Fixed Points Results for Various Types of Tricyclic Contractions

Abstract

In this paper, we introduce four new types of contractions called in this order Kanan-S-type tricyclic contraction, Chattergea-S-type tricyclic contraction, Riech-S-type tricyclic contraction, Cirić-S-type tricyclic contraction, and we prove the existence and uniqueness for a fixed point for each situation.

1. Introduction

It is well known that the Banach contraction principle was published in 1922 by S. Banach as follows:

Theorem 1.

Let $\left(X,d\right)$ be a complete metric space and a self mapping $T:X⟶X$. If there exists $k\in [0,1)$ such that, for all $x,y\in X$, $d(Tx,Ty)\le kd(x,y)$, then T has a unique fixed point in $X.$

The Banach contraction principle has been extensively studied and different generalizations were obtained.

In 1968 [1], Kannan established his famous extension of this contraction.

Theorem 2.

Ref. [1] Let $\left(X,d\right)$ be a complete metric space and a self mapping $T:X⟶X$. If T satisfies the following condition:

$$d(Tx,Ty)\le k\left[d(x,Tx)+d(y,Ty)\right]forallx,y\in Xwhere0<k<\frac{1}{2},$$

then T has a fixed point in $X.$

A similar contractive condition has been introduced by Chattergea in 1972 [2] as follows:

Theorem 3.

Ref. [2] Let $T:X⟶X$, where $\left(X,d\right)$ is a complete metric space. If there exists $0<k<\frac{1}{2}$ such that

$$d(Tx,Ty)\le k\left[d(y,Tx)+d(Ty,x)\right]forallx,y\in X,$$

then T has a fixed point in $X.$

We can also find another extension of the Banach contraction principle obtained by S. Reich, Kannan in 1971 [3].

Theorem 4.

Ref. [3] Let $T:X⟶X$, where $\left(X,d\right)$ is a complete metric space. If there exists $0<k<\frac{1}{3}$ such that

$$d(Tx,Ty)\le k\left[d(x,y)+d(x,Tx)+d(y,Ty)\right]forallx,y\in X,$$

then T has a fixed point in $X.$

In addition, in the same year, Cirić gave the following extension [4].

Theorem 5.

Ref. [4] Let $T:X⟶X$, where $\left(X,d\right)$ a complete metric space. If there exists $k\in [0,1)$ such that

$$d(Tx,Ty)\le kMax\left[d(x,y),d(x,Tx),d(y,Ty),d(y,Tx),d(Ty,x)\right]forallx,y\in X,$$

then T has a fixed point in $X.$

Many authors have investigated these situations and many results were proved (see [5,6,7,8,9,10,11,12,13]).

In this article, we prove the uniqueness and existence of the fixed points in different types contractions for a self mapping T defined on the union of tree closed subsets of a complete metric space with k in different intervals.

2. Preliminaries

In best approximation theory, the concept of tricyclic mappings extends that of ordinary cyclic mappings. Moreover, in the case where two of the sets, say A and C, coincide, we find a cyclic mapping which is also a self-map, and, hence, a best proximity point result for a tricyclic mappings means also a fixed point and a best proximity point result for a self-map and a cyclic mapping.

Definition 1.

Let A, B be nonempty subsets of a metric space $\left(X,d\right)$. A mapping $T:A\cup B⟶A\cup B$ is said to be cyclic if:

$$T\left(A\right)\subseteq B,T\left(B\right)\subseteq A.$$

In 2003, Kirk et al. [14] proved that, if $T:A\cup B⟶A\cup B$ is cyclic and, for some $k\in \left(0,1\right)$, $d\left(Tx,Ty\right)\le kd\left(x,y\right)$ for all $x\in A,y\in B$, then $A\cap B\ne \varnothing$, and T has a unique fixed point in $A\cap B$.

In 2017, Sabar et al. [15] proved a similar result for tricyclic mappings and introduced the concept of tricyclic contractions.

Theorem 6.

Ref. [15] Let $A,B$ and C be nonempty closed subsets of a complete metric space $\left(X,d\right)$, and let a mapping $T:A\cup B\cup C⟶A\cup B\cup C$. If $T\left(A\right)\subseteq B,T\left(B\right)\subseteq C$ and $T\left(C\right)\subseteq A$ and there exists $k\in \left(0,1\right)$ such that $D\left(Tx,Ty,Tz\right)\le kD\left(x,y,z\right)$ for all $\left(x,y,z\right)\in A\times B\times C$, then $A\cap B\cap C$ is nonempty and T has a unique fixed point in $A\cap B\cap C,$

where $D\left(x,y,z\right)=d(x,y)+d(x,z)+d(y,z).$

Definition 2.

Ref. [15] Let $A,B$ and C be nonempty subsets of a metric space $\left(X,d\right)$. A mapping $T:A\cup B\cup C⟶A\cup B\cup C$ is said to be tricyclic contracton if there exists $0<k<1$ such that:

1.

$T\left(A\right)\subseteq B,T\left(B\right)\subseteq C$ and $T\left(C\right)\subseteq A.$

2.

