# A fixed point theorem for cyclic generalized contractions in metric spaces

## Abstract

In this paper, we extend a recent result of V. Pata (J. Fixed Point Theory Appl. 10:299-305, 2011) in the frame of a cyclic representation of a complete metric space.

## 1 Introduction

One of the fundamental result in fixed point theory is the Banach contraction principle. It has various non-trivial applications in many branches of pure and applied sciences (see, for instance, [2, 7, 14] and references cited therein).

Let $\left(X,d\right)$ be a metric space and $f:X\to X$ be an operator. We say that f is a contraction if there exists $\lambda \in \left[0,1\right)$ such that, for all $x,y\in X$,

$d\left(f\left(x\right),f\left(y\right)\right)\le \lambda d\left(x,y\right).$
(1.1)

In terms of Picard operator theory (see ), Banach contraction principle asserts that if f is a contraction and $\left(X,d\right)$ is complete, then f is a Picard operator. This result has been extended to other important classes of maps. Recently, Pata  proved that if $\left(X,d\right)$ is a complete metric space and $f:X\to X$ is an operator such that there exists fixed constants $\gamma \ge 0$, $\alpha \ge 1$ and $\beta \in \left[0,\alpha \right]$ such that, for every $\epsilon \in \left[0,1\right]$ and every $x,y\in X$,

$d\left(f\left(x\right),f\left(y\right)\right)\le \left(1-\epsilon \right)d\left(x,y\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel x\parallel +\parallel y\parallel \right]}^{\beta }$
(1.2)

(where $\psi :\left[0,1\right]\to \left[0,\mathrm{\infty }\right)$ is an increasing function vanishing with continuity at zero and $\parallel x\parallel :=d\left(x,{x}_{0}\right)$, with arbitrary ${x}_{0}\in X$), then f has a unique fixed point in X.

Remark 1.1 (see )

The condition (1.2) is weaker than the contraction condition (1.1). In fact, if

then it can be verified that, for every $x,y\in X$, we have

where

$\gamma =\gamma \left(\theta ,\lambda \right)=\frac{{\theta }^{\theta }}{{\left(1+\theta \right)}^{1+\theta }}\frac{1}{{\left(1-\lambda \right)}^{\theta }}.$

Remark 1.2 (see )

The function $f:\left[1,\mathrm{\infty }\right)\to \left[1,\mathrm{\infty }\right)$ defined as

$f\left(x\right)=-2+x-2\sqrt{x}+4\sqrt{x}$

has a unique fixed point ${x}^{\ast }=1$, but fails to be a contraction on any neighborhood both of 1 and of ∞.

Kirk, Srinivasan and Veeramani  obtained an extension of Banach’s fixed point theorem for mappings satisfying cyclical contractive conditions. Some generalizations of the results given in , using the setting of so-called fixed point structures, are presented in I. A. Rus . In , Păcurar and Rus established a fixed point theorem for cyclic φ-contractions and they further discussed fixed point theory in metric spaces. In , Karapinar proved a fixed point theorem for cyclic weak φ-contraction mappings. Some other recent results concerning this topic are given in [1, 4, 5, 9, 11].

In the present paper, we obtain a fixed point theorem for a generalized contraction in the sense of the assumption (1.2), defined on a cyclic representation of a complete metric space.

## 2 Main results

We need first to recall a known concept.

Definition 2.1 ()

Let X be a nonempty set, m be a positive integer and $f:X\to X$ an operator. Then, we say that ${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of X with respect to f if:

1. (i)

$X={\bigcup }_{i=1}^{m}{A}_{i}$, where ${A}_{i}$ are nonempty sets for each $i\in \left\{1,\dots ,m\right\}$;

2. (ii)

$f\left({A}_{1}\right)\subset {A}_{2},\dots ,f\left({A}_{m-1}\right)\subset {A}_{m},f\left({A}_{m}\right)\subset {A}_{1}$.

