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Article

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

by
Ovidiu Popescu
*,† and
Gabriel Stan
Department of Mathematics and Computer Science, Faculty of Mathematics and Computer Science, Transilvania University of Brasov, 500036 Brasov, Romania
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Symmetry 2020, 12(12), 1973; https://doi.org/10.3390/sym12121973
Submission received: 11 November 2020 / Revised: 22 November 2020 / Accepted: 25 November 2020 / Published: 29 November 2020
(This article belongs to the Special Issue Fixed Point Theory and Computational Analysis with Applications)

## 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 $Y , δ$ be a compact metric space and let $T : Y → Y$ be a mapping such that $δ T u , T v < δ u , v$ for all $u , v ∈ Y$ with $u ≠ v$. Then, T has a unique fixed point.
Theorem 2.
(Hardy-Rogers [4]). Let $Y , δ$ be a compact metric space and let $T : Y → Y$ be a mapping satisfying inequality
$δ T u , T v < A δ u , v + B δ u , T u + C δ v , T v$
for all $u , v ∈ Y$ and $u ≠ v$, 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 $T : Y → Y$ be a mapping satisfying inequality
$T u − T v ≤ a u − v + b u − T u + c v − T v$
for all $u , v ∈ C$, where $0 < a < 1$, $b ≥ 0$, $c ≥ 0$ 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 $T : Y → Y$ of a metric space Y into itself is said to be a quadratic quasicontraction if there exists a $∈ 0 , 1 2$ such that
$δ 2 T u , T v ≤ a δ 2 u , T u + a δ 2 v , T v + 1 − 2 a δ 2 u , v$
for all $u , v ∈ X$ and a strict quadratic quasicontraction if in Relation (1) we have the strict inequality for all $u , v ∈ Y$ and $u ≠ v$.
Theorem 4.
(Popescu-Stan [6]) Let $Y , δ$ be a compact metric space and let $T : Y → Y$ be a strict quadratic quasicontraction. Then, T has a unique fixed point $w ∈ Y$. Moreover, if T is continuous, then, for each $u ∈ Y$, the sequence of iterates $T n u$ converges to w.
Theorem 5.
(Popescu-Stan [6]) Let Y be a Banach space and C be a closed convex subset of Y. Let $T : C → C$ be a mapping satisfying inequality
$T u − T v 2 ≤ a u − T u 2 + a v − T v 2 + b u − v 2$
for all $u , v ∈ C$, where $0 < a < 1 2$, $b = 1 − 2 a$. 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 $α n n = 0 ∞$ and $β n n = 0 ∞$ be sequences of nonnegative numbers and $0 ≤ k < 1$, so that
$α n + 1 ≤ k α n + β n ,$
for all $n ≥ 0$. If $lim n → ∞ β n = 0$, then $lim n → ∞ α n = 0$.

