

(3)

where $k\in [0,1)$ and $x,y\in X$. Then T has a unique fixed point $x^{\ast }\in X$. Moreover, for all $x\in X$, the sequence ${\left\{T^{n}x\right\}}_{n\in \mathbb{N}}$ converges to $x^{\ast }$.

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 $T:X\to X$ be a self-mapping. T is said to be an $F_C$-Khan type II contraction if there exist $F\in {\mathbb{F}}_k$, $t>0$ and $x_0\in X$ such that for all $x\in X$ if $d(T{x}_0,x_0)+d(Tx,x)\ne 0$ then $Tx\ne x$ and

$$t+F\left(d(Tx,x)\right)\le F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x_0)+d(Tx,x)}\right),$$

where $h\in \left[0,\frac{1}{2}\right)$ and if $d(T{x}_0,x_0)+d(Tx,x)=0$ then $Tx=x$.

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 $T{x}_0\ne x_0$. Then using the hypothesis, we find

$$d(T{x}_0,x_0)+d(Tx,x)\ne 0$$

and

$$\begin{array}{ccc}\hfill t+F\left(d(T{x}_0,x_0)\right)& \le & F\left(h\frac{d(T{x}_0,x_0)d(T{x}_0,x_0)+d(T{x}_0,x_0)d(T{x}_0,x_0)}{2d(T{x}_0,x_0)}\right)\hfill \\ & =& F\left(hd(T{x}_0,x_0)\right)<F\left(d(T{x}_0,x_0)\right),\hfill \end{array}$$

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, $T:X\to X$ be a self-mapping and r be defined as in (2). If T is an $F_C$-Khan type II contraction with $x_0\in X$ then $C_{x_0,r}$ is a fixed circle of T.

Proof. 

Let $x\in C_{x_0,r}$. Assume that $Tx\ne x$. Then using Proposition 2, we get

$$d(T{x}_0,x_0)+d(Tx,x)=d(Tx,x)\ne 0.$$

By the $F_C$-Khan type II contractive condition, we obtain

$$\begin{array}{ccc}\hfill t+F\left(d\right(Tx,x\left)\right)& \le & F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x_0)+d(Tx,x)}\right)\hfill \\ & =& F\left(hr\right)\le F\left(hd\right(Tx,x\left)\right)<F\left(d\right(Tx,x\left)\right),\hfill \end{array}$$

a contradiction since $t>0$. Therefore, we have $Tx=x$ and T fixes the circle $C_{x_0,r}$. □

Corollary 2.

Let $(X,d)$ be a metric space, $T:X\to X$ be a self-mapping and r be defined as in (2). If T is an $F_C$-Khan type II contraction with $x_0\in X$ 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 $T:X\to X$ 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

$$\begin{array}{ccc}\hfill t+F\left(d\right(Tx,x\left)\right)& \le & F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}}\right)\hfill \\ & =& F\left(h\frac{d(Tx,x)d(T{x}_0,x)}{d(Tx,x)}\right)\hfill \\ & =& F\left(hd(T{x}_0,x)\right)\hfill \\ & =& F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x_0)+d(Tx,x)}\right).\hfill \end{array}$$

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 $X=\mathbb{R}$ be the metric space with the usual metric $d(x,y)=\left|x-y\right|$. Let us define the self-mapping $T:\mathbb{R}\to \mathbb{R}$ as

$$Tx=\left\{\begin{array}{ccc}x& if& \left|x\right|<6\\ x+1& if& \left|x\right|\ge 6\end{array}\right.,$$

for all $x\in \mathbb{R}$. The self-mapping T is both of an $F_C$-Khan type I and an $F_C$-Khan type II contraction with $F=lnx$, $t=ln2$, $x_0=0$ and $h=\frac{1}{3}$. Indeed, we get

$$d(Tx,x)=1\ne 0,$$

for all $x\in \mathbb{R}$ such that $\left|x\right|\ge 6$. Then we have

$$\begin{array}{ccc}\hfill ln2& \le & ln\left(\frac{1}{3}\left|x\right|\right)\hfill \\ & ⟹& ln2+ln1\le ln\left(hd(x,0)\right)=ln\left(hd(x,x_0)\right)\hfill \\ & ⟹& t+F\left(d(Tx,x)\right)\le F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{max\left\{d(T{x}_0,x_0),d(Tx,x)\right\}}\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill ln2& \le & ln\left(\frac{1}{3}\left|x\right|\right)\hfill \\ & ⟹& ln2+ln1\le ln\left(hd(x,0)\right)=ln\left(hd(x,x_0)\right)\hfill \\ & ⟹& t+F\left(d(Tx,x)\right)\le F\left(h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x_0)+d(Tx,x)}\right).\hfill \end{array}$$

