*Article* **Fixed Points Results for Various Types of Tricyclic Contractions**

**Mustapha Sabiri, Abdelhafid Bassou \*, Jamal Mouline and Taoufik Sabar**

Laboratory of Algebra, Analysis and Applications (L3A), Departement of Mathematics and Computer Science, Faculty of Sciences Ben M'sik, Hassan II University of Casablanca, Casablanca 20000, Morocco; sabiri10mustapha@gmail.com (M.S.); mouline61@gmail.com (J.M.); sabarsaw@gmail.com (T.S.) **\*** Correspondence: hbassou@gmail.com

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

**Keywords:** fixed points; S-type tricyclic contraction; metric spaces

## **1. Introduction**

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

**Theorem 1.** *Let* (*X*, *d*) *be a complete metric space and a self mapping T* : *X* −→ *X. If there exists k* ∈ [0, 1) *such that, for all x*, *y* ∈ *X, d*(*Tx*, *Ty*) ≤ *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* (*X*, *d*) *be a complete metric space and a self mapping T* : *X* −→ *X. If T satisfies the following condition:*

$$d(Tx, Ty) \le k[d(x, Tx) + d(y, Ty)] \quad \text{for all} \quad x, y \in X \quad \text{where} \quad 0 < k < \frac{1}{2}$$

*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* (*X*, *d*) *is a complete metric space. If there exists* 0 < *k* < <sup>1</sup> <sup>2</sup> *such that*

$$d(Tx, Ty) \le k[d(y, Tx) + d(Ty, x)] \quad \text{for all} \quad x, y \in X\_{\prime}$$

*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* (*X*, *d*) *is a complete metric space. If there exists* 0 < *k* < <sup>1</sup> <sup>3</sup> *such that*

$$d(Tx, Ty) \le k[d(x, y) + d(x, Tx) + d(y, Ty)] \quad \text{for all} \quad x, y \in X\_{\prime}$$

**Citation:** Sabiri, M.; Bassou, A.; Mouline, J.; Sabar, T. Fixed Points Results for Various Types of Tricyclic Contractions. *Axioms* **2021**, *10*, 72. https://doi.org/10.3390/ axioms10020072


Academic Editor: Erdal Karapinar

Received: 8 March 2021 Accepted: 8 April 2021 Published: 20 April 2021

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**Copyright:** © 2021 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

,

*then T has a fixed point in X*.

In addition, in the same year, Ciri´c gave the following extension [4].

**Theorem 5.** *Ref. [4] Let T* : *X* −→ *X, where* (*X*, *d*) *a complete metric space. If there exists k* ∈ [0, 1) *such that*

$$d(Tx, Ty) \le k \text{Max} [d(x, y), d(x, Tx), d(y, Ty), d(y, Tx), d(Ty, x)] \quad \text{for all} \quad x, y \in X\_1$$

*then T has a fixed point in X*.

Many authors have investigated these situations and many results were proved (see [5–13]).

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* (*X*, *d*)*. A mapping T* : *A* ∪ *B* −→ *A* ∪ *B is said to be cyclic if :*

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

In 2003, Kirk et al. [14] proved that, if *T* : *A* ∪ *B* −→ *A* ∪ *B* is cyclic and, for some *k* ∈ (0, 1), *d*(*Tx*, *Ty*) ≤ *kd*(*x*, *y*) for all *x* ∈ *A*, *y* ∈ *B*, then *A* ∩ *B* = ∅, and *T* has a unique fixed point in *A* ∩ *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* (*X*, *d*)*, and let a mapping T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C. If T*(*A*) ⊆ *B*, *T*(*B*) ⊆ *C and T*(*C*) ⊆ *A and there exists k* ∈ (0, 1) *such that D*(*Tx*, *Ty*, *Tz*) ≤ *kD*(*x*, *y*, *z*) *for all* (*x*, *y*, *z*) ∈ *A* × *B* × *C, then A* ∩ *B* ∩ *C is nonempty and T has a unique fixed point in A* ∩ *B* ∩ *C*,

where *D*(*x*, *y*, *z*) = *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* (*X*, *d*)*. A mapping T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C is said to be tricyclic contracton if there exists* 0 < *k* < 1 *such that:*


where *δ*(*A*, *B*, *C*) = inf{*D*(*x*, *y*, *z*) : *x* ∈ *A*, *y* ∈ *B*, *z* ∈ *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* (*X*, *d*)*. A mapping T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C is said to be a Kannan-S-type tricyclic contraction, if there exists k* ∈ 0, <sup>1</sup> 3 *such that*