## 2. Main Results

Definition 2.
Let $p ∈ N$ and $a ∈ 0 , 1 2$. A mapping $T : Y → Y$ of a metric space Y into itself is said to be an $( a − p )$-quasicontraction if
$δ p T u , T v ≤ a δ p u , T u + a δ p v , T v + 1 − 2 a δ p u , v$
for all $u , v ∈ Y$ and a strict $( a − p )$-quasicontraction if we have strict inequality in Relation (2) for all $u , v ∈ Y$ with $u ≠ v$.
Remark 1.
For $p = 2$ we have the notion of quadratic quasicontraction mapping.
Proposition 1.
If p, $q ∈ N$, $p ≤ q$, then every $( a − p )$-quasicontraction is an $( a − q )$-quasicontraction.
Proof.
Suppose T is an $( a − p )$-quasicontraction. Using the convexity of $t → t q p$, we have
$δ q T u , T v = δ p T u , T v q p ≤ a δ p u , T u + a δ p v , T v + 1 − 2 a δ p u , v q p ≤ a δ q u , T u + a δ q v , T v + 1 − 2 a δ q u , v$
for all $u , v ∈ Y$. 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 $a ≠ 1 3$.
Example 1.
Let $c 1 : = 1 − 2 a a 1 q − 1 > 0$ and $c 2 : = 1 − 2 a 1 + c 1 q − 1 ∈ 0 , 1$. Let τ be the unique real number satisfying
$0 < τ < 1 and c 2 + a 1 − τ q = τ q .$
Let $Y = 0 , 1$, $δ u , v = u − v$ and $T : Y → Y$ such that
$T u = 0 i f u ∈ 0 , 1 τ i f u = 1 .$
Then, T is an $( a − q )$-quasicontraction, but T is not an $( a − p )$-quasicontraction for any $p ∈ N$ with $p < q$.
Proof.
Let $f : 0 , 1 → R$ defined by
$f t = c 2 + a 1 − t q − t q .$
Obviously, f is continuous on $0 , 1$ and differentiable on $0 , 1$. We have
$f ′ t = − a q 1 − t q − 1 − q t q − 1 < 0$
for any $t ∈ 0 , 1$, so f is strictly decreasing. Since $f 0 = c 2 + a > 0$ and $f 1 = c 2 − 1 < 0$, we note that there exists a unique real number $τ ∈ 0 , 1$ satisfying Relation (3). Now, let g be a function from $0 , 1$ into $R$ defined by
$g u = a u q + 1 − 2 a 1 − u q + a 1 − τ q .$
Then, g is continuous on $0 , 1$ and differentiable on $0 , 1$. We have
$g ′ u = a q u q − 1 − 1 − 2 a q 1 − u q − 1 ,$
so $g ′ u = 0 ⇒ u 1 − u q − 1 = 1 − 2 a a ⇒ u 1 − u = 1 − 2 a a 1 q − 1 = c 1 ⇒ u = c 1 1 + c 1$.
Putting $v : = c 1 1 + c 1 ∈ 0 , 1$, we obtain
$g ′ u < 0 if u < v and g ′ u > 0 if u > v .$
Hence
$min g u : 0 ≤ u ≤ 1 = g v ,$
where
$g v = a c 1 1 + c 1 q + 1 − 2 a 1 1 + c 1 q + a 1 − τ q = a c 1 q 1 + c 1 · 1 1 + c 1 q − 1 + 1 1 + c 1 · 1 − 2 a 1 + c 1 q − 1 + a 1 − τ q = a c 1 q 1 + c 1 · c 2 1 − 2 a + c 2 1 + c 1 + a 1 − τ q = c 1 c 2 1 + c 1 + c 2 1 + c 1 + a 1 − τ q = c 2 + a 1 − τ q = τ q .$
For $u < 1$, we have
$δ q T u , T 1 = τ q = g v ≤ g u = a u q + 1 − 2 a 1 − u q + a 1 − τ q = a δ q u , T u + a δ q 1 , T 1 + 1 − 2 a δ q u , 1 .$
Therefore, T is an $( a − q )$-quasicontraction.
Since $a ≠ 1 3$ we have $c 1 ≠ 1$, $v ≠ 1 2$ and $v ≠ 1 − v$. Using the strict convexity of $t → t q p$, we get
$a δ p v , T v + a δ p 1 , T 1 + 1 − 2 a δ p v , 1 = a v p + 1 − 2 a 1 − v p + a 1 − τ p = a v p + 1 − 2 a 1 − v p + a 1 − τ p q p p q < a v q + 1 − 2 a 1 − v q + a 1 − τ q p q = τ q p q = τ p = δ p T v , T 1 .$