Let $\left(X,d\right)$ be a complete metric space. Selecting an arbitrary ${x}_{1}\in X$, we denote

Our main result is as follows.

Theorem 2.2 Let $\left(X,d\right)$ be a complete metric space, m be a positive integer, ${A}_{1},\dots ,{A}_{m}$ be closed nonempty subsets of X, $Y:={\bigcup }_{i=1}^{m}{A}_{i}$, $\psi :\left[0,1\right]\to \left[0,\mathrm{\infty }\right)$ be an increasing function vanishing with continuity at zero, and $f:Y\to Y$ be an operator. Assume that:

1. 1.

${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of Y with respect to f;

2. 2.

For every $\epsilon \in \left[0,1\right]$, $x\in {A}_{i}$, and $y\in {A}_{i+1}$ ($i\in \left\{1,\dots ,m\right\}$, where ${A}_{m+1}={A}_{1}$), we have

$d\left(f\left(x\right),f\left(y\right)\right)\le \left(1-\epsilon \right)d\left(x,y\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel x\parallel +\parallel y\parallel \right]}^{\beta },$
(2.1)

where $\gamma \ge 0$, $\alpha \ge 1$ and $\beta \in \left[0,\alpha \right]$ are fixed constants.

Then, we have the following conclusions:

1. (i)

f is a Picard operator, i.e., f has a unique fixed point ${x}^{\ast }\in {\bigcap }_{i=1}^{m}{A}_{i}$ and the Picard iteration sequence ${\left\{{f}^{n}\left(x\right)\right\}}_{n\in \mathbb{N}}$ converges to ${x}^{\ast }$, for any initial point $x\in Y$;

2. (ii)

the following estimates hold:

$\begin{array}{c}d\left({x}_{n},{x}^{\ast }\right)\le \parallel {x}^{\ast }\parallel ,\phantom{\rule{1em}{0ex}}n\ge 2;\hfill \\ d\left({x}_{n},{x}_{1}\right)\le 2\parallel {x}^{\ast }\parallel ,\phantom{\rule{1em}{0ex}}n\ge 2.\hfill \end{array}$

Proof (i) For convenience of notation, if $j>m$, define ${A}_{j}={A}_{i}$ where $i=jmodm$ and $1\le i\le m$. Let ${x}_{1}\in {A}_{1}$. Starting from ${x}_{1}$, let ${\left\{{x}_{n}\right\}}_{n\ge 1}$ be the Picard iteration defined by the sequence

${x}_{n}=f\left({x}_{n-1}\right)={f}^{n-1}\left({x}_{1}\right),\phantom{\rule{1em}{0ex}}n\ge 2,$

and set ${c}_{n}=\parallel {x}_{n}\parallel$. Assume ${x}_{n}\ne {x}_{n+1}$ for all n. By (2.1), we have

$d\left({x}_{n},{x}_{n+1}\right)\le d\left({x}_{n-1},{x}_{n}\right)\le \cdots \le d\left({x}_{1},{x}_{2}\right)={c}_{2}.$
(2.2)

First, we prove that the sequence ${\left({c}_{n}\right)}_{n\in {\mathbb{N}}^{\ast }}$ is bounded. By (2.2) we get that

$\begin{array}{rcl}{c}_{n}& \le & d\left({x}_{n},{x}_{n+1}\right)+d\left({x}_{n+1},{x}_{2}\right)+d\left({x}_{2},{x}_{1}\right)\le d\left({x}_{n+1},{x}_{2}\right)+2{c}_{2}\\ =& d\left(f\left({x}_{n}\right),f\left({x}_{1}\right)\right)+2{c}_{2}.\end{array}$