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* <sup>R</sup><sup>2</sup> *normed by the norm* (*x*, *<sup>y</sup>*) <sup>=</sup> <sup>|</sup>*x*<sup>|</sup> <sup>+</sup> <sup>|</sup>*y*|, *and A* = [1, 2] × {0}, *<sup>B</sup>* <sup>=</sup> {0} × [−2, −1], *C* = [−2, −1] × {0}, *then*

$$\delta(A, B, \mathbb{C}) = D((1, 0), (0, -1), (-1, 0)) = 6.$$

*Put T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C such that*

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

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

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

*We have T*(*A*) ⊆ *B*, *T*(*B*) ⊆ *C and T*(*C*) ⊆ *A, and*

$$\begin{aligned} D(T(\mathbf{x},0), T(0,y), T(z,0)) &= \, ^0D( (0, -\frac{\mathbf{x}+2}{3}), (\frac{y-2}{3}, 0), (-\frac{z-2}{3}, 0) )\\ &= \, ^2\frac{2}{3}(\mathbf{x}-y-z) + 4\\ &= \, ^1\frac{1}{3}D( (\mathbf{x},0), (0,y), (z,0) ) + 4\\ &= \, ^1\frac{1}{3}D( (\mathbf{x},0), (0,y), (z,0) ) + (1-\frac{1}{3})\delta(A,B,C) \end{aligned}$$

*for all* (*x*, 0) ∈ *A*,(0, *y*) ∈ *B*,(*z*, 0) ∈ *C*. *On the other hand,*

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

*and*

$$d((2,0),T(2,0)) + d((0,-2),T(0,-2)) + d((-2,0),T(-2,0)) = 10\sqrt{}$$

*which implies that D*(*T*(2, 0), *T*(0, −2), *T*(−2, 0))

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

*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* = R *with the usual metric. Let A* = *B* = *C* = [0, 1], *then δ*(*A*, *B*, *C*) = 0. *Put T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C such that*

$$T\mathbf{x} = \frac{1}{6} \text{ if } 0 \le \mathbf{x} < 1, \quad T\mathbf{x} = \frac{1}{4} \text{ if } \mathbf{x} = 1$$

*For x* = 1, *y* = 1 *and z* = <sup>23</sup> <sup>24</sup> , *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*

$$D(T(1), T(1), T(1)) = 0 \le \frac{9}{4}k^2$$

*for all k* <sup>≥</sup> <sup>0</sup>*, then for* <sup>0</sup> <sup>≤</sup> *<sup>k</sup>* <sup>&</sup>lt; <sup>1</sup> 3 .

• *If x* ∈ [0, 1)*, y* ∈ [0, 1) *and z* ∈ [0, 1)*, we have*

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

$$\begin{array}{l}\text{for all } k \ge 0 \text{, then for } 0 \le k < \frac{1}{3}.\\\text{0} \quad \text{If } \mathbf{x} = \mathbf{1}, \mathbf{y} \in \left[0, 1\right) \text{ and } \mathbf{z} \in \left[0, 1\right) \text{, we have}\end{array}$$

$$D(T\_{1\prime}T\mathcal{Y}\_{\prime}Tz) = D(\frac{1}{4}, \frac{1}{6}, \frac{1}{6}) = \frac{1}{6}$$