Also we obtain

$$r=min\left\{d(Tx,x):Tx\ne x\right\}=1.$$

Consequently, T fixes the circle $C_{0,1}=\{-1,1\}$ and the disc $D_{0,1}=\left\{x\in X:\left|x\right|\le 1\right\}$. 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 $T:X\to X$ is a self-mapping, then for all $x,y\in X$,

$$x\ne y⟹d(Ty,x)+d(Tx,y)\ne 0.$$

Definition 7.

Let $(X,d)$ be a metric space and $T:X\to X$ be a self-mapping. Then T is called a C-Khan type contraction if there exists $x_0\in X$ such that

$$d(Tx,x)\le h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x)+d(Tx,x_0)},$$

(4)

where $h\in \left[0,1\right)$ for all $x\in X-\left\{x_0\right\}$.

We can give the following fixed-circle result.

Theorem 5.

Let $(X,d)$ be a metric space, $T:X\to X$ 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 $x\in C_{x_0,r}$ with $T{x}_0=x_0,$ then T fixes the circle $C_{x_0,r}$.

Proof. 

Let $x\in C_{x_0,r}$. Suppose that $Tx\ne x$. Using the C-Khan type contractive condition with $T{x}_0=x_0$, we find

$$\begin{array}{ccc}\hfill d(Tx,x)& \le & h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x)+d(Tx,x_0)}\hfill \\ & =& \frac{hrd(Tx,x)}{r+d(Tx,x_0)}\hfill \\ & \le & \frac{hrd(Tx,x)}{r}=hd(Tx,x),\hfill \end{array}$$

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, $x_0\in X$ and $T:X\to X$ be a self-mapping. If T is a C-Khan type contraction for all $x\in X-\left\{x_0\right\}$ with $T{x}_0=x_0,$ then T is the identity map $I_X$ on X.

Proof. 

Let $x\in X-\left\{x_0\right\}$ be any point. If $Tx\ne x$ then using the C-Khan type contractive condition (4) with $T{x}_0=x_0$, we find

$$\begin{array}{ccc}\hfill d(Tx,x)& \le & h\frac{d(Tx,x)d(T{x}_0,x)+d(T{x}_0,x_0)d(Tx,x_0)}{d(T{x}_0,x)+d(Tx,x_0)}\hfill \\ & =& h\frac{d(Tx,x)d(x_0,x)}{d(x_0,x)+d(Tx,x_0)}\hfill \\ & \le & h\frac{d(Tx,x)d(x_0,x)+d(Tx,x)d(Tx,x_0)}{d(x_0,x)+d(Tx,x_0)}\hfill \\ & =& h\frac{d(Tx,x)\left[d(x_0,x)+d(Tx,x_0)\right]}{d(x_0,x)+d(Tx,x_0)}\hfill \\ & =& hd(Tx,x),\hfill \end{array}$$

which is a contradiction since $h<1$. Consequently, we have $Tx=x$ and hence T is the identity map $I_X$ on X. □

Example 2.

Let $X=\mathbb{R}$ be the usual metric space and consider the circle $C_{0,3}=\{-3,3\}$. Let us define the self-mapping $T:\mathbb{R}\to \mathbb{R}$ as

$$Tx=\left\{\begin{array}{ccc}\frac{-9x+8}{2x-9}& if& x\in \{-3,3\}\\ 0& if& x\in \mathbb{R}-\{-3,3\}\end{array}\right.,$$

for all $x\in \mathbb{R}$. Then the self-mapping T satisfies the C-Khan type contractive condition for all $x\in C_{0,3}$ and $T0=0$. 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 $CC$. At first, we give the following definition.

Definition 8.

Let $(X,d)$ be a metric space and $T,S:X\to X$ be two self-mappings. A pair of self-mappings $(T,S)$ is called a Khan type $F_{T,S}$-contraction if there exist $F\in {\mathbb{F}}_k$, $t>0$ and $x_0\in X$ such that for all $x\in X$ if the following condition holds

$$max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}\ne 0,$$

then

$$t+F\left(d(Tx,Sx)\right)\le F\left(h\frac{d(Tx,Sx)d(Tx,x_0)+d(T{x}_0,S{x}_0)d(Sx,x_0)}{max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}}\right),$$

where $h\in \left[0,\frac{1}{2}\right)$ and if $max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}=0$ then $Tx=Sx$.