Hence, T is not an $( a − p )$-quasicontraction. □
The folowing example shows that for $a = 1 3$ not every $( a − p )$-quasicontraction $p ≥ 3$ is a quadratic quasicontraction.
Example 2.
(see Example 1 [6]) Let $Y = − 1 , 1$, $δ u , v = u − v$ and
$T u = 0 , u ∈ − 1 , 1 2 , − 1 , u ∈ 1 2 , 1 .$
Then, T satisfies Inequality (2) with $a = 1 3$ and $p ≥ 3$, but T is not a quadratic quasicontraction.
Proof.
Let
$E p u , v = a δ p u , T u + a δ p v , T v + 1 − 2 a δ p u , v .$
For $u ∈ − 1 , 1 2$ and $v ∈ 1 2 , 1$ we have $δ T u , T v = 1$ and
$E p u , v = 1 3 δ p u , T u + 1 3 δ p v , T v + 1 3 δ p u , v = 1 3 u p + 1 3 v + 1 p + 1 3 v − u p > 1 3 1 2 + 1 p = 3 p − 1 2 p > 1 , because p ≥ 3 .$
Therefore, T is an $( 1 3 − p )$-quasicontraction with $p ≥ 3$.
For $u = 0$ and $v = 3 5$ we have $δ T 0 , T 3 5 = δ 0 , − 1 = 1$ and
$E 2 0 , 3 5 = 1 3 δ 2 0 , T 0 + 1 3 δ 2 3 5 , T 3 5 + 1 3 δ 2 0 , 3 5 = 1 3 · 64 25 + 1 3 · 9 25 = 73 75 < δ T 0 , T 3 5 ,$
so T is not a quadratic quasicontraction with $a = 1 3$.
Moreover, since
$E 1 0 , 3 5 = a δ 0 , T 0 + a δ 3 5 , T 3 5 + 1 − 2 a δ 0 , 3 5 = a · 8 5 + 1 − 2 a · 3 5 = 2 a + 3 5 < 1 ,$
we get that T is not an $( a − 1 )$-quasicontraction for any $a ∈ 0 , 1 2$. □
The following theorem is a generalization of Theorem 4 and implicitly a generalization of Theorems 1 and 2.
Theorem 6.
Let $Y , δ$ be a compact metric space and let $T : Y → Y$ be a strict $( a − p )$-quasicontraction. Then, T has a unique fixed point $w ∈ Y$. Moreover, if T is continuous, then, for each $u ∈ Y$, the sequence of iterates $T n u$ converges to w.
Proof.
Letting $v = T u$ in Inequality (2), we have for all $u ∈ Y$ with $u ≠ T u$
$δ p T u , T 2 u < a δ p u , T u + a δ p T u , T 2 u + 1 − 2 a δ p u , T u .$
This implies $δ T u , T 2 u < δ u , T u$.
Now, let $β = inf δ u , T u : u ∈ Y$. Since Y is compact there exists a sequence $u n ⊂ Y$ such that $u n → w ∈ Y$, $T u n → z$ and $β = lim n → ∞ δ u n , T u n = δ w , z$.
If there exists a subsequence $u n k$ of $u n$ such that $u n k = z$ for every $k ∈ N$, then we get $w = z = T z$, so T has a fixed point. Otherwise, we suppose that there exists $N ∈ N$ such that $u n ≠ z$ for all $n ≥ N$. Taking $u = u n$ and $v = z$ in Inequality (2), we obtain
$δ p T u n , T z < a δ p u n , T u n + a δ p z , T z + 1 − 2 a δ p u n , z .$
Letting $n → ∞$, we get
$δ p z , T z ≤ a δ p w , z + a δ p z , T z + 1 − 2 a δ p w , z .$
Thus, $δ z , T z ≤ δ w , z = β$. By definition of $β$, we obtain $δ z , T z = β$.
If $β > 0$, since $δ T z , T 2 z < δ z , T z = β$, we have a contradiction. Therefore, we get $β = 0$ and so $w = z = T z$, 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
$δ p T z , T t ≤ a δ p z , T z + a δ p t , T t + 1 − 2 a δ p z , t ,$