*and*

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

6

*then, for k* = <sup>2</sup> <sup>9</sup> *, 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* ∈ [0, 1), *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* = <sup>2</sup> <sup>9</sup> *, we have*

$$D(T(1), T(1), Tz) \le k(d(1, T(1)) + d(1, T(1)) + d(z, Tz)).$$

*Consequently, for k* = <sup>2</sup> <sup>9</sup> *, we have :*

$$D(T\mathbf{x}, T\mathbf{y}, T\mathbf{z}) \le k(d(\mathbf{x}, T\mathbf{x}) + d(\mathbf{y}, T\mathbf{y}) + d(\mathbf{z}, T\mathbf{z})) \text{ for all } (\mathbf{x}, \mathbf{y}, \mathbf{z}) \in A \times B \times \mathbb{C} \dots$$

**Theorem 7.** *Let A*, *B and C be nonempty closed subsets of a complete metric space* (*X*, *d*), *and let T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be a Kannan-S-type tricyclic contraction. Then, T has a unique fixed point in A* ∩ *B* ∩ *C*.

**Proof.** Fix *x* ∈ *A*. We have

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

Then,

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

which implies

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

Similarly, we have

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

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

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

Then,

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

which implies

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

Consequently,

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

implies that {*Tnx*} is a Cauchy sequence in (*X*, *<sup>d</sup>*). Hence, there exists *<sup>z</sup>* <sup>∈</sup> *<sup>A</sup>* <sup>∪</sup> *<sup>B</sup>* <sup>∪</sup> *<sup>C</sup>* such that *<sup>T</sup>nx* −→ *<sup>z</sup>*. Notice that {*T*<sup>3</sup>*nx*} is a sequence in *<sup>A</sup>*, {*T*3*n*−1*x*} is a sequence in *<sup>C</sup>* and {*T*3*n*−2*x*} is a sequence in *<sup>B</sup>* and that both sequences tend to the same limit *<sup>z</sup>*. Regarding the fact that *A*, *B* and *C* are closed, we conclude *z* ∈ *A* ∩ *B* ∩ *C*, hence *A* ∩ *B* ∩ *C* = ∅. To show that *z* is a fixed point, we must show that *Tz* = *z*. Observe that

$$\begin{aligned} d(Tz, z) &= \lim\_{\varepsilon \to 0} d\left(Tz, T^{3n}\mathbf{x}\right) \le \lim\_{\varepsilon \to 0} D\left(T^{3n}\mathbf{x}, T^{3n-1}\mathbf{x}, T\mathbf{z}\right) \\ &\le \lim\_{\varepsilon \to 0} k[d\left(T^{3n-1}\mathbf{x}, T^{3n}\mathbf{x}\right) + d\left(T^{3n-2}\mathbf{x}, T^{3n-1}\mathbf{x}\right) + d(z, Tz)] \\ &\le \quad kd(Tz, z), \end{aligned}$$

which is equivalent to

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

Since *k* ∈ 0, <sup>1</sup> 3 , then *d*(*Tz*, *z*) = 0, which implies *Tz* = *z*.

To prove the uniqueness of *z*,, assume that there exists *w* ∈ *A* ∪ *B* ∪ *C* such that *w* = *z* and *Tw* = *w*. Taking into account that *T* is tricyclic, we get *w* ∈ *A* ∩ *B* ∩ *C*. We have

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

which implies *d*(*z*, *w*) = 0. We get that *z* = *w* and hence *z* is the unique fixed point of *T*.

**Example 3.** *Let <sup>X</sup> be* <sup>R</sup><sup>2</sup> *normed by the norm* (*x*, *<sup>y</sup>*) <sup>=</sup> <sup>|</sup>*x*<sup>|</sup> <sup>+</sup> <sup>|</sup>*y*|, *let <sup>A</sup>* <sup>=</sup> {0} × [0, <sup>+</sup>1], *<sup>B</sup>* <sup>=</sup> [0, +1] × {0}*, C* = {0} × [−1, 0] *and let T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be defined by*

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

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

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

*We have*

$$T(A) \subseteq B, T(B) \subseteq \mathbb{C} \text{ and } T(\mathbb{C}) \subseteq A.$$

*In addition, for all* (0, *x*) ∈ *A*,(*y*, 0) ∈ *B*,(0, *z*) ∈ *C, we have*

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

*In addition, we have*

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

*This implies*

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

*Then, T is a Kannan-S-type tricyclic contraction, and T has a unique fixed point* (0, 0) *in A* ∩ *B* ∩ *C*.

