Some Fixed Point Theorems for (a−p)-Quasicontractions

Abstract

In this paper, we introduced the notion of $(a-p)$-quasicontraction and proved two generalizations of some classical fixed point theorems. Furthermore, we present some examples to support our results.

1. Introduction

Based probably on ideas of Cauchy and Liouville, Picard [1] developed the method of succesive approximations to prove the existence of solutions of initial value problems for ordinary differential equations. In his famous dissertation from 1922, Banach [2] formulated and proved the Contraction Mapping Principle, which is considered the starting point of fixed point theory. The result was extended and generalized by many researchers in a very dynamical field of research. In what follows, we recall some classical results of this theory.

Theorem 1.

(Edelstein [3]). Let $\left(Y,\delta \right)$ be a compact metric space and let $T:Y\to Y$ be a mapping such that $\delta \left(Tu,Tv\right)<\delta \left(u,v\right)$ for all $u,v\in Y$ with $u\ne v$. Then, T has a unique fixed point.

Theorem 2.

(Hardy-Rogers [4]). Let $\left(Y,\delta \right)$ be a compact metric space and let $T:Y\to Y$ be a mapping satisfying inequality

$$\delta \left(Tu,Tv\right)<A\delta \left(u,v\right)+B\delta \left(u,Tu\right)+C\delta \left(v,Tv\right)$$

for all $u,v\in Y$ and $u\ne v$, where A, B, C are positive and $A+B+C=1$. Then, T has a unique fixed point.

Theorem 3.

(Greguš [5]). Let Y be a Banach space and C a closed convex subset of Y. Let $T:Y\to Y$ be a mapping satisfying inequality

$$∥Tu-Tv∥\le a∥u-v∥+b∥u-Tu∥+c∥v-Tv∥$$

for all $u,v\in C$, where $0<a<1$, $b\ge 0$, $c\ge 0$ and $a+b+c=1$. Then, T has a unique fixed point.

Very recently, the authors [6] have introduced the concept of quadratic quasicontraction mapping and proved two generalizations of Theorems 1–3.

Definition 1.

A mapping $T:Y\to Y$ of a metric space Y into itself is said to be a quadratic quasicontraction if there exists a $\in \left(0,\frac{1}{2}\right)$ such that

$${\delta }^2\left(Tu,Tv\right)\le a{\delta }^2\left(u,Tu\right)+a{\delta }^2\left(v,Tv\right)+\left(1-2a\right){\delta }^2\left(u,v\right)$$

(1)

for all $u,v\in X$ and a strict quadratic quasicontraction if in Relation (1) we have the strict inequality for all $u,v\in Y$ and $u\ne v$.

Theorem 4.

(Popescu-Stan [6]) Let $\left(Y,\delta \right)$ be a compact metric space and let $T:Y\to Y$ be a strict quadratic quasicontraction. Then, T has a unique fixed point $w\in Y$. Moreover, if T is continuous, then, for each $u\in Y$, the sequence of iterates $\left\{T^{n}u\right\}$ converges to w.

Theorem 5.

(Popescu-Stan [6]) Let Y be a Banach space and C be a closed convex subset of Y. Let $T:C\to C$ be a mapping satisfying inequality

$${∥Tu-Tv∥}^2\le a{∥u-Tu∥}^2+a{∥v-Tv∥}^2+b{∥u-v∥}^2$$

for all $u,v\in C$, where $0<a<\frac{1}{2}$, $b=1-2a$. Then, T has a unique fixed point.

In this paper, we extend the notion of quadratic quasicontraction mapping and prove the analogues of Theorems 4 and 5 for a new type of mappings.

The following lemma will be necessary in the proof of the main result:

Lemma 1.

(see Lemma 1.6 [7]) Let ${\left\{{\alpha }_n\right\}}_{n=0}^{\infty }$ and ${\left\{{\beta }_n\right\}}_{n=0}^{\infty }$ be sequences of nonnegative numbers and $0\le k<1$, so that

$${\alpha }_{n+1}\le k{\alpha }_n+{\beta }_n,$$

for all $n\ge 0$. If ${\displaystyle \underset{n\to \infty }{lim}}{\beta }_n=0$, then ${\displaystyle \underset{n\to \infty }{lim}}{\alpha }_n=0$.

2. Main Results

Definition 2.

Let $p\in \mathbb{N}$ and $a\in \left(0,\frac{1}{2}\right)$. A mapping $T:Y\to Y$ of a metric space Y into itself is said to be an $(a-p)$-quasicontraction if

$${\delta }^p\left(Tu,Tv\right)\le a{\delta }^p\left(u,Tu\right)+a{\delta }^p\left(v,Tv\right)+\left(1-2a\right){\delta }^p\left(u,v\right)$$

(2)

for all $u,v\in Y$ and a strict $(a-p)$-quasicontraction if we have strict inequality in Relation (2) for all $u,v\in Y$ with $u\ne v$.

Remark 1.

For $p=2$ we have the notion of quadratic quasicontraction mapping.

Proposition 1.

If p, $q\in \mathbb{N}$, $p\le q$, then every $(a-p)$-quasicontraction is an $(a-q)$-quasicontraction.

Proof. 

Suppose T is an $(a-p)$-quasicontraction. Using the convexity of $t\to t^{\frac{q}{p}}$, we have

$$\begin{array}{ccc}\hfill {\delta }^q\left(Tu,Tv\right)& =& {\left({\delta }^p\left(Tu,Tv\right)\right)}^{\frac{q}{p}}\hfill \\ & \le & {\left(a{\delta }^p\left(u,Tu\right)+a{\delta }^p\left(v,Tv\right)+\left(1-2a\right){\delta }^p\left(u,v\right)\right)}^{\frac{q}{p}}\hfill \\ & \le & a{\delta }^q\left(u,Tu\right)+a{\delta }^q\left(v,Tv\right)+\left(1-2a\right){\delta }^q\left(u,v\right)\hfill \end{array}$$

for all $u,v\in Y$. Hence T is an $(a-q)$-quasicontraction. □

Inspired by Example 3.3 ([8]) we give an example which shows that not every $(a-q)$-quasicontraction is an $(a-p)$-quasicontraction if $p<q$, where $a\ne \frac{1}{3}$.