Since ${x}_{1}\in {A}_{1}$ and ${x}_{n}\in {A}_{n}$, from (2.1), we obtain that

$\begin{array}{rcl}{c}_{n}& \le & \left(1-\epsilon \right)d\left({x}_{n},{x}_{1}\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel {x}_{n}\parallel +\parallel {x}_{1}\parallel \right]}^{\beta }+2{c}_{2}\\ =& \left(1-\epsilon \right){c}_{n}+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+{c}_{n}\right]}^{\beta }+2{c}_{2}\\ \le & \left(1-\epsilon \right){c}_{n}+a{\epsilon }^{\alpha }\psi \left(\epsilon \right){c}_{n}^{\alpha }+b,\end{array}$

where ${c}_{1}=\parallel {x}_{1}\parallel =d\left({x}_{1},{x}_{1}\right)=0$, $\beta \le \alpha$, and for some $a,b>0$. Thus,

$\epsilon {c}_{n}\le a{\epsilon }^{\alpha }\psi \left(\epsilon \right){c}_{n}^{\alpha }+b.$

If there is a subsequence ${\left({c}_{{n}_{k}}\right)}_{k\in {\mathbb{N}}^{\ast }}\to \mathrm{\infty }$, the choice $\epsilon ={\epsilon }_{k}=\frac{\left(1+b\right)}{{c}_{{n}_{k}}}$ leads to the contradiction

$1\le a{\left(1+b\right)}^{\alpha }\psi \left({\epsilon }_{k}\right)\to 0.$

Therefore, the sequence $\left({c}_{n}\right)$ is bounded.

From (2.2) we obtain that the sequence $\left\{d\left({x}_{n},{x}_{n+1}\right)\right\}$ is nonincreasing and then it is convergent to the real number

$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{n+1}\right)=r=inf\left\{d\left({x}_{n-1},{x}_{n}\right):n=2,3,\dots \right\}.$

Now we show that $r=0$. Assume that $r>0$. Let ${x}_{n}\in {A}_{n}$ and ${x}_{n+1}\in {A}_{n+1}$. By (2.1), we have

$\begin{array}{rcl}r& \le & d\left({x}_{n},{x}_{n+1}\right)=d\left(f\left({x}_{n-1}\right),f\left({x}_{n}\right)\right)\\ \le & \left(1-\epsilon \right)d\left({x}_{n-1},{x}_{n}\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel {x}_{n-1}\parallel +\parallel {x}_{n}\parallel \right]}^{\beta }\\ \le & \left(1-\epsilon \right)d\left({x}_{n-1},{x}_{n}\right)+K\epsilon \psi \left(\epsilon \right),\end{array}$

for some $K>0$. Letting $n\to \mathrm{\infty }$, we obtain

which implies $r=0$. This leads to a contradiction, therefore

$\underset{n\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{n+1}\right)=0.$

For $p\ge 1$, suppose there exists j, $0\le j\le m-1$, such that $\left(n+p\right)-n+j=1modm$, i.e., $p+j=1modm$. Now, let p be fixed, $j=0$ and let

${q}_{n}={n}^{\alpha }d\left({x}_{n},{x}_{n+p}\right).$

So, we have

${q}_{n+1}={\left(n+1\right)}^{\alpha }d\left({x}_{n+1},{x}_{n+1+p}\right)={\left(n+1\right)}^{\alpha }d\left(f\left({x}_{n}\right),f\left({x}_{n+p}\right)\right).$

Since $p=1modm$, ${x}_{n}$ and ${x}_{n+p}$ lie in different sets ${A}_{i}$ and ${A}_{i+1}$, for some $1\le i\le m$. Then by (2.1) we have

${q}_{n+1}={\left(n+1\right)}^{\alpha }\left(1-\epsilon \right)d\left({x}_{n},{x}_{n+p}\right)+C{\left(n+1\right)}^{\alpha }{\epsilon }^{\alpha }\psi \left(\epsilon \right),$
(2.3)

where $C=sup\gamma {\left(1+2{c}_{n}\right)}^{\beta }<\mathrm{\infty }$. Choosing for each n