**Corollary 1.** *Let* (*X*, *d*) *be a complete metric space and a self mapping T* : *X* −→ *X. If there exists k* ∈ 0, <sup>1</sup> 3 *such that*

$$D(T\mathbf{x}, T\mathbf{y}, T\mathbf{z}) \le k[d(\mathbf{x}, T\mathbf{x}) + d(\mathbf{y}, T\mathbf{y}) + d(\mathbf{z}, T\mathbf{z})],$$

*for all* (*x*, *<sup>y</sup>*, *<sup>z</sup>*) <sup>∈</sup> *<sup>X</sup>*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* (*X*, *d*)*. A mapping T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C is said to be a Chattergea-S-type tricyclic contraction if T*(*A*) ⊆ *B*, *T*(*B*) ⊆ *C*, *T*(*C*) ⊆ *A, and there exist k* ∈ 0, <sup>1</sup> 3 *such that D*(*Tx*, *Ty*, *Tz*) ≤ *k*[*d*(*y*, *Tx*) + *d*(*z*, *Ty*) + *d*(*x*, *Tz*)] *for all* (*x*, *y*, *z*) ∈ *A* × *B* × *C*.

**Theorem 8.** *Let A*, *B and C be nonempty closed subsets of a complete metric space* (*X*, *d*)*, and let T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be a Chattergea-S-type tricyclic contraction. Then, T has a unique fixed point in A* ∩ *B* ∩ *C*.

**Proof.** Fix *x* ∈ *A*. We have

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

which implies

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

so

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

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

and

$$d\left(T^2\mathbf{x}, T\mathbf{x}\right) \le D\left(T^3\mathbf{x}, T^2\mathbf{x}, T\mathbf{x}\right) \le \frac{k}{(1-k)} \left[d\left(T^3\mathbf{x}, T^2\mathbf{x}\right) + d(\mathbf{x}, T\mathbf{x})\right],\tag{182}$$

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

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

Then,

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

which implies

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

for all *n* ≥ 1. Consequently,

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

implies that {*Tnx*} is a Cauchy sequence in (*X*, *<sup>d</sup>*). Hence, there exists *<sup>z</sup>* <sup>∈</sup> *<sup>A</sup>* <sup>∪</sup> *<sup>B</sup>* <sup>∪</sup> *<sup>C</sup>* such that *<sup>T</sup>nx* −→ *<sup>z</sup>*. Notice that {*T*<sup>3</sup>*nx*} is a sequence in *<sup>A</sup>*, {*T*3*n*−1*x*} is a sequence in *<sup>C</sup>*, and {*T*3*n*−2*x*} is a sequence in *<sup>B</sup>* and that both sequences tend to the same limit *<sup>z</sup>*. Regarding that *A*, *B* and *C* are closed, we conclude *z* ∈ *A* ∩ *B* ∩ *C*, hence *A* ∩ *B* ∩ *C* = ∅.

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

$$\begin{aligned} d(Tz, z) &= \lim\_{\varepsilon \to 0} d\left(Tz, T^{3n}\mathbf{x}\right) \le \lim\_{\varepsilon \to 0} D\left(Tz, T^{3n}\mathbf{x}, T^{3n-1}\mathbf{x}\right) \\ &\le \lim\_{\varepsilon \to 0} k \left[ d\left(T^{3n-1}\mathbf{x}, T\mathbf{z}\right) + \left(T^{3n-2}\mathbf{x}, T^{3n}\mathbf{x}\right) + d(z, T^{3n-1}\mathbf{x}) \right] \le k d(Tz, z), \end{aligned}$$

which is equivalent to (1 − *k*)*d*(*Tz*, *z*) = 0. Since *k* ∈ 1, <sup>1</sup> 3 , then *d*(*Tz*, *z*) = 0, which implies *Tz* = *z*.

To prove the uniqueness of *z*, assume that there exists *w* ∈ *A* ∪ *B* ∪ *C* such that *w* = *z* and *Tw* = *w*. Taking into account that *T* is tricyclic, we get *w* ∈ *A* ∩ *B* ∩ *C*.