$D\left(Tx,Ty,Tz\right)\le kD\left(x,y,z\right)+(1-k)\delta \left(A,B,C\right)$ for all $\left(x,y,z\right)\in A\times B\times C.$

where $\delta \left(A,B,C\right)=inf\{D\left(x,y,z\right):x\in A,y\in B,z\in C\}$

Very Recently, Sabiri et al. introduced an extension of the aforementioned mappings and called them p-cyclic contractions [16].

3. Main Results

Definition 3.

Let $A,B$ and C be nonempty subsets of a metric space $\left(X,d\right)$. A mapping $T:A\cup B\cup C⟶A\cup B\cup C$ is said to be a Kannan-S-type tricyclic contraction, if there exists $k\in \left(0,\frac{1}{3}\right)$ such that

1.

$T\left(A\right)\subseteq B,T\left(B\right)\subseteq C,T\left(C\right)\subseteq A.$

2.

$D\left(Tx,Ty,Tz\right)\le k\left[d\left(x,Tx\right)+d\left(y,Ty\right)+d\left(z,Tz\right)\right]$ for all $\left(x,y,z\right)\in A\times B\times C.$

We give an example to show that a map can be a tricyclic contraction but not a Kannan-S-type tricyclic contraction.

Example 1.

Let X be ${\mathbb{R}}^2$ normed by the norm $\Vert (x,y)\Vert =\left|x\right|+\left|y\right|,$ and $A=[1,2]\times \left\{0\right\},B=\left\{0\right\}\times [-2,-1],$ $C=[-2,-1]\times \left\{0\right\},$ then

$$\delta \left(A,B,C\right)=D((1,0),(0,-1),(-1,0))=6.$$

Put $T:A\cup B\cup C⟶A\cup B\cup C$ such that

$$T\left(x,0\right)=\left(0,-\frac{x+2}{3}\right)if\left(x,0\right)\in A,$$

$$T\left(0,y\right)=\left(\frac{y-2}{3},0\right)if\left(0,y\right)\in B,$$

$$T\left(z,0\right)=\left(-\frac{z-2}{3},0\right)if\left(z,0\right)\in C,$$

We have $T\left(A\right)\subseteq B,T\left(B\right)\subseteq C$ and $T\left(C\right)\subseteq A$, and

$$\begin{array}{ccc}\hfill D(T\left(x,0\right),T\left(0,y\right),T\left(z,0\right))& =& D\left((0,-\frac{x+2}{3}),(\frac{y-2}{3},0),(-\frac{z-2}{3},0)\right)\hfill \\ & =& \frac{2}{3}(x-y-z)+4\hfill \\ & =& \frac{1}{3}D(\left(x,0\right),\left(0,y\right),\left(z,0\right))+4\hfill \\ & =& \frac{1}{3}D(\left(x,0\right),\left(0,y\right),\left(z,0\right))+(1-\frac{1}{3})\delta \left(A,B,C\right)\hfill \end{array}$$

for all $\left(x,0\right)\in A,\left(0,y\right)\in B,\left(z,0\right)\in C.$

On the other hand,

$$D(T\left(2,0\right),T\left(0,-2\right),T\left(-2,0\right))=D\left((0,-\frac{4}{3}),(\frac{-4}{3},0),(\frac{4}{3},0)\right)=8$$

and

$$d(\left(2,0\right),T\left(2,0\right))+d(\left(0,-2\right),T\left(0,-2\right))+d(\left(-2,0\right),T\left(-2,0\right))=10,$$

which implies that

$D(T\left(2,0\right),T\left(0,-2\right),T\left(-2,0\right))$

$$>\frac{1}{3}\left[d(\left(2,0\right),T\left(2,0\right))+d(\left(0,-2\right),T\left(0,-2\right))+d(\left(-2,0\right),T\left(-2,0\right))\right]$$

Then, T is tricyclic contraction but not a Kannan-S-type tricyclic contraction.

Now, we give an example for which T is a Kannan-S-type tricyclic contraction but not a tricyclic contraction.

Example 2.

Let $X=\mathbb{R}$ with the usual metric. Let $A=B=C=\left[0,1\right],$ then $\delta \left(A,B,C\right)=0.$ Put $T:A\cup B\cup C⟶A\cup B\cup C$ such that

$$Tx=\frac{1}{6}if0\le x<1,Tx=\frac{1}{4}ifx=1$$

For $x=1,y=1$ and $z=\frac{23}{24},$ we have

$$D(T(1),T(1),T(\frac{23}{24}))=D(\frac{1}{4},\frac{1}{4},\frac{1}{6})=2d(\frac{1}{4},\frac{1}{6})=\frac{1}{6}.$$

and

$$D(1,1,\frac{23}{24})=2d(1,\frac{23}{24})=\frac{1}{12}.$$

Then, T is not tricyclic contraction.