$\epsilon =1-{\left(\frac{n}{n+1}\right)}^{\alpha }\le \frac{\alpha }{n+1},$

the relation (2.3) becomes

${q}_{n+1}\le {n}^{\alpha }d\left({x}_{n},{x}_{n+p}\right)+C{\alpha }^{\alpha }\psi \left(\frac{\alpha }{n+1}\right)={q}_{n}+C{\alpha }^{\alpha }\psi \left(\frac{\alpha }{n+1}\right).$

Since ${q}_{0}=0$, it follows that

${q}_{n}=\sum _{k=1}^{n}\left({q}_{k}-{q}_{k-1}\right)\le \sum _{k=1}^{n}C{\alpha }^{\alpha }\psi \left(\frac{\alpha }{k}\right)=C{\alpha }^{\alpha }\sum _{k=1}^{n}\psi \left(\frac{\alpha }{k}\right).$

Consequently,

$d\left({x}_{n},{x}_{n+p}\right)\le C{\left(\frac{\alpha }{n}\right)}^{\alpha }\sum _{k=1}^{n}\psi \left(\frac{\alpha }{k}\right).$

This shows that $\left\{{x}_{n}\right\}$ is a Cauchy sequence in the complete metric space $\left(Y,d\right)$ and, thus, it is convergent to a point $y\in Y={\bigcup }_{i=1}^{m}{A}_{i}$. The case $j\ne 0$ similar.

On the other hand, the sequence $\left\{{x}_{n}\right\}$ has an infinite number of terms in each ${A}_{i}$, for every $i\in \left\{1,\dots ,m\right\}$. Since $\left(Y,d\right)$ is complete, in each ${A}_{i}$, $i\in \left\{1,\dots ,m\right\}$ we can construct a subsequence of $\left\{{x}_{n}\right\}$ which converges to y. Since each ${A}_{i}$ is closed for $i\in \left\{1,\dots ,m\right\}$, we get that $y\in {\bigcap }_{i=1}^{m}{A}_{i}$. Then ${\bigcap }_{i=1}^{m}{A}_{i}\ne \mathrm{\varnothing }$ and we can consider the restriction which satisfies the conditions of Theorem 1 in , since ${\bigcap }_{i=1}^{m}{A}_{i}$ is also closed and complete. From this result, it follows that g has a unique fixed point, say ${x}^{\ast }\in {\bigcap }_{i=1}^{m}{A}_{i}$.

We claim now that for any initial value $x\in Y$, we get the same limit point ${x}^{\ast }\in {\bigcap }_{i=1}^{m}{A}_{i}$. Indeed, for $x\in Y={\bigcup }_{i=1}^{m}{A}_{i}$, by repeating the above process, the corresponding iterative sequence yields that g has a unique fixed point, say $z\in {\bigcap }_{i=1}^{m}{A}_{i}$. Since ${x}^{\ast }$, $z\in {\bigcap }_{i=1}^{m}{A}_{i}$, we have ${x}^{\ast }$, $z\in {A}_{i}$ for all $i\in \left\{1,\dots ,m\right\}$ and, hence, $d\left({x}^{\ast },z\right)$ and $d\left(f\left({x}^{\ast }\right),f\left(z\right)\right)$ are well defined. We can write (2.1) in the form

$d\left({x}^{\ast },z\right)=d\left(f\left({x}^{\ast }\right),f\left(z\right)\right)\le \left(1-\epsilon \right)d\left({x}^{\ast },z\right)+K\epsilon \psi \left(\epsilon \right),$

for some $K>0$. Suppose that $\epsilon =0$. Then we have

$d\left(f\left({x}^{\ast }\right),f\left(z\right)\right)\le d\left({x}^{\ast },z\right).$

If equality occurs, the relation

$d\left({x}^{\ast },z\right)\le K\psi \left(\epsilon \right)$

is valid for every $\epsilon \in \left[0,1\right]$, which implies $d\left({x}^{\ast },z\right)=0$. Thus, ${x}^{\ast }$ is the unique fixed point of f for any initial value $x\in Y$.