We have

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

Then, *d*(*z*, *w*) = 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* ∈ 0, <sup>1</sup> 3 *such that*

$$D(T\mathbf{x}, T\mathbf{y}, Tz) \le k[d(\mathbf{y}, T\mathbf{x}) + d(z, Ty) + d(\mathbf{x}, Tz)]$$

*for all* (*x*, *<sup>y</sup>*, *<sup>z</sup>*) <sup>∈</sup> *<sup>X</sup>*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* (*X*, *d*)*. A mapping T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C is said to be a Reich-S-type tricyclic contraction if there exists k* ∈ 0, <sup>1</sup> 7 *such that:*

*1. T*(*A*) ⊆ *B*, *T*(*B*) ⊆ *C*, *T*(*C*) ⊆ *A*.

*2. D*(*Tx*, *Ty*, *Tz*) ≤ *k*[*D*(*x*, *y*, *z*) + *d*(*x*, *Tx*) + *d*(*y*, *Ty*) + *d*(*z*, *Tz*)] *for all* (*x*, *y*, *z*) ∈ *A* × *B* × *C*.

**Theorem 9.** *Let A*, *B and C be nonempty closed subsets of a complete metric space* (*X*, *d*), *and let T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be a Reich-S-type tricyclic contraction. Then, T has a unique fixed point in A* ∩ *B* ∩ *C*.

**Proof.** Fix *x* ∈ *A*. We have

*d T*2*x*, *T*3*x* ≤ *D Tx*, *T*2*x*, *T*3*x* ≤ *k D*(*x*, *Tx*, *T*2*x*) + *d T*2*x*, *T*3*x* + *d Tx*, *T*2*x* + *d*(*x*, *Tx*) =⇒ *d T*2*x*, *T*3*x* (1 − *k*) ≤ *k*[2*d <sup>T</sup>*2*x*, *Tx* + 2*d*(*x*, *Tx*) + *d T*2*x*, *x* ] =⇒ *d T*2*x*, *T*3*x* ≤ *k* 1 − *k* 2*d <sup>T</sup>*2*x*, *Tx* + 2*d*(*x*, *Tx*) + *d T*2*x*, *x* ≤ *k* 1 − *k* 2*d <sup>T</sup>*2*x*, *Tx* + 2*d*(*x*, *Tx*) + *d <sup>T</sup>*2*x*, *Tx* + *d*(*Tx*, *x*) ≤ *k* 1 − *k* 3*d <sup>T</sup>*2*x*, *Tx* + 3*d*(*x*, *Tx*) =⇒ *d T*2*x*, *T*3*x* <sup>≤</sup> <sup>3</sup>*<sup>k</sup>* 1 − *k* [*d <sup>T</sup>*2*x*, *Tx* + *d*(*x*, *Tx*)]