Example 1.

Let $c_1:={\left(\frac{1-2a}{a}\right)}^{\frac{1}{q-1}}>0$ and $c_2:=\frac{1-2a}{{\left(1+c_1\right)}^{q-1}}\in \left(0,1\right)$. Let τ be the unique real number satisfying

$$0<\tau <1\mathit{and}{c}_2+a{\left(1-\tau \right)}^q={\tau }^q.$$

(3)

Let $Y=\left[0,1\right]$, $\delta \left(u,v\right)=\left|u-v\right|$ and $T:Y\to Y$ such that

$$Tu=\left\{\begin{array}{cc}0& ifu\in \left[0,1\right)\hfill \\ \tau & ifu=1\hfill \end{array}\right..$$

Then, T is an $(a-q)$-quasicontraction, but T is not an $(a-p)$-quasicontraction for any $p\in \mathbb{N}$ with $p<q$.

Proof. 

Let $f:\left[0,1\right]\to \mathbb{R}$ defined by

$$f\left(t\right)=c_2+a{\left(1-t\right)}^q-t^q.$$

Obviously, f is continuous on $\left[0,1\right]$ and differentiable on $\left(0,1\right)$. We have

$$f^{\prime }\left(t\right)=-aq{\left(1-t\right)}^{q-1}-q{t}^{q-1}<0$$

for any $t\in \left(0,1\right)$, so f is strictly decreasing. Since $f\left(0\right)=c_2+a>0$ and $f\left(1\right)=c_2-1<0$, we note that there exists a unique real number $\tau \in \left(0,1\right)$ satisfying Relation (3). Now, let g be a function from $\left[0,1\right]$ into $\mathbb{R}$ defined by

$$g\left(u\right)=a{u}^q+\left(1-2a\right){\left(1-u\right)}^q+a{\left(1-\tau \right)}^q.$$

Then, g is continuous on $\left[0,1\right]$ and differentiable on $\left(0,1\right)$. We have

$$g^{\prime }\left(u\right)=aq{u}^{q-1}-\left(1-2a\right)q{\left(1-u\right)}^{q-1},$$

so $g^{\prime }\left(u\right)=0\Rightarrow {\left(\frac{u}{1-u}\right)}^{q-1}=\frac{1-2a}{a}\Rightarrow \frac{u}{1-u}={\left(\frac{1-2a}{a}\right)}^{\frac{1}{q-1}}=c_1\Rightarrow u=\frac{c_1}{1+c_1}$.

Putting $v:=\frac{c_1}{1+c_1}\in \left(0,1\right)$, we obtain

$$\begin{array}{ccc}\hfill g^{\prime }\left(u\right)& <& 0\mathrm{if}u<v\mathrm{and}\hfill \\ \hfill g^{\prime }\left(u\right)& >& 0\mathrm{if}u>v.\hfill \end{array}$$

Hence

$$min\left\{g\left(u\right):0\le u\le 1\right\}=g\left(v\right),$$

where

$$\begin{array}{ccc}\hfill g\left(v\right)& =& a{\left(\frac{c_1}{1+c_1}\right)}^q+\left(1-2a\right)\frac{1}{{\left(1+c_1\right)}^q}+a{\left(1-\tau \right)}^q\hfill \\ & =& a\frac{c_1^q}{1+c_1}·\frac{1}{{\left(1+c_1\right)}^{q-1}}+\frac{1}{1+c_1}·\frac{1-2a}{{\left(1+c_1\right)}^{q-1}}+a{\left(1-\tau \right)}^q\hfill \\ & =& a\frac{c_1^q}{1+c_1}·\frac{c_2}{1-2a}+\frac{c_2}{1+c_1}+a{\left(1-\tau \right)}^q\hfill \\ & =& \frac{c_1{c}_2}{1+c_1}+\frac{c_2}{1+c_1}+a{\left(1-\tau \right)}^q\hfill \\ & =& c_2+a{\left(1-\tau \right)}^q\hfill \\ & =& {\tau }^q.\hfill \end{array}$$

For $u<1$, we have

$$\begin{array}{ccc}\hfill {\delta }^q\left(Tu,T1\right)& =& {\tau }^q=g\left(v\right)\hfill \\ & \le & g\left(u\right)\hfill \\ & =& a{u}^q+\left(1-2a\right){\left(1-u\right)}^q+a{\left(1-\tau \right)}^q\hfill \\ & =& a{\delta }^q\left(u,Tu\right)+a{\delta }^q\left(1,T1\right)+\left(1-2a\right){\delta }^q\left(u,1\right).\hfill \end{array}$$

Therefore, T is an $(a-q)$-quasicontraction.