However T is a Kannan-S-type tricyclic contraction. Indeed:

• If $x=y=z=1$, we have

  $$D(T(1),T(1),T(1))=0\le \frac{9}{4}k$$

  for all $k\ge 0$, then for 0 $\le k<\frac{1}{3}.$
• If $x\in$$\left[0,1\right)$, $y\in$$\left[0,1\right)$ and $z\in$$\left[0,1\right)$, we have

  $$D(Tx,Ty,Tz)=0\le k(d(x,\frac{1}{6})+d(y,\frac{1}{6})+d(z,\frac{1}{6})$$

  for all $k\ge 0$, then for 0 $\le k<\frac{1}{3}.$
• If $x=1,y\in$$\left[0,1\right)$ and $z\in$$\left[0,1\right)$, we have

  $$D(T_1,Ty,Tz)=D(\frac{1}{4},\frac{1}{6},\frac{1}{6})=\frac{1}{6}$$

  and

  $$d(1,T(1))+d(y,Ty)+d(z,Tz)=\frac{3}{4}+d(y,\frac{1}{6})+d(z,\frac{1}{6}),$$

  then, for $k=\frac{2}{9}$, we have

  $$D(T(1),T(y,Tz)\le k(d(1,T(1))+d(y,Ty)+d(z,Tz)).$$
• If $x=1,y=1$ and $z\in$$\left[0,1\right),$ we have

  $$D(T(1),T(1),Tz)=D(\frac{1}{4},\frac{1}{4},\frac{1}{6})=\frac{1}{6}$$

  and

  $$d(1,T(1))+d(1,T(1))+d(z,Tz)=\frac{3}{2}+d(z,\frac{1}{6}).$$
  Then, for $k=\frac{2}{9}$, we have

  $$D(T(1),T(1),Tz)\le k(d(1,T(1))+d(1,T(1))+d(z,Tz)).$$
  Consequently, for $k=\frac{2}{9}$, we have:

  $$D(Tx,Ty,Tz)\le k(d(x,Tx)+d(y,Ty)+d(z,Tz))forall\left(x,y,z\right)\in A\times B\times C.$$

Theorem 7.

Let $A,B$ and C be nonempty closed subsets of a complete metric space $\left(X,d\right),$ and let $T:A\cup B\cup C⟶A\cup B\cup C$ be a Kannan-S-type tricyclic contraction. Then, T has a unique fixed point in $A\cap B\cap C.$

Proof. 

Fix $x\in A$. We have

$$d\left(T^{3}x,T^{2}x\right)\le D\left(T^{3}x,T^{2}x,Tx\right)\le k\left[d\left(T^{2}x,T^{3}x\right)+d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right].$$

Then,

$$d\left(T^{3}x,T^{2}x\right)\le k\left[d\left(T^{2}x,T^{3}x\right)+d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right],$$

which implies

$$d\left(T^{3}x,T^{2}x\right)\le \frac{k}{(1-k)}\left[d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right].$$

Similarly, we have

$$d\left(T^{2}x,Tx\right)\le \frac{k}{(1-k)}\left[d\left(T^{3}x,T^{2}x\right)+d\left(x,Tx\right)\right]$$

$$d\left(T^{2}x,Tx\right)\le \frac{k}{(1-k)}\left[\frac{k}{(1-k)}\left[d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right]+d\left(x,Tx\right)\right]$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{k}{1-2k}(d\left(x,Tx\right)).$$

Then,

$$d\left(T^{2}x,Tx\right)\le td\left(x,Tx\right)\mathrm{where}t=\frac{k}{1-2k}\mathrm{and}t\in \left(0,1\right),$$

which implies

$$d\left(T^{n+1}x,T^{n}x\right)\le t^{n}d\left(x,Tx\right),\mathrm{for}\mathrm{all}n\ge 1$$

Consequently,

$$\underset{n=1}{\sum ^{+\infty }}d\left(T^{n+1}x,T^{n}x\right)\le \underset{n=1}{\stackrel{+\infty }{(\sum }}{t}^n)d\left(x,Tx\right)<+\infty$$

implies that $\left\{T^{n}x\right\}$ is a Cauchy sequence in $\left(X,d\right).$ Hence, there exists $z\in A\cup B\cup C$ such that $T^{n}x⟶z.$ Notice that $\left\{T^{3n}x\right\}$ is a sequence in $A,\left\{T^{3n-1}x\right\}$ is a sequence in C and $\left\{T^{3n-2}x\right\}$ is a sequence in B and that both sequences tend to the same limit z. Regarding the fact that $A,B$ and C are closed, we conclude $z\in A\cap B\cap C$, hence $A\cap B\cap C\ne ⌀.$

To show that z is a fixed point, we must show that $Tz=z$. Observe that

$$\begin{array}{ccc}\hfill d\left(Tz,z\right)& =& limd\left(Tz,T^{3n}x\right)\le limD\left(T^{3n}x,T^{3n-1}x,Tz\right)\hfill \\ & \le & limk[d\left(T^{3n-1}x,T^{3n}x\right)+d\left(T^{3n-2}x,T^{3n-1}x\right)+d(z,Tz)]\hfill \\ & \le & kd\left(Tz,z\right),\hfill \end{array}$$

which is equivalent to

$$\left(1-k\right)d\left(Tz,z\right)=0.$$

Since $k\in \left(0,\frac{1}{3}\right)$, $\mathrm{then}d\left(Tz,z\right)=0$, which implies $Tz=z$.