by where
$δ p z , t < 1 − 2 a δ p z , t ,$
Now assume T is continuous. Let $u 0 ∈ Y$ and define a sequence $u n ⊂ Y$ by $u n : = T u n − 1 = T n u 0$. Suppose there exists $N ∈ N ∪ 0$ such that $u N = z$. Then, we have $u n = z$ for all $n ≥ N$, so $u n → z$. Otherwise, we suppose $u n ≠ z$ for all $n ≥ N$. By uniqueness of z, we get $u n ≠$$u n + 1$ for every $n ∈ N ∪ 0$. Hence, we have $δ u n , u n + 1 = δ T u n − 1 , T u n < δ u n − 1 , u n$ for every $n ∈ N$, so the sequence $δ u n , u n + 1 n ∈ N$ is decreasing and positive. Therefore, there exists $b = lim n → ∞ δ u n − 1 , u n$. We claim that $b = 0$. Since Y is compact, there exists a subsequence $u n k$ of $u n$ such that $u n k → z ∈ Y$ as $k → ∞$. If $b > 0$, we have
$b = lim k → ∞ δ u n k , u n k + 1 = δ z , T z > 0 ,$
$b = lim k → ∞ δ u n k + 1 , u n k + 2 = δ T z , T 2 z > 0 .$
Hence, we get $δ z , T z = δ T z , T 2 z = b > 0$, which is a contradiction. Thus, $b = 0$. Now, taking $u = u n$ and $v = z$ in Inequality (2), we obtain
$δ p u n + 1 , z = δ p T u n , z < a δ p u n , T u n + a δ p z , T z + 1 − 2 a δ p u n , z .$
This implies
$α n + 1 < 1 − 2 a α n + β n ,$
where $α n : = δ p x n , z$ and $β n = a δ p u n , u n + 1$. Since $lim n → ∞ β n = 0$, by Lemma 1, we get $lim n → ∞ α n = 0$, i.e., $lim n → ∞ u n = z$. □
Example 3.
Let $Y = − 2 , 2$, $δ u , v = u − v$, $T : Y → Y$ defined by
$T u = 1 − u 2 , i f u ∈ − 2 , − 1 − 2 5 , i f u = − 1 0 , i f u ∈ − 1 , 1 2 5 , i f u = 1 − 1 − u 2 i f u ∈ 1 , 2 .$
Then, T is an $( 1 3 − p )$-quasicontraction for $p ≥ 4$, but T is not an $( 1 3 − 2 )$-quasicontraction (quadratic quasicontraction). Moreover, T is not asymptotic regular.
Proof.
We distinguish 13 cases:
• $1 0$ If $u , v ∈ − 2 , − 1$, $u < v$, we have:
$δ T u , T v = v − u 2 ,$
$E 1 u , v = 1 − 3 u 6 + 1 − 3 v 6 + v − u 3 = 2 − 5 u − v 6 ,$
$δ T u , T v < E 1 u , v ⟺ 2 v + u < 1 ,$
which is obvious.
• $2 0$ By symmetry, the case $u , v ∈ 1 , 2$, $u < v$ is similar.
• $3 0$ If $u ∈ − 2 , − 1$, $v ∈ − 1 , 1$, we have:
$δ T u , T v = 1 − u 2 ,$
$E 1 u , v = 1 − 3 u 6 + v 3 + v − u 3 = 1 − 5 u + 2 v + 2 v 6 ,$
$δ T u , T v < E 1 u , v ⟺ 2 + 2 u < 2 v + 2 v ,$
which is obvious ($2 + 2 u < 0 ≤ 2 v + 2 v$).
• $4 0$ By symmetry, the case $u ∈ 1 , 2$, $v ∈ − 1 , 1$ is similar.
• $5 0$ If $u ∈ − 2 , − 1$, $v ∈ 1 , 2$, we have:
$δ T u , T v = 2 − u + v 2 ,$
$E 1 u , v = 1 − 3 u 6 + 1 + 3 v 6 + v − u 3 = 2 − 5 u + 5 v 6 ,$
$δ T u , T v < E 1 u , v ⟺ 2 < v − u ,$
which is obvious.
• $6 0$ If $u , v ∈ − 1 , 1$, $u ≠ v$, we have:
$δ T u , T v = 0 < u 3 + v 3 + u − v 3 = E 1 u , v .$
• $7 0$ If $u = − 1$, $v = 1$, we have:
$δ T u , T v = 4 5 , E 1 u , v = 1 3 · 3 5 + 1 3 · 3 5 + 1 3 · 2 = 16 15$
$δ T u , T v < E 1 u , v$
• $8 0$ For $u = 1$, $v ∈ − 1 , 1$, we get:
$δ T u , T v = 2 5 , E 1 u , v = 1 3 · 3 5 + v 3 + 1 − v 3 = 8 + 5 v − 5 v 15$
$δ T u , T v < E 1 u , v ⟺ 0 < 2 + 5 v − 5 v ,$
which is obvious.
• $9 0$ By symmetry, the case $u = − 1$, $v ∈ − 1 , 1$ is similar.
• $10 0$ For $u = 1$, $v ∈ − 2 , − 1$, we have:
$δ T u , T v = 1 − 5 v 10 , E 1 u , v = 1 3 · 1 − 3 v 2 + 1 3 · 3 5 + 1 − v 3 = 21 − 25 v 30$
$δ T u , T v < E 1 u , v ⟺ 10 v < 18 ,$
which is obvious.
• $11 0$ By symmetry, the case $u = − 1$, $v ∈ 1 , 2$ is similar.
• $12 0$ For $u = 1$, $v ∈ 1 , 2$, we have:
$δ p T u , T v = 9 + 5 v 10 p , E p u , v = 1 3 · 3 5 p + 1 3 · 3 v + 1 2 p + 1 3 · 1 − v p$
Since the function $v → 15 v + 5 9 + 5 v$ is increasing and $p ≥ 4$, we obtain
$15 v + 5 9 + 5 v p > 15 + 5 9 + 5 p = 10 7 p ≥ 10 7 4 ≥ 3 ,$
so
$9 + 5 v 10 p < 1 3 · 3 v + 1 2 p ,$
by where $δ p T u , T v < E p u , v$.
• $13 0$ By symmetry, the case $u = − 1$, $v ∈ − 2 , − 1$ is similar.
By Proposition 1 is obvious that $δ T u , T v < E 1 u , v ⇒ δ p T u , T v < E p u , v$. Therefore we get that in all cases we have $δ p T u , T v < E p u , v$. Hence, T is an $( 1 3 − p )$- quasicontraction for $p ≥ 4$.
If $u = 1$, $v = 4 3$, we have:
$δ 2 T u , T v = 6627 2700 ,$
$E 2 u , v = 1 3 · 9 25 + 1 3 · 25 4 + 1 3 · 1 9 = 5949 2700 ,$
so $δ 2 T u , T v > E 2 u , v$. This implies that T is not a quadratic quasicontraction.
It is easy to prove that $T n u = − 1 n · u + 2 n − 1 2 n$ for $u ∈ 1 , 2$, so $T n 2 = − 1 n · 2 n + 1 2 n$, for all $n ∈ N$, and $δ T n 2 , T n + 1 2 = 2 n + 1 2 n + 2 n + 1 + 1 2 n + 1 > 2$ 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 $Y , δ$ and let $T : C → C$ be an $( a − p )$-quasicontraction mapping. Assume that there exist constants $c , b ∈ R$ such that $0 ≤ c < 1$ and $b > 0$. If for every $u ∈ C$ there exists $v ∈ C$ such that $δ v , T v ≤ c δ u , T u$ and $δ v , u ≤ b δ u , T u$, then T has a unique fixed point.
Proof.
Let $u 0 ∈ C$. We construct the sequence $u n ⊂ C$ such that
$δ u n + 1 , T u n + 1 ≤ c δ u n , T u n ,$
$δ u n + 1 , u n ≤ b δ u n , T u n , n = 0 , 1 , 2 , . . .$
Since
$δ u n + 1 , u n ≤ b δ u n , T u n ≤ b c δ u n − 1 , T u n − 1 ≤ . . . ≤ b c n δ u 0 , T u 0$
it is easy to prove that $u n$ is a Cauchy sequence. By completeness of C, there exists $w ∈ C$ such that $lim n → ∞ u n = w$. From the above inequalities and sandwich theorem, we get $lim n → ∞ δ u n , T u n = 0$, so $lim n → ∞ T u n = w$ and
$δ p T u n , T w ≤ a δ p u n , T u n + a δ p w , T w + 1 − 2 a δ p u n , w .$
Letting $n → ∞$, we obtain
$δ p w , T w ≤ a δ p w , T w .$
This implies $δ w , T w = 0$, i.e., $w = T w$.
If t is another fixed point of T, taking $u = w$, $v = t$ in Inequality (2) we obtain
$δ p T w , T t ≤ a δ p w , T w + a δ p t , T t + 1 − 2 a δ p w , t ,$
hence
$δ p w , t ≤ 1 − 2 a δ p w , t .$
Therefore, $δ w , t = 0$, 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 $T : C → C$ be a mapping satisfying the inequality:
$T u − T v p ≤ a u − T u p + a v − T v p + b u − v p$
for all $u , v ∈ C$, where $0 < a < 1 2$, $b = 1 − 2 a$, $b ≤ 2 p + 1 − 1 p − 1 p − 1 2 p − 1$, $p ∈ N$, $p ≥ 2$. Then, T has a unique fixed point.
Proof.
Taking $v = T u$ in (4), we have:
$T u − T 2 u p ≤ a u − T u p + a T u − T 2 u p + b u − T u p .$
This implies
$T u − T 2 u ≤ u − T u , for all u ∈ C .$
Now, let $u ∈ C$ arbitrary fixed and $z = 1 2 T 2 u + 1 2 T 3 u$. Since C is convex, we have $z ∈ C$. Then, by Inequalities (4) and (5), we get
$T u − T 3 u p ≤ a u − T u p + a T 2 u − T 3 u p + b u − T 2 u p ≤ 2 a u − T u p + b u − T u + T u − T 2 u p ≤ 2 a + 2 p b u − T u p ,$
so
$T u − T 3 u ≤ 2 a + 2 p b p u − T u .$
Therefore,
$T u − z = 1 2 T u − T 2 u + 1 2 T u − T 3 u ≤ 1 2 T u − T 2 u + 1 2 T u − T 3 u ≤ 1 2 u − T u + 1 2 2 a + 2 p b p u − T u = 1 2 1 + 2 a + 2 p b p u − T u .$
$T 2 u − z = 1 2 T 2 u − T 3 u ≤ 1 2 u − T u .$
Now, by Inequalities (4), (5) and (7), we obtain
$T 2 u − T z p ≤ a T u − T 2 u p + a z − T z p + b T u − z p ≤ a u − T u p + a z − T z p + b 1 + 2 a + 2 p b p 2 p u − T u p = a z − T z p + a + b 1 + 2 a + 2 p b p 2 p u − T u p .$
By Inequalities (4), (5) and (8), we obtain
$T 3 u − T z p ≤ a T 2 u − T 3 u p + a z − T z p + b T 2 u − z p ≤ a u − T u p + a z − T z p + b 2 p u − T u p = a z − T z p + a + b 2 p u − T u p .$
Since
$z − T z = 1 2 T 2 u − T z + 1 2 T 3 u − T z ≤ 1 2 T 2 u − T z + 1 2 T 3 u − T z ,$
by Inequalities (9) and (10), we get
$z − T z ≤ 1 2 a z − T z p + a + b 1 + 2 a + 2 p b p 2 p u − T u p 1 p + 1 2 a z − T z p + a + b 2 p u − T u p 1 p .$
If $u = T u$, then u is a fixed point of T. Otherwise, dividing Inequality (11) by $1 2 u − T u$, we get
$2 · z − T z u − T u ≤ a z − T z p u − T u p + a + b 1 + 2 a + 2 p b p 2 p 1 p + a z − T z p u − T u p + a + b 2 p 1 p .$
Denoting $t : = z − T z p u − T u p$, we obtain
$2 t 1 p ≤ a t + a + b 1 + 2 a + 2 p b p 2 p 1 p + a t + a + b 2 p 1 p ,$
which implies $z = T z$ or $f t ≥ 2$, where
$f t = a + a t + b t 1 + 2 a + 2 p b p 2 p 1 p + a + a t + b 2 p · t 1 p$
Clearly, f is a decreasing function and
$f 1 = 2 a + b 1 + 2 a + 2 p b p 2 p 1 p + a + b 2 p 1 p = 1 − b + b 1 + 1 − b + 2 p b p 2 p 1 p + 1 − b + b 2 p 1 p .$
We claim that $f 1 < 2$. Since $b ≤ 2 p + 1 − 1 p − 1 p − 1 2 p − 1$, we have $1 + 1 − b + 2 p b p 2 p ≤ 2 p + 1 − 1 2 p ≤ 2 − 1 2 p$ and then
$f 1 < 1 − b + b 2 − 1 2 p 1 p + 1 − b + b 2 p 1 p = 1 + b − b 2 p 1 p + 1 − b + b 2 p 1 p .$
Using strict concavity of $t → t 1 p$, we obtain
$f 1 < 2 1 + b − b 2 p + 1 − b + b 2 p 2 1 p = 2 .$
Since f is a decreasing function and $f t ≥ 2$, there exists $c < 1$ such that $t ≤ c$. Therefore, $z − T z ≤ c u − T u$.
Now, since
$z − u ≤ 1 2 T 2 u − u + 1 2 T 3 u − u ≤ 1 2 T 2 u − T u + T u − u + 1 2 T 3 u − T 2 u + T 2 u − T u + T u − u ≤ 5 2 T u − u ,$