Since $a\ne \frac{1}{3}$ we have $c_1\ne 1$, $v\ne \frac{1}{2}$ and $v\ne 1-v$. Using the strict convexity of $t\to t^{\frac{q}{p}}$, we get

$$\begin{array}{ccc}& & a{\delta }^p\left(v,Tv\right)+a{\delta }^p\left(1,T1\right)+\left(1-2a\right){\delta }^p\left(v,1\right)\hfill \\ & =& a{v}^p+\left(1-2a\right){\left(1-v\right)}^p+a{\left(1-\tau \right)}^p\hfill \\ & =& {\left(a{v}^p+\left(1-2a\right){\left(1-v\right)}^p+a{\left(1-\tau \right)}^p\right)}^{\left(\frac{q}{p}\right)\left(\frac{p}{q}\right)}\hfill \\ & <& {\left(a{v}^q+\left(1-2a\right){\left(1-v\right)}^q+a{\left(1-\tau \right)}^q\right)}^{\frac{p}{q}}\hfill \\ & =& {\left({\tau }^q\right)}^{\frac{p}{q}}={\tau }^p={\delta }^p\left(Tv,T1\right).\hfill \end{array}$$

Hence, T is not an $(a-p)$-quasicontraction. □

The folowing example shows that for $a=\frac{1}{3}$ not every $(a-p)$-quasicontraction $\left(p\ge 3\right)$ is a quadratic quasicontraction.

Example 2.

(see Example 1 [6]) Let $Y=\left[-1,1\right]$, $\delta \left(u,v\right)=\left|u-v\right|$ and

$$Tu=\left\{\begin{array}{cc}0,& u\in \left[-1,\frac{1}{2}\right],\\ -1,& u\in \left(\frac{1}{2},1\right]\end{array}\right..$$

Then, T satisfies Inequality (2) with $a=\frac{1}{3}$ and $p\ge 3$, but T is not a quadratic quasicontraction.

Proof. 

Let

$$E_p\left(u,v\right)=a{\delta }^p\left(u,Tu\right)+a{\delta }^p\left(v,Tv\right)+\left(1-2a\right){\delta }^p\left(u,v\right).$$

For $u\in \left[-1,\frac{1}{2}\right]$ and $v\in \left(\frac{1}{2},1\right]$ we have $\delta \left(Tu,Tv\right)=1$ and

$$\begin{array}{ccc}\hfill E_p\left(u,v\right)& =& \frac{1}{3}{\delta }^p\left(u,Tu\right)+\frac{1}{3}{\delta }^p\left(v,Tv\right)+\frac{1}{3}{\delta }^p\left(u,v\right)\hfill \\ & =& \frac{1}{3}{\left|u\right|}^p+\frac{1}{3}{\left(v+1\right)}^p+\frac{1}{3}{\left(v-u\right)}^p\hfill \\ & >& \frac{1}{3}{\left(\frac{1}{2}+1\right)}^p=\frac{3^{p-1}}{2^p}>1,\mathrm{because}p\ge 3.\hfill \end{array}$$

Therefore, T is an $(\frac{1}{3}-p)$-quasicontraction with $p\ge 3$.

For $u=0$ and $v=\frac{3}{5}$ we have $\delta \left(T0,T\frac{3}{5}\right)=\delta \left(0,-1\right)=1$ and

$$\begin{array}{ccc}\hfill E_2\left(0,\frac{3}{5}\right)& =& \frac{1}{3}{\delta }^2\left(0,T0\right)+\frac{1}{3}{\delta }^2\left(\frac{3}{5},T\frac{3}{5}\right)+\frac{1}{3}{\delta }^2\left(0,\frac{3}{5}\right)\hfill \\ & =& \frac{1}{3}·\frac{64}{25}+\frac{1}{3}·\frac{9}{25}=\frac{73}{75}<\delta \left(T0,T\frac{3}{5}\right),\hfill \end{array}$$

so T is not a quadratic quasicontraction with $a=\frac{1}{3}$.

Moreover, since

$$\begin{array}{ccc}\hfill E_1\left(0,\frac{3}{5}\right)& =& a\delta \left(0,T0\right)+a\delta \left(\frac{3}{5},T\frac{3}{5}\right)+\left(1-2a\right)\delta \left(0,\frac{3}{5}\right)\hfill \\ & =& a·\frac{8}{5}+\left(1-2a\right)·\frac{3}{5}=\frac{2a+3}{5}<1,\hfill \end{array}$$

we get that T is not an $(a-1)$-quasicontraction for any $a\in \left(0,\frac{1}{2}\right)$. □

The following theorem is a generalization of Theorem 4 and implicitly a generalization of Theorems 1 and 2.

Theorem 6.

Let $\left(Y,\delta \right)$ be a compact metric space and let $T:Y\to Y$ be a strict $(a-p)$-quasicontraction. Then, T has a unique fixed point $w\in Y$. Moreover, if T is continuous, then, for each $u\in Y$, the sequence of iterates $\left\{T^{n}u\right\}$ converges to w.

Proof. 

Letting $v=Tu$ in Inequality (2), we have for all $u\in Y$ with $u\ne Tu$

$${\delta }^p\left(Tu,T^{2}u\right)<a{\delta }^p\left(u,Tu\right)+a{\delta }^p\left(Tu,T^{2}u\right)+\left(1-2a\right){\delta }^p\left(u,Tu\right).$$

This implies $\delta \left(Tu,T^{2}u\right)<\delta \left(u,Tu\right)$.

Now, let $\beta =inf\left\{\delta \left(u,Tu\right):u\in Y\right\}$. Since Y is compact there exists a sequence $\left\{u_n\right\}\subset Y$ such that $u_n\to w\in Y$, $T{u}_n\to z$ and $\beta ={\displaystyle \underset{n\to \infty }{lim}}\delta \left(u_n,T{u}_n\right)=\delta \left(w,z\right)$.