To prove the uniqueness of $z,$ assume that there exists $w\in A\cup B\cup C$ such that $w\ne z$ and $Tw=w$. Taking into account that T is tricyclic, we get $w\in A\cap B\cap C.$ We have

$$d\left(z,w\right)=d\left(Tz,Tw\right)\le D\left(Tz,Tw,Tw\right)\le k[d\left(z,Tz\right)+d\left(w,Tw\right)+d\left(w,Tw\right)]=0$$

which implies $d\left(z,w\right)=0$. We get that $z=w$ and hence z is the unique fixed point of $T.$ □

Example 3.

Let X be ${\mathbb{R}}^2$ normed by the norm $\Vert \left(x,y\right)\Vert =\left|x\right|+\left|y\right|,$ let $A=\left\{0\right\}\times [0,+1],B=[0,+1]\times \left\{0\right\}$, $C=\left\{0\right\}\times [-1,0]$ and let $T:A\cup B\cup C⟶A\cup B\cup C$ be defined by

$$T\left(0,x\right)=\left(\frac{x}{6},0\right)if\left(0,x\right)\in A,$$

$$T\left(y,0\right)=\left(0,\frac{-y}{6}\right)if\left(y,0\right)\in B,$$

$$T\left(0,z\right)=\left(0,\frac{-z}{6}\right)if\left(0,z\right)\in C.$$

We have

$$T\left(A\right)\subseteq B,T\left(B\right)\subseteq C\mathrm{and}T\left(C\right)\subseteq A$$

In addition, for all $\left(0,x\right)\in A,\left(y,0\right)\in B,\left(0,z\right)\in C$, we have

$$D\left(T\left(0,x\right),T\left(y,0\right),T\left(0,z\right)\right)=D\left(\left(\frac{x}{6},0\right),\left(0,\frac{-y}{6}\right),\left(0,\frac{-z}{6}\right)\right)=\frac{1}{3}(x+y-z)$$

In addition, we have

$$d\left(\left(0,x\right),T\left(0,x\right)\right)+d\left(\left(y,0\right),T\left(y,0\right)\right)+d\left(\left(0,z\right),T\left(0,z\right)\right)=\frac{7}{6}(x+y-z)$$

This implies

$$D\left(T\left(0,x\right),T\left(y,0\right),T\left(0,z\right)\right)=\frac{2}{7}\left[d\left(\left(0,x\right),T\left(0,x\right)\right)+d\left(\left(y,0\right),T\left(y,0\right)\right)+d\left(\left(0,z\right),T\left(0,z\right)\right)\right].$$

Then, T is a Kannan-S-type tricyclic contraction, and T has a unique fixed point $\left(0,0\right)$ in $A\cap B\cap C.$

Corollary 1.

Let $(X,d)$ be a complete metric space and a self mapping $T:X⟶X$. If there exists $k\in \left(0,\frac{1}{3}\right)$ such that

$$D\left(Tx,Ty,Tz\right)\le k\left[d\left(x,Tx\right)+d\left(y,Ty\right)+d\left(z,Tz\right)\right]$$

for all $\left(x,y,z\right)\in X^3$, then T has a unique fixed point.

Now, we shall define another type of a tricyclic contraction.

Definition 4.

Let $A,B$ and C be nonempty subsets of a metric space $\left(X,d\right)$. A mapping $T:A\cup B\cup C⟶A\cup B\cup C$ is said to be a Chattergea-S-type tricyclic contraction if $T\left(A\right)\subseteq B,T\left(B\right)\subseteq C,T\left(C\right)\subseteq A$, and there exist $k\in \left(0,\frac{1}{3}\right)$ such that$D\left(Tx,Ty,Tz\right)\le k\left[d\left(y,Tx\right)+d\left(z,Ty\right)+d\left(x,Tz\right)\right]$ for all $\left(x,y,z\right)\in A\times B\times C.$

Theorem 8.

Let $A,B$ and C be nonempty closed subsets of a complete metric space $\left(X,d\right)$, and let $T:A\cup B\cup C⟶A\cup B\cup C$ be a Chattergea-S-type tricyclic contraction. Then, T has a unique fixed point in $A\cap B\cap C.$

Proof. 