and

*d <sup>T</sup>*2*x*, *Tx* ≤ *D Tx*, *T*2*x*, *T*3*x* ≤ *k D*(*x*, *Tx*, *T*2*x*) + *d T*2*x*, *T*3*x* + *d Tx*, *T*2*x* + *d*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx* ≤ *k* 3*d <sup>T</sup>*2*x*, *Tx* + 3*d*(*x*, *Tx*) + *d T*2*x*, *T*3*x* =⇒ *d <sup>T</sup>*2*x*, *Tx* (1 − 3*k*) ≤ *k*[*d T*2*x*, *T*3*x* + 3*d*(*x*, *Tx*)] =⇒ *d <sup>T</sup>*2*x*, *Tx* ≤ *k* 1 − 3*k d T*2*x*, *T*3*x* + 3*k* 1 − 3*k d*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx* ≤ *k* 1 − 3*k* 3*k* 1 − *k* [*d <sup>T</sup>*2*x*, *Tx* <sup>+</sup> *<sup>d</sup>*(*x*, *Tx*)] + <sup>3</sup>*<sup>k</sup>* 1 − 3*k d*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx* <sup>≤</sup> <sup>3</sup>*k*<sup>2</sup> (1 − 3*k*)(1 − *k*) *d <sup>T</sup>*2*x*, *Tx* + ( <sup>3</sup>*k*<sup>2</sup> (<sup>1</sup> <sup>−</sup> <sup>3</sup>*k*)(<sup>1</sup> <sup>−</sup> *<sup>k</sup>*) <sup>+</sup> 3*k* (1 − 3*k*) )*d*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx* <sup>1</sup> <sup>−</sup> <sup>3</sup>*k*<sup>2</sup> (1 − 3*k*)(1 − *k*) <sup>≤</sup> <sup>3</sup>*k*<sup>2</sup> <sup>+</sup> <sup>3</sup>*k*(<sup>1</sup> <sup>−</sup> *<sup>k</sup>*) (1 − 3*k*)(1 − *k*) *d*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx*(<sup>1</sup> <sup>−</sup> <sup>3</sup>*k*)(<sup>1</sup> <sup>−</sup> *<sup>k</sup>*) <sup>−</sup> <sup>3</sup>*k*<sup>2</sup> <sup>≤</sup> (3*k*<sup>2</sup> <sup>+</sup> <sup>3</sup>*k*(<sup>1</sup> <sup>−</sup> *<sup>k</sup>*))*d*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx* (1 − 4*k*) ≤ 3*kd*(*x*, *Tx*) =⇒ *d <sup>T</sup>*2*x*, *Tx* <sup>≤</sup> <sup>3</sup>*<sup>k</sup>* (1 − 4*k*) *d*(*x*, *Tx*).

Then,

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

which implies

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

consequently

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

This implies that {*Tnx*} is a Cauchy sequence in (*X*, *<sup>d</sup>*). Hence, there exists *<sup>z</sup>* <sup>∈</sup> *<sup>A</sup>* <sup>∪</sup> *<sup>B</sup>* <sup>∪</sup> *<sup>C</sup>* such that *<sup>T</sup>nx* −→ *<sup>z</sup>*. Notice that {*T*<sup>3</sup>*nx*} is a sequence in *<sup>A</sup>*, {*T*3*n*−1*x*} is a sequence in *<sup>C</sup>* and {*T*3*n*−2*x*} is a sequence in *<sup>B</sup>* and that both sequences tend to the same limit *z*. Regarding the fact that *A*, *B* and *C* are closed, we conclude that *z* ∈ *A* ∩ *B* ∩ *C*, hence *A* ∩ *B* ∩ *C* = ∅.

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

$$\begin{aligned} d(Tz, z) &= \quad \lim\_{} d\left(Tz, T^{3n}x\right) \\ &\le \quad \lim\_{} D\left(T^{3n}x, T^{3n-1}x, Tz\right) \\ &\le \quad \lim\_{} k[d\left(T^{3n-1}x, T^{3n-2}x\right) + d\left(T^{3n-1}x, z\right) + d\left(T^{3n-2}x, z\right)] \\ &+ d\left(T^{3n-1}x, T^{3n}x\right) + d\left(T^{3n-2}x, T^{3n-1}x\right) + d(z, Tz)] \\ &\le \quad kd(Tz, z)\_{\prime} \end{aligned}$$

which is equivalent to (1 − *k*)*d*(*Tz*, *z*) = 0.

Since *k* ∈ 0, <sup>1</sup> 7 , then *d*(*Tz*, *z*) = 0, which implies *Tz* = *z*.

To prove the uniqueness of *z*, assume that there exists *w* ∈ *A* ∪ *B* ∪ *C* such that *w* = *z* and *Tw* = *w*. Taking into account that *T* is tricyclic, we get *w* ∈ *A* ∩ *B* ∩ *C*.

$$\begin{aligned} d(z, w) &= \quad d(T z, Tw) \\ &\le \quad D(T z, Tw, Tw) \\ &\le \quad k[2d(z, w) + d(w, w) + d(z, Tz) + d(Tw, w) + d(Tw, w)] \\ &\le \quad 2kd(z, w) \end{aligned}$$

implies *d*(*z*, *w*) = 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* <sup>R</sup><sup>2</sup> *normed by the norm* (*x*, *<sup>y</sup>*) <sup>=</sup> <sup>|</sup>*x*<sup>|</sup> <sup>+</sup> <sup>|</sup>*y*|*,*