If there exists a subsequence $\left\{u_{n\left(k\right)}\right\}$ of $\left\{u_n\right\}$ such that $u_{n\left(k\right)}=z$ for every $k\in \mathbb{N}$, then we get $w=z=Tz$, so T has a fixed point. Otherwise, we suppose that there exists $N\in \mathbb{N}$ such that $u_n\ne z$ for all $n\ge N$. Taking $u=u_n$ and $v=z$ in Inequality (2), we obtain

$${\delta }^p\left(T{u}_n,Tz\right)<a{\delta }^p\left(u_n,T{u}_n\right)+a{\delta }^p\left(z,Tz\right)+\left(1-2a\right){\delta }^p\left(u_n,z\right).$$

Letting $n\to \infty$, we get

$${\delta }^p\left(z,Tz\right)\le a{\delta }^p\left(w,z\right)+a{\delta }^p\left(z,Tz\right)+\left(1-2a\right){\delta }^p\left(w,z\right).$$

Thus, $\delta \left(z,Tz\right)\le \delta \left(w,z\right)=\beta$. By definition of $\beta$, we obtain $\delta \left(z,Tz\right)=\beta$.

If $\beta >0$, since $\delta \left(Tz,T^{2}z\right)<\delta \left(z,Tz\right)=\beta$, we have a contradiction. Therefore, we get $\beta =0$ and so $w=z=Tz$, i.e., z is a fixed point of T.

If t is another fixed point of T, by Inequality (2), taking $u=z$ and $v=t$, we obtain

$${\delta }^p\left(Tz,Tt\right)\le a{\delta }^p\left(z,Tz\right)+a{\delta }^p\left(t,Tt\right)+\left(1-2a\right){\delta }^p\left(z,t\right),$$

by where

$${\delta }^p\left(z,t\right)<\left(1-2a\right){\delta }^p\left(z,t\right),$$

which is a contradiction.

Now assume T is continuous. Let $u_0\in Y$ and define a sequence $\left\{u_n\right\}\subset Y$ by $u_n:=T{u}_{n-1}=T^n{u}_0$. Suppose there exists $N\in \mathbb{N}\cup \left\{0\right\}$ such that $u_N=z$. Then, we have $u_n=z$ for all $n\ge N$, so $u_n\to z$. Otherwise, we suppose $u_n\ne z$ for all $n\ge N$. By uniqueness of z, we get $u_n\ne$$u_{n+1}$ for every $n\in \mathbb{N}\cup \left\{0\right\}$. Hence, we have $\delta \left(u_n,u_{n+1}\right)=\delta \left(T{u}_{n-1},T{u}_n\right)<\delta \left(u_{n-1},u_n\right)$ for every $n\in \mathbb{N}$, so the sequence ${\left\{\delta \left(u_n,u_{n+1}\right)\right\}}_{n\in \mathbb{N}}$ is decreasing and positive. Therefore, there exists $b={\displaystyle \underset{n\to \infty }{lim}}\delta \left(u_{n-1},u_n\right)$. We claim that $b=0$. Since Y is compact, there exists a subsequence $\left\{u_{n\left(k\right)}\right\}$ of $\left\{u_n\right\}$ such that $u_{n\left(k\right)}\to z\in Y$ as $k\to \infty$. If $b>0$, we have

$$b=\underset{k\to \infty }{lim}\delta \left(u_{n\left(k\right)},u_{n\left(k\right)+1}\right)=\delta \left(z,Tz\right)>0,$$

$$b=\underset{k\to \infty }{lim}\delta \left(u_{n\left(k\right)+1},u_{n\left(k\right)+2}\right)=\delta \left(Tz,T^{2}z\right)>0.$$

Hence, we get $\delta \left(z,Tz\right)=\delta \left(Tz,T^{2}z\right)=b>0$, which is a contradiction. Thus, $b=0$. Now, taking $u=u_n$ and $v=z$ in Inequality (2), we obtain

$${\delta }^p\left(u_{n+1},z\right)={\delta }^p\left(T{u}_n,z\right)<a{\delta }^p\left(u_n,T{u}_n\right)+a{\delta }^p\left(z,Tz\right)+\left(1-2a\right){\delta }^p\left(u_n,z\right).$$

This implies

$${\alpha }_{n+1}<\left(1-2a\right){\alpha }_n+{\beta }_n,$$

where ${\alpha }_n:={\delta }^p\left(x_n,z\right)$ and ${\beta }_n=a{\delta }^p\left(u_n,u_{n+1}\right)$. Since ${\displaystyle \underset{n\to \infty }{lim}}{\beta }_n=0$, by Lemma 1, we get ${\displaystyle \underset{n\to \infty }{lim}}{\alpha }_n=0$, i.e., ${\displaystyle \underset{n\to \infty }{lim}}{u}_n=z$. □

Example 3.

Let $Y=\left[-2,2\right]$, $\delta \left(u,v\right)=\left|u-v\right|$, $T:Y\to Y$ defined by

$$Tu=\left\{\begin{array}{cc}\frac{1-u}{2},& ifu\in \left[-2,-1\right)\hfill \\ -\frac{2}{5},& ifu=-1\hfill \\ 0,& ifu\in \left(-1,1\right)\hfill \\ \frac{2}{5},& ifu=1\hfill \\ \frac{-1-u}{2}& ifu\in \left(1,2\right]\hfill \end{array}\right..$$

Then, T is an $(\frac{1}{3}-p)$-quasicontraction for $p\ge 4$, but T is not an $(\frac{1}{3}-2)$-quasicontraction (quadratic quasicontraction). Moreover, T is not asymptotic regular.

Proof. 