An immediate consequence of this definition is the following proposition.

Proposition 3.

Let $(X,d)$ be a metric space and $T,S:X\to X$ be two self-mappings. If the pair of self-mappings $(T,S)$ is a Khan type $F_{T,S}$-contraction with $x_0\in X$ 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 $S{x}_0\ne x_0$. Then using the hypothesis, we get

$$max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}=d(S{x}_0,x_0)\ne 0$$

and so

$$\begin{array}{ccc}\hfill t+F\left(d(T{x}_0,S{x}_0)\right)& \le & F\left(h\frac{d(T{x}_0,S{x}_0)d(T{x}_0,x_0)+d(T{x}_0,S{x}_0)d(S{x}_0,x_0)}{d(S{x}_0,x_0)}\right)\hfill \\ & =& F\left(hd(T{x}_0,S{x}_0)\right),\hfill \end{array}$$

which is a contradiction since $h\in \left[0,\frac{1}{2}\right)$ and $t>0$.

Case 2: Let $T{x}_0\ne x_0$ 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 $T{x}_0\ne x_0$, $S{x}_0\ne x_0$ and $T{x}_0\ne S{x}_0$. Using the hypothesis, we obtain

$$max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}\ne 0$$

and so

$$t+F\left(d(T{x}_0,S{x}_0)\right)\le F\left(h\frac{d(T{x}_0,S{x}_0)d(T{x}_0,x_0)+d(T{x}_0,S{x}_0)d(S{x}_0,x_0)}{max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}}\right).$$

(5)

Assume that $d(T{x}_0,x_0)>d(S{x}_0,x_0)$. Using the inequality (5), we get

$$\begin{array}{ccc}\hfill t+F\left(d(T{x}_0,S{x}_0)\right)& \le & F\left(h\frac{d(T{x}_0,S{x}_0)d(T{x}_0,x_0)+d(T{x}_0,S{x}_0)d(S{x}_0,x_0)}{d(T{x}_0,x_0)}\right)\hfill \\ & =& F\left(hd(T{x}_0,S{x}_0)+h\frac{d(T{x}_0,S{x}_0)d(S{x}_0,x_0)}{d(T{x}_0,x_0)}\right)\hfill \\ & <& F\left(2hd(T{x}_0,S{x}_0)\right)<F\left(d(T{x}_0,S{x}_0)\right),\hfill \end{array}$$

which is a contradiction. Suppose that $d(T{x}_0,x_0)<d(S{x}_0,x_0)$. Using the inequality (5), we find

$$t+F\left(d(T{x}_0,S{x}_0)\right)<F\left(d(T{x}_0,S{x}_0)\right),$$

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):

$$M(x,y)=max\left\{d(Sx,Sy),d(Sx,Tx),d(Sy,Ty),\frac{d(Sx,Ty)+d(Sy,Tx)}{2}\right\}.$$

(6)

We give the following definition.

Definition 9.

Let $(X,d)$ be a metric space and $T,S:X\to X$ be two self-mappings. A pair of self-mappings $(T,S)$ is called a Ćirić type $F_{T,S}$-contraction if there exist $F\in {\mathbb{F}}_k$, $t>0$ and $x_0\in X$ such that for all $x\in X$

$$d(Tx,x)>0⟹t+F\left(d(Tx,x)\right)\le F\left(M(x,x_0)\right).$$

We get the following proposition.

Proposition 4.

Let $(X,d)$ be a metric space and $T,S:X\to X$ 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 $x_0\in X$ 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 $T{x}_0\ne x_0$. Then using the Ćirić type $F_{T,S}$-contractive condition, we get

$$\begin{array}{ccc}\hfill t+F\left(d(T{x}_0,x_0)\right)& \le & F\left(M(x_0,x_0)\right)\hfill \\ & =& F\left(max\left\{\begin{array}{c}d(S{x}_0,S{x}_0),d(S{x}_0,T{x}_0),d(S{x}_0,T{x}_0),\\ \frac{d(S{x}_0,T{x}_0)+d(S{x}_0,T{x}_0)}{2}\end{array}\right\}\right)\hfill \\ & =& F\left(d(S{x}_0,T{x}_0)\right)=F\left(0\right),\hfill \end{array}$$

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 $CC$.