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

*and let T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be defined by*

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

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

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

*We have T is tricyclic and for all* (0, *x*) ∈ *A*,(*y*, 0) ∈ *B*,(0, *z*) ∈ *C ,*

$$\begin{aligned} D(T(0,x),T(y,0),T(0,z)) &&= \, \_0D\left( \left( \frac{x}{6},0 \right), \left( 0, \frac{-y}{6} \right), \left( 0, \frac{-z}{6} \right) \right) \\ &= \, \_1\frac{1}{3}(x+y-z). \end{aligned}$$

*In addition, we have*

$$D((0, \mathbf{x}), (y, 0), (0, z)) + d((0, \mathbf{x}), T(0, \mathbf{x})) + d((y, 0), T(y, 0)) + d((0, z), T(0, z))$$

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

*Then,*

$$\begin{aligned} D(T(0, \mathbf{x}), T(y, 0), T(0, z)) &= \frac{2}{19} (D((0, \mathbf{x}), (y, 0), (0, z)) + d((0, \mathbf{x}), T(0, \mathbf{x}))) \\ &+ d((y, 0), T(y, 0)) + d((0, z), T(0, z))) \\ &\leq \frac{1}{7} (D((0, \mathbf{x}), (y, 0), (0, z)) + d((0, \mathbf{x}), T(0, \mathbf{x}))) \\ &+ d((y, 0), T(y, 0)) + d((0, z), T(0, z))) \end{aligned}$$

*This implies that T is a Reich-S-type tricyclic contraction, and T has a unique fixed point* (0, 0) *in A* ∩ *B* ∩ *C*.

**Corollary 3.** *Let* (*X*, *d*) *a complete metric space and a self mapping T* : *X* −→ *X. If there exists k* ∈ 0, <sup>1</sup> 7 *such that*

$$D(Tx, Ty, Tz) \le k[D(\mathbf{x}, y, z) + d(\mathbf{x}, Tx) + d(y, Ty) + d(z, Tz)]$$

*for all* (*x*, *<sup>y</sup>*, *<sup>z</sup>*) <sup>∈</sup> *<sup>X</sup>*3*, then T has a unique fixed point in X.*

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

**Definition 6.** *Let A*, *B and C be nonempty subsets of a metric space* (*X*, *d*), *T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be a Cirié-S-type tricyclic contraction, if there exists k* ∈ (0, 1) *such that*

*1. T*(*A*) ⊆ *B*, *T*(*B*) ⊆ *C*, *T*(*C*) ⊆ *A 2. D*(*Tx*, *Ty*, *Tz*) ≤ *kM*(*x*, *y*, *z*) *for all* (*x*, *y*, *z*) ∈ *A* × *B* × *C*. *where M*(*x*, *y*, *z*) = max{*D*(*x*, *y*, *z*), *d*(*x*, *Tx*), *d*(*y*, *Ty*), *d*(*z*, *Tz*)}

The fixed point theorem of the Ciri´c-S-type tricyclic contraction reads as follows.

**Theorem 10.** *Let A*, *B and C be nonempty closed subsets of a complete metric space* (*X*, *d*), *and let T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C be a Ciri´c-S- type tricyclic contraction, then T has a unique fixed point in A* ∩ *B* ∩ *C*.

**Proof.** Taking *x* ∈ *A*, we have *D*(*Tx*, *Ty*, *Tz*) ≤ *kM*(*x*, *y*, *z*) for all (*x*, *y*, *z*) ∈ *A* × *B* × *C*. If *M*(*x*, *y*, *z*) = *D*(*x*, *y*, *z*), Theorem 7 implies the desired result.