We distinguish 13 cases:

• $1^0$ If $u,v\in \left[-2,-1\right)$, $u<v$, we have:

  $$\delta \left(Tu,Tv\right)=\frac{v-u}{2},$$

  $$E_1\left(u,v\right)=\frac{1-3u}{6}+\frac{1-3v}{6}+\frac{v-u}{3}=\frac{2-5u-v}{6},$$

  $$\delta \left(Tu,Tv\right)<E_1\left(u,v\right)⟺2v+u<1,$$

  which is obvious.
• $2^0$ By symmetry, the case $u,v\in \left(1,2\right]$, $u<v$ is similar.
• $3^0$ If $u\in \left[-2,-1\right)$, $v\in \left(-1,1\right)$, we have:

  $$\delta \left(Tu,Tv\right)=\frac{1-u}{2},$$

  $$E_1\left(u,v\right)=\frac{1-3u}{6}+\frac{\left|v\right|}{3}+\frac{v-u}{3}=\frac{1-5u+2v+2\left|v\right|}{6},$$

  $$\delta \left(Tu,Tv\right)<E_1\left(u,v\right)⟺2+2u<2v+2\left|v\right|,$$

  which is obvious ($2+2u<0\le 2v+2\left|v\right|$).
• $4^0$ By symmetry, the case $u\in \left(1,2\right]$, $v\in \left(-1,1\right)$ is similar.
• $5^0$ If $u\in \left[-2,-1\right)$, $v\in \left(1,2\right]$, we have:

  $$\delta \left(Tu,Tv\right)=\frac{2-u+v}{2},$$

  $$E_1\left(u,v\right)=\frac{1-3u}{6}+\frac{1+3v}{6}+\frac{v-u}{3}=\frac{2-5u+5v}{6},$$

  $$\delta \left(Tu,Tv\right)<E_1\left(u,v\right)⟺2<v-u,$$

  which is obvious.
• $6^0$ If $u,v\in \left(-1,1\right)$, $u\ne v$, we have:

  $$\delta \left(Tu,Tv\right)=0<\frac{\left|u\right|}{3}+\frac{\left|v\right|}{3}+\frac{\left|u-v\right|}{3}=E_1\left(u,v\right).$$
• $7^0$ If $u=-1$, $v=1$, we have:

  $$\begin{array}{c}\delta \left(Tu,Tv\right)=\frac{4}{5},\\ E_1\left(u,v\right)=\frac{1}{3}·\frac{3}{5}+\frac{1}{3}·\frac{3}{5}+\frac{1}{3}·2=\frac{16}{15}\end{array}$$

  $$\delta \left(Tu,Tv\right)<E_1\left(u,v\right)$$
• $8^0$ For $u=1$, $v\in \left(-1,1\right)$, we get:

  $$\begin{array}{c}\delta \left(Tu,Tv\right)=\frac{2}{5},\\ E_1\left(u,v\right)=\frac{1}{3}·\frac{3}{5}+\frac{\left|v\right|}{3}+\frac{1-v}{3}=\frac{8+5\left|v\right|-5v}{15}\end{array}$$

  $$\delta \left(Tu,Tv\right)<E_1\left(u,v\right)⟺0<2+5\left|v\right|-5v,$$

  which is obvious.
• $9^0$ By symmetry, the case $u=-1$, $v\in \left(-1,1\right)$ is similar.
• ${10}^0$ For $u=1$, $v\in \left[-2,-1\right)$, we have:

  $$\begin{array}{c}\delta \left(Tu,Tv\right)=\frac{1-5v}{10},\\ E_1\left(u,v\right)=\frac{1}{3}·\frac{1-3v}{2}+\frac{1}{3}·\frac{3}{5}+\frac{1-v}{3}=\frac{21-25v}{30}\end{array}$$

  $$\delta \left(Tu,Tv\right)<E_1\left(u,v\right)⟺10v<18,$$

  which is obvious.
• ${11}^0$ By symmetry, the case $u=-1$, $v\in \left(1,2\right]$ is similar.
• ${12}^0$ For $u=1$, $v\in \left(1,2\right]$, we have:

  $$\begin{array}{c}{\delta }^p\left(Tu,Tv\right)={\left(\frac{9+5v}{10}\right)}^p,\\ E_p\left(u,v\right)=\frac{1}{3}·{\left(\frac{3}{5}\right)}^p+\frac{1}{3}·{\left(\frac{3v+1}{2}\right)}^p+\frac{1}{3}·{\left(1-v\right)}^p\end{array}$$
  Since the function $v\to \frac{15v+5}{9+5v}$ is increasing and $p\ge 4$, we obtain

  $${\left(\frac{15v+5}{9+5v}\right)}^p>{\left(\frac{15+5}{9+5}\right)}^p={\left(\frac{10}{7}\right)}^p\ge {\left(\frac{10}{7}\right)}^4\ge 3,$$

  so

  $${\left(\frac{9+5v}{10}\right)}^p<\frac{1}{3}·{\left(\frac{3v+1}{2}\right)}^p,$$

  by where ${\delta }^p\left(Tu,Tv\right)<E_p\left(u,v\right)$.
• ${13}^0$ By symmetry, the case $u=-1$, $v\in \left[-2,-1\right)$ is similar.

By Proposition 1 is obvious that $\delta \left(Tu,Tv\right)<E_1\left(u,v\right)\Rightarrow {\delta }^p\left(Tu,Tv\right)<E_p\left(u,v\right)$. Therefore we get that in all cases we have ${\delta }^p\left(Tu,Tv\right)<E_p\left(u,v\right)$. Hence, T is an $(\frac{1}{3}-p)$- quasicontraction for $p\ge 4$.

If $u=1$, $v=\frac{4}{3}$, we have:

$${\delta }^2\left(Tu,Tv\right)=\frac{6627}{2700},$$

$$E_2\left(u,v\right)=\frac{1}{3}·\frac{9}{25}+\frac{1}{3}·\frac{25}{4}+\frac{1}{3}·\frac{1}{9}=\frac{5949}{2700},$$

so ${\delta }^2\left(Tu,Tv\right)>E_2\left(u,v\right)$. This implies that T is not a quadratic quasicontraction.