Fix $x\in A$. We have

$$D\left(Tx,T^{2}x,T^{3}x\right)\le k\left[d\left(Tx,Tx\right)+d\left(T^{2}x,T^{2}x\right)+d\left(T^{3}x,x\right)\right]$$

which implies

$$D\left(T^{3}x,T^{2}x,Tx\right)\le kd\left(T^{3}x,x\right)$$

so

$$d\left(T^{3}x,T^{2}x\right)\le k\left[d\left(T^{3}x,T^{2}x\right)+d\left(T^{2}x,Tx\right)+d\left(Tx,x\right)\right](\mathrm{by}\mathrm{the}\mathrm{triangular}\mathrm{inequality})$$

$$⟹d\left(T^{3}x,T^{2}x\right)\le \frac{k}{(1-k)}\left[d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right]$$

and

$$d\left(T^{2}x,Tx\right)\le D\left(T^{3}x,T^{2}x,Tx\right)\le \frac{k}{(1-k)}\left[d\left(T^{3}x,T^{2}x\right)+d\left(x,Tx\right)\right]$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{k}{(1-k)}\left[\frac{k}{(1-k)}\left[d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right]+d\left(x,Tx\right)\right]$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{k}{1-2k}(d\left(x,Tx\right))$$

Then,

$$d\left(T^{2}x,Tx\right)\le td\left(x,Tx\right)\mathrm{where}t=\frac{k}{1-2k}\mathrm{and}t\in \left(0,1\right),$$

which implies

$$d\left(T^{n+1}x,T^{n}x\right)\le t^{n}d\left(x,Tx\right)$$

for all $n\ge 1$. Consequently,

$$\underset{n=1}{\sum ^{+\infty }}d\left(T^{n+1}x,T^{n}x\right)\le \underset{n=1}{\stackrel{+\infty }{(\sum }}{t}^n)d\left(x,Tx\right)<+\infty$$

implies that $\left\{T^{n}x\right\}$ is a Cauchy sequence in $\left(X,d\right)$. Hence, there exists $z\in A\cup B\cup C$ such that $T^{n}x⟶z.$ Notice that $\left\{T^{3n}x\right\}$ is a sequence in $A,\left\{T^{3n-1}x\right\}$ is a sequence in C, and $\left\{T^{3n-2}x\right\}$ is a sequence in B and that both sequences tend to the same limit z. Regarding that $A,B$ and C are closed, we conclude $z\in A\cap B\cap C,$ hence $A\cap B\cap C\ne ⌀.$

To show that z is a fixed point, we must show that $Tz=z.$ Observe that

$$\begin{array}{ccc}\hfill d\left(Tz,z\right)& =& limd\left(Tz,T^{3n}x\right)\le limD\left(Tz,T^{3n}x,T^{3n-1}x\right)\hfill \\ & \le & limk[d\left(T^{3n-1}x,Tz\right)+\left(T^{3n-2}x,T^{3n}x\right)+d(z,T^{3n-1}x)]\le kd\left(Tz,z\right),\hfill \end{array}$$

which is equivalent to $\left(1-k\right)d\left(Tz,z\right)=0.$ Since $k\in \left(1,\frac{1}{3}\right),$ then $d\left(Tz,z\right)=0$, which implies $Tz=z.$

To prove the uniqueness of $z,$ assume that there exists $w\in A\cup B\cup C$ such that $w\ne z$ and $Tw=w.$ Taking into account that T is tricyclic, we get $w\in A\cap B\cap C.$

We have

$$\begin{array}{ccc}\hfill d\left(z,w\right)& =& d\left(Tz,Tw\right)\le D\left(Tz,Tw,Tw\right)\hfill \\ & \le & k[d\left(Tz,w\right)+d\left(Tw,w\right)+d\left(Tw,z\right)]\hfill \\ & \le & 2kd\left(z,w\right).\hfill \end{array}$$

Then, $d\left(z,w\right)=0.$ We conclude that $z=w$ and hence z is the unique fixed point of $T.$ □

Corollary 2.

Let $(X,d)$ be a complete metric space and a self mapping $T:X⟶X$. If there exists $k\in \left(0,\frac{1}{3}\right)$ such that

$$D\left(Tx,Ty,Tz\right)\le k\left[d\left(y,Tx\right)+d\left(z,Ty\right)+d\left(x,Tz\right)\right]$$

for all $\left(x,y,z\right)\in X^3$, then T has a unique fixed point.

In this step, we define a Reich-S-type tricyclic contraction.

Definition 5.

Let $A,B$ and C be nonempty subsets of a metric space $\left(X,d\right)$.

A mapping $T:A\cup B\cup C⟶A\cup B\cup C$ is said to be a Reich-S-type tricyclic contraction if there exists $k\in \left(0,\frac{1}{7}\right)$ such that:

1.

$T\left(A\right)\subseteq B,T\left(B\right)\subseteq C,T\left(C\right)\subseteq A.$

2.

$D\left(Tx,Ty,Tz\right)\le k\left[D\left(x,y,z\right)+d\left(x,Tx\right)+d\left(y,Ty\right)+d\left(z,Tz\right)\right]$ for all $\left(x,y,z\right)\in A\times B\times C.$

Theorem 9.

Let $A,B$ and C be nonempty closed subsets of a complete metric space $\left(X,d\right),$ and let $T:A\cup B\cup C⟶A\cup B\cup C$ be a Reich-S-type tricyclic contraction. Then, T has a unique fixed point in $A\cap B\cap C.$

Proof. 