Consider the case *M*(*x*, *y*, *z*) = *d*(*x*, *Tx*). We have:

$$\begin{aligned} D\left(T\mathbf{x}, T^2\mathbf{x}, T^3\mathbf{x}\right) \le k d(\mathbf{x}, T\mathbf{x}) &\implies d\left(T\mathbf{x}, T^2\mathbf{x}\right) \le k d(\mathbf{x}, T\mathbf{x})\\ \implies d\left(T^n\mathbf{x}, T^{n+1}\mathbf{x}\right) \le k^n d(\mathbf{x}, T\mathbf{x}) \end{aligned}$$

Consequently,

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

which implies that {*Tnx*} is a Cauchy sequence in (*X*, *<sup>d</sup>*). Hence, there exists *<sup>z</sup>* <sup>∈</sup> *<sup>A</sup>* <sup>∪</sup> *<sup>B</sup>* <sup>∪</sup> *<sup>C</sup>* such that *<sup>T</sup>nx* −→ *<sup>z</sup>*. Notice that {*T*<sup>3</sup>*nx*} is a sequence in *<sup>A</sup>*, {*T*3*n*−1*x*} is a sequence in *<sup>C</sup>*, and {*T*3*n*−2*x*} is a sequence in *<sup>B</sup>* and that both sequences tend to the same limit *<sup>z</sup>*; regarding the fact that *A*, *B* and *C* are closed, we conclude *z* ∈ *A* ∩ *B* ∩ *C*, hence *A* ∩ *B* ∩ *C* = ∅.

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

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

which is equivalent to (1 − *k*)*d*(*Tz*, *z*) = 0. Since *k* ∈ (0, 1), then *d*(*Tz*, *z*) = 0, which implies *Tz* = *z*.

To prove the uniqueness of *z*, assume that there exists *w* ∈ *A* ∪ *B* ∪ *C* such that *w* = *z* and *Tw* = *w*.

Taking into account that *T* is tricyclic, we get *w* ∈ *A* ∩ *B* ∩ *C*.

*d*(*z*, *w*) = *d*(*Tz*, *Tw*) ≤ *D*(*Tz*, *Tw*, *Tw*) ≤ *kd*(*z*, *Tz*) = 0 implies *d*(*z*, *w*) = 0. We conclude that *z* = *w* and hence *z* is the unique fixed point of *T*. Consider the case *M*(*x*, *y*, *z*) = *d*(*y*, *Ty*). We have :

$$d\left(T\mathbf{x}, T^2\mathbf{x}, T^3\mathbf{x}\right) \le k d\left(T\mathbf{x}, T^2\mathbf{x}\right) \implies d\left(T\mathbf{x}, T^2\mathbf{x}\right) \le k d\left(T\mathbf{x}, T^2\mathbf{x}\right) < d\left(T\mathbf{x}, T^2\mathbf{x}\right) \dots$$

which is impossible since *k* ∈ (0, 1)

Consider the case *M*(*x*, *y*, *z*) = *d*(*z*, *Tz*). We have:

$$D\left(T\mathbf{x}, T^2\mathbf{x}, T^3\mathbf{x}\right) \le k d\left(T^2\mathbf{x}, T^3\mathbf{x}\right) \implies d\left(T^2\mathbf{x}, T^3\mathbf{x}\right) \le k d\left(T^2\mathbf{x}, T^3\mathbf{x}\right) < d\left(T^2\mathbf{x}, T^3\mathbf{x}\right),$$

which is impossible since *k* ∈ (0, 1).

**Corollary 4.** *Let A*, *B and C be a nonempty subset of a complete metric space* (*X*, *d*) *and let a mapping T* : *A* ∪ *B* ∪ *C* −→ *A* ∪ *B* ∪ *C. If there exists k* ∈ (0, 1) *such that*


*Then, T has a unique fixed point in A* ∩ *B* ∩ *C*.

**Author Contributions:** Conceptualization, M.S. and A.B.; validation, M.S., J.M. and A.B.; writing original draft preparation, M.S. and J.M.; writing—review and editing, A.B. and T.S.; supervision, J.M. and A.B.; project administration, M.S. and J.M. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Institutional Review Board Statement:** Not applicable.

**Informed Consent Statement:** Not applicable.

**Data Availability Statement:** Not applicable.

**Conflicts of Interest:** The authors declare no conflict of interest.

#### **References**