It is easy to prove that $T^{n}u={\left(-1\right)}^n·\frac{u+2^n-1}{2^n}$ for $u\in \left(1,2\right]$, so $T^{n}2={\left(-1\right)}^n·\frac{2^n+1}{2^n}$, for all $n\in \mathbb{N}$, and $\delta \left(T^{n}2,T^{n+1}2\right)=\frac{2^n+1}{2^n}+\frac{2^{n+1}+1}{2^{n+1}}>2$ i.e., T is not asymptotic regular. □

The following lemma plays a very important role in the next theorem.

Lemma 2.

Let C be a nonempty closed subset of a complete metric space $\left(Y,\delta \right)$ and let $T:C\to C$ be an $(a-p)$-quasicontraction mapping. Assume that there exist constants $c,b\in \mathbb{R}$ such that $0\le c<1$ and $b>0$. If for every $u\in C$ there exists $v\in C$ such that $\delta \left(v,Tv\right)\le c\delta \left(u,Tu\right)$ and $\delta \left(v,u\right)\le b\delta \left(u,Tu\right)$, then T has a unique fixed point.

Proof. 

Let $u_0\in C$. We construct the sequence $\left\{u_n\right\}\subset C$ such that

$$\delta \left(u_{n+1},T{u}_{n+1}\right)\le c\delta \left(u_n,T{u}_n\right),$$

$$\delta \left(u_{n+1},u_n\right)\le b\delta \left(u_n,T{u}_n\right),n=0,1,2,...$$

Since

$$\delta \left(u_{n+1},u_n\right)\le b\delta \left(u_n,T{u}_n\right)\le bc\delta \left(u_{n-1},T{u}_{n-1}\right)\le ...\le b{c}^n\delta \left(u_0,T{u}_0\right)$$

it is easy to prove that $\left\{u_n\right\}$ is a Cauchy sequence. By completeness of C, there exists $w\in C$ such that ${\displaystyle \underset{n\to \infty }{lim}}{u}_n=w$. From the above inequalities and sandwich theorem, we get ${\displaystyle \underset{n\to \infty }{lim}}\delta \left(u_n,T{u}_n\right)=0$, so ${\displaystyle \underset{n\to \infty }{lim}}T{u}_n=w$ and

$${\delta }^p\left(T{u}_n,Tw\right)\le a{\delta }^p\left(u_n,T{u}_n\right)+a{\delta }^p\left(w,Tw\right)+\left(1-2a\right){\delta }^p\left(u_n,w\right).$$

Letting $n\to \infty$, we obtain

$${\delta }^p\left(w,Tw\right)\le a{\delta }^p\left(w,Tw\right).$$

This implies $\delta \left(w,Tw\right)=0$, i.e., $w=Tw$.

If t is another fixed point of T, taking $u=w$, $v=t$ in Inequality (2) we obtain

$${\delta }^p\left(Tw,Tt\right)\le a{\delta }^p\left(w,Tw\right)+a{\delta }^p\left(t,Tt\right)+\left(1-2a\right){\delta }^p\left(w,t\right),$$

hence

$${\delta }^p\left(w,t\right)\le \left(1-2a\right){\delta }^p\left(w,t\right).$$

Therefore, $\delta \left(w,t\right)=0$, which is a contradiction. ☐

The next theorem is a partial generalization of Theorem 5.

Theorem 7.

Let Y be a Banach space and C be a closed convex subset of X. Let $T:C\to C$ be a mapping satisfying the inequality:

$${∥Tu-Tv∥}^p\le a{∥u-Tu∥}^p+a{∥v-Tv∥}^p+b{∥u-v∥}^p$$

(4)

for all $u,v\in C$, where $0<a<\frac{1}{2}$, $b=1-2a$, $b\le \frac{{\left(\sqrt[p]{2^{p+1}-1}-1\right)}^p-1}{2^p-1}$, $p\in \mathbb{N}$, $p\ge 2$. Then, T has a unique fixed point.

Proof. 

Taking $v=Tu$ in (4), we have:

$${∥Tu-T^{2}u∥}^p\le a{∥u-Tu∥}^p+a{∥Tu-T^{2}u∥}^p+b{∥u-Tu∥}^p.$$

This implies

$$∥Tu-T^{2}u∥\le ∥u-Tu∥,\mathrm{for}\mathrm{all}u\in C.$$

(5)

Now, let $u\in C$ arbitrary fixed and $z=\frac{1}{2}{T}^{2}u+\frac{1}{2}{T}^{3}u$. Since C is convex, we have $z\in C$. Then, by Inequalities (4) and (5), we get

$$\begin{array}{ccc}\hfill {∥Tu-T^{3}u∥}^p& \le & a{∥u-Tu∥}^p+a{∥T^{2}u-T^{3}u∥}^p+b{∥u-T^{2}u∥}^p\hfill \\ & \le & 2a{∥u-Tu∥}^p+b{\left(∥u-Tu∥+∥Tu-T^{2}u∥\right)}^p\hfill \\ & \le & \left(2a+2^{p}b\right){∥u-Tu∥}^p,\hfill \end{array}$$

so

$$∥Tu-T^{3}u∥\le \sqrt[p]{2a+2^{p}b}∥u-Tu∥.$$

(6)

Therefore,

$$\begin{array}{ccc}\hfill ∥Tu-z∥& =& ∥\frac{1}{2}\left(Tu-T^{2}u\right)+\frac{1}{2}\left(Tu-T^{3}u\right)∥\hfill \\ & \le & \frac{1}{2}∥Tu-T^{2}u∥+\frac{1}{2}∥Tu-T^{3}u∥\hfill \\ & \le & \frac{1}{2}∥u-Tu∥+\frac{1}{2}\sqrt[p]{2a+2^{p}b}∥u-Tu∥\hfill \\ & =& \frac{1}{2}\left(1+\sqrt[p]{2a+2^{p}b}\right)∥u-Tu∥.\hfill \end{array}$$