Fix $x\in A$. We have

$$\begin{array}{ccc}\hfill d\left(T^{2}x,T^{3}x\right)& \le & D\left(Tx,T^{2}x,T^{3}x\right)\hfill \\ & \le & k\left[D(x,Tx,T^{2}x)+d\left(T^{2}x,T^{3}x\right)+d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right]\hfill \end{array}$$

$$⟹d\left(T^{2}x,T^{3}x\right)\left(1-k\right)\le k[2d\left(T^{2}x,Tx\right)+2d\left(x,Tx\right)+d\left(T^{2}x,x\right)]$$

$$\begin{array}{ccc}\hfill ⟹d\left(T^{2}x,T^{3}x\right)& \le & \frac{k}{1-k}\left[2d\left(T^{2}x,Tx\right)+2d\left(x,Tx\right)+d\left(T^{2}x,x\right)\right]\hfill \\ & \le & \frac{k}{1-k}\left[2d\left(T^{2}x,Tx\right)+2d\left(x,Tx\right)+d\left(T^{2}x,Tx\right)+d\left(Tx,x\right)\right]\hfill \\ & \le & \frac{k}{1-k}\left[3d\left(T^{2}x,Tx\right)+3d\left(x,Tx\right)\right]\hfill \end{array}$$

$$⟹d\left(T^{2}x,T^{3}x\right)\le \frac{3k}{1-k}[d\left(T^{2}x,Tx\right)+d\left(x,Tx\right)]$$

and

$$d\left(T^{2}x,Tx\right)\le D\left(Tx,T^{2}x,T^{3}x\right)\le k\left[D(x,Tx,T^{2}x)+d\left(T^{2}x,T^{3}x\right)+d\left(Tx,T^{2}x\right)+d\left(x,Tx\right)\right]$$

$$⟹d\left(T^{2}x,Tx\right)\le k\left[3d\left(T^{2}x,Tx\right)+3d\left(x,Tx\right)+d\left(T^{2}x,T^{3}x\right)\right]$$

$$⟹d\left(T^{2}x,Tx\right)\left(1-3k\right)\le k[d\left(T^{2}x,T^{3}x\right)+3d\left(x,Tx\right)]$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{k}{1-3k}d\left(T^{2}x,T^{3}x\right)+\frac{3k}{1-3k}d\left(x,Tx\right)$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{k}{1-3k}\frac{3k}{1-k}[d\left(T^{2}x,Tx\right)+d\left(x,Tx\right)]+\frac{3k}{1-3k}d\left(x,Tx\right)$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{3k^2}{\left(1-3k\right)\left(1-k\right)}d\left(T^{2}x,Tx\right)+(\frac{3k^2}{\left(1-3k\right)\left(1-k\right)}+\frac{3k}{\left(1-3k\right)})d\left(x,Tx\right)$$

$$⟹d\left(T^{2}x,Tx\right)\left(1-\frac{3k^2}{\left(1-3k\right)\left(1-k\right)}\right)\le \frac{3k^2+3k\left(1-k\right)}{\left(1-3k\right)\left(1-k\right)}d\left(x,Tx\right)$$

$$⟹d\left(T^{2}x,Tx\right)\left(\left(1-3k\right)\left(1-k\right)-3k^2\right)\le (3k^2+3k\left(1-k\right))d\left(x,Tx\right)$$

$$⟹d\left(T^{2}x,Tx\right)\left(1-4k\right)\le 3kd\left(x,Tx\right)$$

$$⟹d\left(T^{2}x,Tx\right)\le \frac{3k}{\left(1-4k\right)}d\left(x,Tx\right).$$

Then,

$$d\left(T^{2}x,Tx\right)\le td\left(x,Tx\right)\mathrm{where}t=\frac{3k}{\left(1-4k\right)}\mathrm{and}t\in \left(0,1\right),$$

which implies

$$d\left(T^{n+1}x,T^{n}x\right)\le t^{n}d\left(x,Tx\right),$$

consequently

$$\underset{n=1}{\sum ^{+\infty }}d\left(T^{n+1}x,T^{n}x\right)\le \underset{n=1}{\stackrel{+\infty }{(\sum }}{t}^n)d\left(x,Tx\right)<+\infty$$

This implies that $\left\{T^{n}x\right\}$ is a Cauchy sequence in $\left(X,d\right)$. Hence, there exists $z\in A\cup B\cup C$ such that $T^{n}x⟶z.$ Notice that $\left\{T^{3n}x\right\}$ is a sequence in $A,\left\{T^{3n-1}x\right\}$ is a sequence in C and $\left\{T^{3n-2}x\right\}$ is a sequence in B and that both sequences tend to the same limit z. Regarding the fact that $A,B$ and C are closed, we conclude that $z\in A\cap B\cap C,$ hence $A\cap B\cap C\ne ⌀.$

To show that z is a fixed point, we must show that $Tz=z$. Observe that

$$\begin{array}{ccc}\hfill d\left(Tz,z\right)& =& limd\left(Tz,T^{3n}x\right)\hfill \\ & \le & limD\left(T^{3n}x,T^{3n-1}x,Tz\right)\hfill \\ & \le & limk[d\left(T^{3n-1}x,T^{3n-2}x\right)+d\left(T^{3n-1}x,z\right)+d(T^{3n-2}x,z)\hfill \\ & +& d\left(T^{3n-1}x,T^{3n}x\right)+d\left(T^{3n-2}x,T^{3n-1}x\right)+d\left(z,Tz\right)]\hfill \\ & \le & kd\left(Tz,z\right),\hfill \end{array}$$

which is equivalent to $\left(1-k\right)d\left(Tz,z\right)=0$.