To prove that the Picard iteration converges to ${x}^{\ast }$, let us consider ${x}_{1}\in Y={\bigcup }_{i=1}^{m}{A}_{i}$. Then there exists ${i}_{0}\in \left\{1,\dots ,m\right\}$ such that ${x}_{n}\in {A}_{{i}_{0}}$. As ${x}^{\ast }\in {\bigcap }_{i=1}^{m}{A}_{i}$ it follows that ${x}^{\ast }\in {A}_{{i}_{0}+1}$ as well. By the continuity of f, we obtain

$d\left({f}^{n-1}\left({x}_{1}\right),{x}^{\ast }\right)=d\left(f\left({x}_{n-1}\right),{x}^{\ast }\right)=d\left({x}_{n},{x}^{\ast }\right)=\underset{p\to \mathrm{\infty }}{lim}d\left({x}_{n},{x}_{n+p}\right)\le C{\left(\frac{\alpha }{n}\right)}^{\alpha }\sum _{k=1}n\psi \left(\frac{\alpha }{k}\right).$

Letting $n\to \mathrm{\infty }$, it follows that $\left({x}_{n}\right)\to {x}^{\ast }$, i.e., the Picard iteration converges to the unique fixed point of f for any initial point ${x}_{1}\in Y$.

1. (ii)

Since ${x}^{\ast }$ is a fixed point and ${x}^{\ast }\in {\bigcap }_{i=1}^{m}{A}_{i}$, we obtain that

$d\left({x}_{n},{x}^{\ast }\right)=d\left(f\left({x}_{n-1}\right),f\left({x}^{\ast }\right)\right)\le d\left({x}_{n-1},{x}^{\ast }\right)\le \cdots \le d\left({x}_{1},{x}^{\ast }\right)=\parallel {x}^{\ast }\parallel .$
(2.4)

By (2.4), it follows that

$d\left({x}_{n},{x}_{1}\right)\le d\left({x}_{n},{x}^{\ast }\right)+d\left({x}^{\ast },{x}_{1}\right)\le \parallel {x}^{\ast }\parallel +d\left({x}^{\ast },{x}_{1}\right)\le 2\parallel {x}^{\ast }\parallel .$

□

In view of Remark 1.1, we immediately obtain the following corollary.

Corollary 2.3 (Kirk, Srinivasan, Veeramani , Theorem 1.3])

Let $\left(X,d\right)$ be a complete metric space, m be a positive integer, ${A}_{1},\dots ,{A}_{m}$ be closed nonempty subsets of X, $Y:={\bigcup }_{i=1}^{m}{A}_{i}$ and $f:Y\to Y$ be an operator. Assume that:

1. (i)

${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of Y with respect to f;

2. (ii)

there exists $\lambda \in \left[0,1\right)$ such that, for any $x\in {A}_{i}$, $y\in {A}_{i+1}$, where ${A}_{m+1}={A}_{1}$, we have

$d\left(f\left(x\right),f\left(y\right)\right)\le \lambda d\left(x,y\right).$

Then f has a unique fixed point ${x}^{\ast }\in {\bigcap }_{i=1}^{m}{A}_{i}$.

Finally, we will prove a periodic point theorem. For this purpose, notice first that if f satisfies (1.2) with constants α, β, γ and function ψ, and if $\parallel f\left(x\right)\parallel \le \parallel x\parallel$ for each $x\in X$, then its m-iterate ${f}^{m}$ also satisfies the condition (1.2) with constants α, β, and function ψ. Indeed, let us suppose that f satisfies (1.2) with constants α, β, γ. Then, for every $\epsilon \in \left[0,1\right]$, we have