(7)

In addition,

$$∥T^{2}u-z∥=\frac{1}{2}∥T^{2}u-T^{3}u∥\le \frac{1}{2}∥u-Tu∥.$$

(8)

Now, by Inequalities (4), (5) and (7), we obtain

$$\begin{array}{ccc}& & {∥T^{2}u-Tz∥}^p\hfill \\ & \le & a{∥Tu-T^{2}u∥}^p+a{∥z-Tz∥}^p+b{∥Tu-z∥}^p\hfill \\ & \le & a{∥u-Tu∥}^p+a{∥z-Tz∥}^p+b{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p{∥u-Tu∥}^p\hfill \\ & =& a{∥z-Tz∥}^p+\left[a+b{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p\right]{∥u-Tu∥}^p.\hfill \end{array}$$

(9)

By Inequalities (4), (5) and (8), we obtain

$$\begin{array}{ccc}\hfill {∥T^{3}u-Tz∥}^p& \le & a{∥T^{2}u-T^{3}u∥}^p+a{∥z-Tz∥}^p+b{∥T^{2}u-z∥}^p\hfill \\ & \le & a{∥u-Tu∥}^p+a{∥z-Tz∥}^p+\frac{b}{2^p}{∥u-Tu∥}^p\hfill \\ & =& a{∥z-Tz∥}^p+\left(a+\frac{b}{2^p}\right){∥u-Tu∥}^p.\hfill \end{array}$$

(10)

Since

$$\begin{array}{ccc}\hfill ∥z-Tz∥& =& ∥\frac{1}{2}\left(T^{2}u-Tz\right)+\frac{1}{2}\left(T^{3}u-Tz\right)∥\hfill \\ & \le & \frac{1}{2}∥T^{2}u-Tz∥+\frac{1}{2}∥T^{3}u-Tz∥,\hfill \end{array}$$

by Inequalities (9) and (10), we get

$$\begin{array}{ccc}\hfill ∥z-Tz∥& \le & \frac{1}{2}{\left\{a{∥z-Tz∥}^p+\left[a+b{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p\right]{∥u-Tu∥}^p\right\}}^{\frac{1}{p}}\hfill \\ & & +\frac{1}{2}{\left[a{∥z-Tz∥}^p+\left(a+\frac{b}{2^p}\right){∥u-Tu∥}^p\right]}^{\frac{1}{p}}.\hfill \end{array}$$

(11)

If $u=Tu$, then u is a fixed point of T. Otherwise, dividing Inequality (11) by $\frac{1}{2}∥u-Tu∥$, we get

$$\begin{array}{ccc}\hfill 2·\frac{∥z-Tz∥}{∥u-Tu∥}& \le & {\left[a\frac{{∥z-Tz∥}^p}{{∥u-Tu∥}^p}+a+b{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p\right]}^{\frac{1}{p}}\hfill \\ & & +{\left[a\frac{{∥z-Tz∥}^p}{{∥u-Tu∥}^p}+a+\frac{b}{2^p}\right]}^{\frac{1}{p}}.\hfill \end{array}$$

Denoting $t:=\frac{{∥z-Tz∥}^p}{{∥u-Tu∥}^p}$, we obtain

$$2t^{\frac{1}{p}}\le {\left[at+a+b{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p\right]}^{\frac{1}{p}}+{\left[at+a+\frac{b}{2^p}\right]}^{\frac{1}{p}},$$

which implies $z=Tz$ or $f\left(t\right)\ge 2$, where

$$f\left(t\right)={\left[a+\frac{a}{t}+\frac{b}{t}{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p\right]}^{\frac{1}{p}}+{\left(a+\frac{a}{t}+\frac{b}{2^p·t}\right)}^{\frac{1}{p}}$$

Clearly, f is a decreasing function and

$$\begin{array}{ccc}\hfill f\left(1\right)& =& {\left[2a+b{\left(\frac{1+\sqrt[p]{2a+2^{p}b}}{2}\right)}^p\right]}^{\frac{1}{p}}+{\left(a+\frac{b}{2^p}\right)}^{\frac{1}{p}}\hfill \\ & =& {\left[1-b+b{\left(\frac{1+\sqrt[p]{1-b+2^{p}b}}{2}\right)}^p\right]}^{\frac{1}{p}}+{\left(1-b+\frac{b}{2^p}\right)}^{\frac{1}{p}}.\hfill \end{array}$$

We claim that $f\left(1\right)<2$. Since $b\le \frac{{\left(\sqrt[p]{2^{p+1}-1}-1\right)}^p-1}{2^p-1}$, we have ${\left(\frac{1+\sqrt[p]{1-b+2^{p}b}}{2}\right)}^p\le \frac{2^{p+1}-1}{2^p}\le 2-\frac{1}{2^p}$ and then

$$\begin{array}{ccc}\hfill f\left(1\right)& <& {\left[1-b+b\left(2-\frac{1}{2^p}\right)\right]}^{\frac{1}{p}}+{\left(1-b+\frac{b}{2^p}\right)}^{\frac{1}{p}}\hfill \\ & =& {\left(1+b-\frac{b}{2^p}\right)}^{\frac{1}{p}}+{\left(1-b+\frac{b}{2^p}\right)}^{\frac{1}{p}}.\hfill \end{array}$$

Using strict concavity of $t\to t^{\frac{1}{p}}$, we obtain

$$f\left(1\right)<2{\left(\frac{1+b-\frac{b}{2^p}+1-b+\frac{b}{2^p}}{2}\right)}^{\frac{1}{p}}=2.$$

Since f is a decreasing function and $f\left(t\right)\ge 2$, there exists $c<1$ such that $t\le c$. Therefore, $∥z-Tz∥\le \sqrt{c}∥u-Tu∥$.