applying Lemma 2, we get that T has a unique fixed point. □
Example 4.
Let $Y = l ∞ R$ be the set of bounded sequence of real numbers and $u = sup n ∈ N u n$, where $u = u n n ∈ N$. It is well known that $Y , ·$ is a Banach space. Let $C = u ∈ Y : u ≤ 1$ and $T : C → C$ defined by
$T u = 1 3 , i f u = − 1 − 1 , i f u n ∈ 1 3 , 1 f o r e v e r y n ∈ N 0 , o t h e r w i s e ,$
where $u = u n n ∈ N$, $c = c , c , c , . . .$. Clearly, C is closed, convex and not compact. Since $T n − 1 = 1 3$ if n is odd and $T n − 1 = − 1$ if n is even, we note that T is not asymptotic regular.
If $u = − 1$ and $v = v n n ∈ N$, where $v n ∈ 1 3 , 1$ for every $n ∈ N$, then $T u − T v = 4 3$ and
$E 1 u , v = 4 3 a + a sup n ∈ N 1 + v n + b sup n ∈ N 1 + v n ≥ 4 3 a + 4 3 a + b = 4 3 .$
Hence $T u − T v ≤ E 1 u , v$, and then $T u − T v 3 ≤ E 3 u , v$.
If $u = − 1$ and $v ≠ u$, $v = v n n ∈ N$, where there exists $n 0$ such that $v n 0 ∉ 1 3 , 1$, then $T u − T v = 1 3$ and
$E 1 u , v = 4 3 a + a sup n ∈ N v n + b sup n ∈ N 1 + v n ≥ 4 3 a ≥ 1 3 i f a ≥ 1 4 .$
If $u = u n n ∈ N$ where $u n ∈ 1 3 , 1$ for every $n ∈ N$ and $v = v n n ∈ N$, where there exists $n 0$ such that $v n 0 ∉ 1 3 , 1$, then $T u − T v = 1$ and
$E 3 u , v = a sup n ∈ N 1 + u n 3 + a sup n ∈ N v n 3 + b sup n ∈ N u n − v n 3 ≥ 4 3 3 a = 64 27 a .$
If $a ≥ 27 64$ we have $T u − T v 3 ≤ E 3 u , v$.
Therefore, T satisfies Inequality (4) with $a = 28 64 = 7 16$, $b = 1 8 ≤ 15 3 − 1 3 − 1 7 ≃ 0.3074$, $p = 3$.
However, T does not satisfy Inequality (4) if $u = 1 3$, $v = 0$ and $p = 2$: $T u − T v = 1$ and
$E 2 u , v = 16 9 a + a · 0 + 1 9 b = 16 a + b 9 = 1 + 14 a 9 < 1 .$

## 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.

## Author Contributions

Conceptualization, O.P. and G.S.; methodology, O.P. and G.S.; investigation, O.P. and G.S.; writing—original draft preparation, O.P. and G.S.; writing—review and editing, O.P and G.S. All authors have read and agreed to the published version of the manuscript.

## Funding

The author declares that there is no funding for the present paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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Popescu, O.; Stan, G. Some Fixed Point Theorems for (ap)-Quasicontractions. Symmetry 2020, 12, 1973. https://doi.org/10.3390/sym12121973

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Popescu O, Stan G. Some Fixed Point Theorems for (ap)-Quasicontractions. Symmetry. 2020; 12(12):1973. https://doi.org/10.3390/sym12121973

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Popescu, Ovidiu, and Gabriel Stan. 2020. "Some Fixed Point Theorems for (ap)-Quasicontractions" Symmetry 12, no. 12: 1973. https://doi.org/10.3390/sym12121973

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