$\begin{array}{c}d\left({f}^{2}\left(x\right),{f}^{2}\left(y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-\epsilon \right)d\left(f\left(x\right),f\left(y\right)\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel f\left(x\right)\parallel +\parallel f\left(y\right)\parallel \right]}^{\beta }\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-\epsilon \right)\left[\left(1-\epsilon \right)d\left(x,y\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left(1+\parallel x\parallel +\parallel y\parallel \right)}^{\beta }\right]\hfill \\ \phantom{\rule{2em}{0ex}}+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel f\left(x\right)\parallel +\parallel f\left(y\right)\parallel \right]}^{\beta }\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-\epsilon \right)\left[\left(1-\epsilon \right)d\left(x,y\right)+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left(1+\parallel x\parallel +\parallel y\parallel \right)}^{\beta }\right]\hfill \\ \phantom{\rule{2em}{0ex}}+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel x\parallel +\parallel y\parallel \right]}^{\beta }\hfill \\ \phantom{\rule{1em}{0ex}}={\left(1-\epsilon \right)}^{2}d\left(x,y\right)+\left(1-\epsilon \right)\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left(1+\parallel x\parallel +\parallel y\parallel \right)}^{\beta }\hfill \\ \phantom{\rule{2em}{0ex}}+\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left[1+\parallel x\parallel +\parallel y\parallel \right]}^{\beta }\hfill \\ \phantom{\rule{1em}{0ex}}={\left(1-\epsilon \right)}^{2}d\left(x,y\right)+\left(2-\epsilon \right)\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left(1+\parallel x\parallel +\parallel y\parallel \right)}^{\beta }\hfill \\ \phantom{\rule{1em}{0ex}}\le \left(1-\epsilon \right)d\left(x,y\right)+2\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left(1+\parallel x\parallel +\parallel y\parallel \right)}^{\beta }.\hfill \end{array}$

Thus, we immediately get that, for $m\in \mathbb{N}$ with $m\ge 2$, we have

$d\left({f}^{m}\left(x\right),{f}^{m}\left(y\right)\right)\le \left(1-\epsilon \right)d\left(x,y\right)+m\gamma {\epsilon }^{\alpha }\psi \left(\epsilon \right){\left(1+\parallel x\parallel +\parallel y\parallel \right)}^{\beta }.$

Notice also that if ${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of X with respect to f, then each ${A}_{i}$ ($i\in \left\{1,2,\dots ,m\right\}$) is an invariant set with respect to ${f}^{m}$. Using these two remarks, we get the following periodic point theorem.

Theorem 2.4 Let $\left(X,d\right)$ be a complete metric space, m be a positive integer, ${A}_{1},\dots ,{A}_{m}$ be nonempty subsets of X, $Y:={\bigcup }_{i=1}^{m}{A}_{i}$, $\psi :\left[0,1\right]\to \left[0,\mathrm{\infty }\right)$ be an increasing function vanishing with continuity at zero and $f:Y\to Y$ be an operator such that $\parallel f\left(x\right)\parallel \le \parallel x\parallel$ for each $x\in Y$. Assume that:

1. 1.

${\bigcup }_{i=1}^{m}{A}_{i}$ is a cyclic representation of Y with respect to f.

2. 2.

There exists ${i}_{0}\in \left\{1,\dots ,m\right\}$ such that ${A}_{{i}_{0}}$ is closed.

3. 3.

For every $\epsilon \in \left[0,1\right]$ and each $x,y\in {A}_{{i}_{0}}$, we have where $\gamma \ge 0$, $\alpha \ge 1$ and $\beta \in \left[0,\alpha \right]$ are fixed constants.

Then, ${f}^{m}$ has a fixed point.

Proof Notice that, by the above considerations, ${f}^{m}$ is a self mapping on ${A}_{{i}_{0}}$ and it satisfies the condition (1.2) with constants α, β, and function ψ. Thus, by Theorem 1 in  we get the conclusion. □

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Alghamdi, M.A., Petruşel, A. & Shahzad, N. A fixed point theorem for cyclic generalized contractions in metric spaces. Fixed Point Theory Appl 2012, 122 (2012). https://doi.org/10.1186/1687-1812-2012-122

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### Keywords

• Point Theorem
• Fixed Point Theorem
• Nonempty Subset
• Fixed Point Theory
• Unique Fixed Point 