Theorem 7.

Let $(X,d)$ be a metric space, $T,S:X\to X$ be two self-mappings and r be defined as in (2). If $d(Tx,x_0)=d(Sx,x_0)=r$ for all $x\in C_{x_0,r}$ 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 $x_0\in X$ then $C_{x_0,r}$ is a common fixed circle of T and S, that is, $Tx=Sx=x$ for all $x\in C_{x_0,r}$.

Proof. 

Let $x\in C_{x_0,r}$. We show that x is a coincidence point of T and S. Using Proposition 4, we get

$$max\left\{d(T{x}_0,x_0),d(S{x}_0,x_0)\right\}=0$$

and so by the definition of the Khan type $F_{T,S}$-contraction we obtain

$$Tx=Sx.$$

Now we prove that $C_{x_0,r}$ is a common fixed circle of T and S. Assume that $Tx\ne x$. Using Proposition 4 and the hypothesis Ćirić type $F_{T,S}$-contractive condition, we find

$$\begin{array}{ccc}\hfill t+F\left(d\right(Tx,x\left)\right)& \le & F\left(M(x,x_0)\right)\hfill \\ & =& F\left(max\left\{\begin{array}{c}d(Sx,S{x}_0),d(Sx,Tx),d(S{x}_0,T{x}_0),\\ \frac{d(Sx,T{x}_0)+d(S{x}_0,Tx)}{2}\end{array}\right\}\right)\hfill \\ & =& F\left(max\left\{d(Sx,x_0),d(Sx,Tx),\frac{d(Sx,x_0)+d(x_0,Tx)}{2}\right\}\right)\hfill \\ & =& F(max\left\{r,d(Sx,Tx),r\right\})=F\left(r\right),\hfill \end{array}$$

which contradicts with the definition of r. Consequently, we have $Tx=x$ and so $C_{x_0,r}$ is a common fixed circle of T and S. □

Corollary 3.

Let $(X,d)$ be a metric space, $T,S:X\to X$ be two self-mappings and r be defined as in (2). If $d(Tx,x_0)=d(Sx,x_0)=r$ for all $x\in C_{x_0,r}$ 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 $x_0\in X$ then T and S fix the disc $D_{x_0,r}$, that is, $Tx=Sx=x$ for all $x\in D_{x_0,r}$.

We give an illustrative example.

Example 3.

Let $X=\left[1,\infty \right)\cup \{-1,0\}$ be the metric space with the usual metric. Let us define the self-mappings $T:X\to X$ and $S:X\to X$ as

$$Tx=\left\{\begin{array}{ccc}{x}^2& if& x\in \{0,1,3\}\\ -1& if& x=-1\\ x+1& & \mathit{otherwise}\end{array}\right.$$

and

$$Sx=\left\{\begin{array}{ccc}\frac{1}{x}& if& x\in \{-1,1\}\\ 3x& if& x\in \{0,3\}\\ x+1& & \mathit{otherwise}\end{array}\right.,$$

for all $x\in X$. 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 $F=lnx$, $t=ln\frac{3}{2}$ and $x_0=0$. Indeed, we get

$$max\left\{d\right(T0,0),d(S0,0\left)\right\}=0$$

and so $Tx=Sx$. Therefore, the pair $(T,S)$ is a Khan type $F_{T,S}$-contraction. Also we get

$$d(T3,3)=6\ne 0,$$

for $x=3$ and

$$d(Tx,x)=1\ne 0,$$

for all $x\in X\backslash \{-1,0,1,3\}$. Then we have

$$\begin{array}{ccc}\hfill ln\frac{3}{2}& \le & ln9\hfill \\ & ⟹& ln\frac{3}{2}+ln6\le ln9\hfill \\ & ⟹& ln\frac{3}{2}+ln\left(d(T3,3)\right)\le ln\left(M(3,0)\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill ln\frac{3}{2}& \le & ln\left|x+1\right|\hfill \\ & ⟹& ln\frac{3}{2}+ln1\le ln\left|x+1\right|\hfill \\ & ⟹& ln\frac{3}{2}+ln\left(d(Tx,x)\right)\le ln\left(M(x,0)\right).\hfill \end{array}$$

Hence the pair $(T,S)$ is a Ćirić type $F_{T,S}$-contraction. Also we obtain

$$r=min\left\{d\right(Tx,x):Tx\ne x\}=min\{1,6\}=1.$$

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?