Now, since

$$\begin{array}{ccc}\hfill ∥z-u∥& \le & \frac{1}{2}∥T^{2}u-u∥+\frac{1}{2}∥T^{3}u-u∥\hfill \\ & \le & \frac{1}{2}\left(∥T^{2}u-Tu∥+∥Tu-u∥\right)\hfill \\ & & +\frac{1}{2}\left(∥T^{3}u-T^{2}u∥+∥T^{2}u-Tu∥+∥Tu-u∥\right)\hfill \\ & \le & \frac{5}{2}∥Tu-u∥,\hfill \end{array}$$

applying Lemma 2, we get that T has a unique fixed point. □

Example 4.

Let $Y=l^{\infty }\left(\mathbb{R}\right)$ be the set of bounded sequence of real numbers and $∥u∥={\displaystyle \underset{n\in \mathbb{N}}{sup}}\left|u_n\right|$, where $u={\left\{u_n\right\}}_{n\in \mathbb{N}}$. It is well known that $\left(Y,∥·∥\right)$ is a Banach space. Let $C=\left\{u\in Y:∥u∥\le 1\right\}$ and $T:C\to C$ defined by

$$Tu=\left\{\begin{array}{cc}\frac{\mathbf{1}}{\mathbf{3}},& ifu=-\mathbf{1}\hfill \\ -\mathbf{1},& if{u}_n\in \left[\frac{1}{3},1\right]foreveryn\in \mathbb{N}\hfill \\ \mathbf{0},& otherwise\hfill \end{array}\right.,$$

where $u={\left\{u_n\right\}}_{n\in \mathbb{N}}$, $\mathbf{c}=\left\{c,c,c,...\right\}$. Clearly, C is closed, convex and not compact. Since $T^n\left(-\mathbf{1}\right)=\frac{\mathbf{1}}{\mathbf{3}}$ if n is odd and $T^n\left(-\mathbf{1}\right)=-\mathbf{1}$ if n is even, we note that T is not asymptotic regular.

If $u=-\mathbf{1}$ and $v={\left\{v_n\right\}}_{n\in \mathbb{N}}$, where $v_n\in \left[\frac{1}{3},1\right]$ for every $n\in \mathbb{N}$, then $∥Tu-Tv∥=\frac{4}{3}$ and

$$\begin{array}{ccc}\hfill E_1\left(u,v\right)& =& \frac{4}{3}a+a\underset{n\in \mathbb{N}}{sup}\left(1+v_n\right)+b\underset{n\in \mathbb{N}}{sup}\left(1+v_n\right)\hfill \\ & \ge & \frac{4}{3}a+\frac{4}{3}\left(a+b\right)=\frac{4}{3}.\hfill \end{array}$$

Hence $∥Tu-Tv∥\le E_1\left(u,v\right)$, and then ${∥Tu-Tv∥}^3\le E_3\left(u,v\right)$.

If $u=-\mathbf{1}$ and $v\ne u$, $v={\left\{v_n\right\}}_{n\in \mathbb{N}}$, where there exists $n_0$ such that $v_{n_0}\notin \left[\frac{1}{3},1\right]$, then $∥Tu-Tv∥=\frac{1}{3}$ and

$$\begin{array}{ccc}\hfill E_1\left(u,v\right)& =& \frac{4}{3}a+a\underset{n\in \mathbb{N}}{sup}\left|v_n\right|+b\underset{n\in \mathbb{N}}{sup}\left(1+v_n\right)\hfill \\ & \ge & \frac{4}{3}a\ge \frac{1}{3}ifa\ge \frac{1}{4}.\hfill \end{array}$$

If $u={\left\{u_n\right\}}_{n\in \mathbb{N}}$ where $u_n\in \left[\frac{1}{3},1\right]$ for every $n\in \mathbb{N}$ and $v={\left\{v_n\right\}}_{n\in \mathbb{N}}$, where there exists $n_0$ such that $v_{n_0}\notin \left[\frac{1}{3},1\right]$, then $∥Tu-Tv∥=1$ and

$$\begin{array}{ccc}\hfill E_3\left(u,v\right)& =& a\underset{n\in \mathbb{N}}{sup}{\left(1+u_n\right)}^3+a\underset{n\in \mathbb{N}}{sup}{\left|v_n\right|}^3+b\underset{n\in \mathbb{N}}{sup}{\left|u_n-v_n\right|}^3\hfill \\ & \ge & {\left(\frac{4}{3}\right)}^{3}a=\frac{64}{27}a.\hfill \end{array}$$

If $a\ge \frac{27}{64}$ we have ${∥Tu-Tv∥}^3\le E_3\left(u,v\right)$.

Therefore, T satisfies Inequality (4) with $a=\frac{28}{64}=\frac{7}{16}$, $b=\frac{1}{8}\le \frac{{\left(\sqrt[3]{15}-1\right)}^3-1}{7}\simeq 0.3074$, $p=3$.

However, T does not satisfy Inequality (4) if $u=\frac{\mathbf{1}}{\mathbf{3}}$, $v=\mathbf{0}$ and $p=2$: $∥Tu-Tv∥=1$ and

$$E_2\left(u,v\right)=\frac{16}{9}a+a·0+\frac{1}{9}b=\frac{16a+b}{9}=\frac{1+14a}{9}<1.$$

3. Conclusions

In this paper, we generalized the notion of quadratic quasicontraction introduced in Popescu and Stan (Symmetry 2019, 11, 1329 [6]). In the context of the new notion of $(a-p)$-quasicontraction we proved two generalizations of some classical fixed point theorems of Edelstein (J. London Math. Soc. 1962, 37, 74–79 [3]) and Greguš (Boll. Un. Mat. Ital. 1980, 17, 193–198 [5]). Furthermore, we present some examples to support our results.