Since $k\in \left(0,\frac{1}{7}\right),$ then $d\left(Tz,z\right)=0$, which implies $Tz=z.$

To prove the uniqueness of $z,$ assume that there exists $w\in A\cup B\cup C$ such that $w\ne z$ and $Tw=w.$ Taking into account that T is tricyclic, we get $w\in A\cap B\cap C.$

$$\begin{array}{ccc}\hfill d\left(z,w\right)& =& d\left(Tz,Tw\right)\hfill \\ & \le & D\left(Tz,Tw,Tw\right)\hfill \\ & \le & k[2d\left(z,w\right)+d\left(w,w\right)+d\left(z,Tz\right)+d\left(Tw,w\right)+d\left(Tw,w\right)]\hfill \\ & \le & 2kd\left(z,w\right)\hfill \end{array}$$

implies $d\left(z,w\right)=0.$ We conclude that $z=w$ and hence z is the unique fixed point of $T.$ □

Example 4.

We take the same example 3.

Let X be ${\mathbb{R}}^2$ normed by the norm $\Vert \left(x,y\right)\Vert =\left|x\right|+\left|y\right|$,

$$A=\left\{0\right\}\times [0,+1],B=[0,+1]\times \left\{0\right\},C=\left\{0\right\}\times [-1,0]$$

and let $T:A\cup B\cup C⟶A\cup B\cup C$ be defined by

$$T\left(0,x\right)=\left(\frac{x}{6},0\right)if\left(0,x\right)\in A,$$

$$T\left(y,0\right)=\left(0,\frac{-y}{6}\right)if\left(y,0\right)\in B,$$

$$T\left(0,z\right)=\left(0,\frac{-z}{6}\right)if\left(0,z\right)\in C,$$

We have T is tricyclic and for all $\left(0,x\right)\in A,\left(y,0\right)\in B,\left(0,z\right)\in C$,

$$\begin{array}{ccc}\hfill D\left(T\left(0,x\right),T\left(y,0\right),T\left(0,z\right)\right)& =& D\left(\left(\frac{x}{6},0\right),\left(0,\frac{-y}{6}\right),\left(0,\frac{-z}{6}\right)\right)\hfill \\ & =& \frac{1}{3}(x+y-z).\hfill \end{array}$$

In addition, we have

$$D(\left(0,x\right),\left(y,0\right),\left(0,z\right))+d\left(\left(0,x\right),T\left(0,x\right)\right)+d\left(\left(y,0\right),T\left(y,0\right)\right)+d\left(\left(0,z\right),T\left(0,z\right)\right)$$

$$=2(x+y-z)+\frac{7}{6}(x+y-z)=\frac{19}{6}(x+y-z).$$

Then,

$$\begin{array}{ccc}\hfill D\left(T\left(0,x\right),T\left(y,0\right),T\left(0,z\right)\right)& =& \frac{2}{19}(D(\left(0,x\right),\left(y,0\right),\left(0,z\right))+d\left(\left(0,x\right),T\left(0,x\right)\right)\hfill \\ & & +d\left(\left(y,0\right),T\left(y,0\right)\right)+d\left(\left(0,z\right),T\left(0,z\right)\right))\hfill \\ & \le & \frac{1}{7}(D(\left(0,x\right),\left(y,0\right),\left(0,z\right))+d\left(\left(0,x\right),T\left(0,x\right)\right)\hfill \\ & & +d\left(\left(y,0\right),T\left(y,0\right)\right)+d\left(\left(0,z\right),T\left(0,z\right)\right))\hfill \end{array}$$

This implies that T is a Reich-S-type tricyclic contraction, and T has a unique fixed point $\left(0,0\right)$ in $A\cap B\cap C.$

Corollary 3.

Let $(X,d)$ a complete metric space and a self mapping $T:X⟶X$. If there exists $k\in \left(0,\frac{1}{7}\right)$ such that

$$D\left(Tx,Ty,Tz\right)\le k\left[D\left(x,y,z\right)+d\left(x,Tx\right)+d\left(y,Ty\right)+d\left(z,Tz\right)\right]$$

for all $\left(x,y,z\right)\in X^3$, then T has a unique fixed point in X.

The next tricyclic contraction considered in this section is the Cirić-S-type tricyclic contraction defined below.

Definition 6.

Let $A,B$ and C be nonempty subsets of a metric space $\left(X,d\right),$ $T:A\cup B\cup C⟶A\cup B\cup C$ be a Cirié-S-type tricyclic contraction, if there exists $k\in \left(0,1\right)$ such that

1.

$T\left(A\right)\subseteq B,T\left(B\right)\subseteq C,T\left(C,\right)\subseteq A$

2.

$D\left(Tx,Ty,Tz\right)\le kM\left(x,y,z\right)$ for all $\left(x,y,z\right)\in A\times B\times C.$

where $M\left(x,y,z\right)=max\{D\left(x,y,z\right),d\left(x,Tx\right),d\left(y,Ty\right),d\left(z,Tz\right)\}$

The fixed point theorem of the Cirić-S-type tricyclic contraction reads as follows.

Theorem 10.

Let $A,B$ and C be nonempty closed subsets of a complete metric space $\left(X,d\right),$ and let $T:A\cup B\cup C⟶A\cup B\cup C$ be a Cirić-S- type tricyclic contraction, then T has a unique fixed point in $A\cap B\cap C.$

Proof. 

Taking $x\in A,$ we have $D\left(Tx,Ty,Tz\right)\le kM\left(x,y,z\right)$ for all $\left(x,y,z\right)\in A\times B\times C$. If $M\left(x,y,z\right)=D\left(x,y,z\right)$, Theorem 7 implies the desired result.

Consider the case $M\left(x,y,z\right)=d\left(x,Tx\right).$ We have:

$$\begin{array}{ccc}\hfill D\left(Tx,T^{2}x,T^{3}x\right)\le kd\left(x,Tx\right)& ⟹& d\left(Tx,T^{2}x\right)\le kd\left(x,Tx\right)\hfill \\ & ⟹& d\left(T^{n}x,T^{n+1}x\right)\le k^{n}d\left(x,Tx\right)\hfill \end{array}$$

Consequently,

$$\underset{n=1}{\sum ^{+\infty }}d\left(T^{n+1}x,T^{n}x\right)\le \underset{n=1}{\stackrel{+\infty }{(\sum }}{k}^n)d\left(x,Tx\right)<+\infty$$

which implies that $\left\{T^{n}x\right\}$ is a Cauchy sequence in $\left(X,d\right)$. Hence, there exists $z\in A\cup B\cup C$ such that $T^{n}x⟶z.$ Notice that $\left\{T^{3n}x\right\}$ is a sequence in $A,\left\{T^{3n-1}x\right\}$ is a sequence in C, and $\left\{T^{3n-2}x\right\}$ is a sequence in B and that both sequences tend to the same limit z; regarding the fact that $A,B$ and C are closed, we conclude $z\in A\cap B\cap C,$ hence $A\cap B\cap C\ne ⌀.$

To show that z is a fixed point, we must show that $Tz=z.$ Observe that

$$d\left(Tz,z\right)=limd\left(Tz,T^{3n}x\right)\le limD\left(T^{3n}x,T^{3n-1}x,Tz\right)\le kd\left(Tz,z\right),$$

which is equivalent to $\left(1-k\right)d\left(Tz,z\right)=0$. Since $k\in \left(0,1\right),$ then $d\left(Tz,z\right)=0$, which implies $Tz=z.$

To prove the uniqueness of $z,$ assume that there exists $w\in A\cup B\cup C$ such that $w\ne z$ and $Tw=w$.

Taking into account that T is tricyclic, we get $w\in A\cap B\cap C.$

$d\left(z,w\right)=d\left(Tz,Tw\right)\le D\left(Tz,Tw,Tw\right)\le kd(z,Tz)=0$ implies $d\left(z,w\right)=0.$ We conclude that $z=w$ and hence z is the unique fixed point of $T.$

Consider the case $M\left(x,y,z\right)=d\left(y,Ty\right)$. We have:

$$D\left(Tx,T^{2}x,T^{3}x\right)\le kd\left(Tx,T^{2}x\right)⟹d\left(Tx,T^{2}x\right)\le kd\left(Tx,T^{2}x\right)<d\left(Tx,T^{2}x\right),$$

which is impossible since $k\in \left(0,1\right)$

Consider the case $M\left(x,y,z\right)=d\left(z,Tz\right)$. We have:

$$D\left(Tx,T^{2}x,T^{3}x\right)\le kd\left(T^{2}x,T^{3}x\right)⟹d\left(T^{2}x,T^{3}x\right)\le kd\left(T^{2}x,T^{3}x\right)<d\left(T^{2}x,T^{3}x\right),$$

which is impossible since $k\in \left(0,1\right).$ □

Corollary 4.

Let $A,B$ and C be a nonempty subset of a complete metric space $\left(X,d\right)$ and let a mapping $T:A\cup B\cup C⟶A\cup B\cup C$. If there exists $k\in \left(0,1\right)$ such that

1.

$T\left(A\right)\subseteq B,T\left(B\right)\subseteq C,T\left(C\right)\subseteq A.$

2.

$D\left(Tx,Ty,Tz\right)\le kmax\{D\left(x,y,z\right),d\left(x,Tx\right)\}$$\forall \left(x,y,z\right)\in A\times B\times C$.

Then, T has a unique fixed point in $A\cap B\cap C.$