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Article

# On the Study of Fixed Points for a New Class of α-Admissible Mappings

by 3 and
1
Department of Mathematics, Faculty of Science, Damanhour University, Damanhour 22511, Egypt
2
Department of Mathematics, College of Science, King Saud University, P.O. Box 2455, Riyadh 11451, Saudi Arabia
3
School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, Galway H91 TK33, Ireland
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(12), 1240; https://doi.org/10.3390/math7121240
Received: 27 October 2019 / Revised: 9 December 2019 / Accepted: 10 December 2019 / Published: 14 December 2019
(This article belongs to the Section Mathematics and Computer Science)

## Abstract

:
In this paper, we discuss the existence of fixed points for new classes of mappings. Some examples are presented to illustrate our results.
MSC:
54H25; 47H10

## 1. Introduction

The Banach contraction principle is one of the most famous and important results in metric fixed point theory. It is a useful tool in establishing existence results in nonlinear analysis. This principle has been extended and generalized by several authors in many directions (see e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15], and the references therein).
In [16], the author introduced the class of F-contractions, and established a fixed point result for this class of mappings, which generalizes the Banach contraction principle. The main result in [16] can be stated as follows.
Theorem 1.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be a mapping satisfying
$τ + F ( d ( T x , T y ) ) ≤ F ( d ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $d ( T x , T y ) > 0$, where $τ > 0$ is a constant and $F : ( 0 , + ∞ ) → R$ is a function satisfying
(a)
F is nondecreasing.
(b)
For every sequence ${ t n } ⊂ ( 0 , + ∞ )$, we have
$lim n → + ∞ F ( t n ) = − ∞ ⟺ lim n → + ∞ t n = 0 .$
(c)
There exists $k ∈ ( 0 , 1 )$ such that $lim t → 0 + t k F ( t ) = 0$.
Then T has a unique fixed point. Moreover, for any $x ∈ X$, the Picard sequence ${ T n x }$ converges to this fixed point.
Observe that, if $T : X → X$ is a q-contraction for some $0 < q < 1$, i.e.,
$d ( T x , T y ) ≤ q d ( x , y ) , ( x , y ) ∈ X × X ,$
then T satisfies (1) with $F ( t ) = ln t$, $t > 0$, and $τ = − ln q$. Therefore, the Banach contraction principle follows from Theorem 1.
For different extensions and generalizations of Theorem 1, we refer the reader to [17,18,19,20,21,22,23,24,25,26,27], and the references therein.
In [5], Ćirić introduced a class of mappings with a non-unique fixed point and he established the following fixed point result.
Theorem 2.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be a continuous mapping satisfying
$min { d ( T x , T y ) , d ( x , T x ) , d ( y , T y ) } − min { d ( x , T y ) , d ( y , T x ) } ≤ q d ( x , y ) ,$
for all $( x , y ) ∈ X × X$, where $0 < q < 1$ is a constant. Then, for any $x ∈ X$, the Picard sequence ${ T n x }$ converges to a fixed point of T.
An example was presented in [5] to show that the set of fixed points of mappings satisfying the condition of Theorem 2 contains in general more than one element.
In this paper, we first introduce the class of generalized Ćirić-contractions by combining the ideas in [5,16]. Next, a fixed point result is established for this class of mappings. Our result generalizes Theorem 2 and extends Theorem 1. Next, we introduce a more general class of mappings using the concept of $α$-admissibility introduced in [28] (see also [29]). Our fixed point result for this class of mappings has several consequences. It is not only a generalization of Theorems 1 and 2, but generalizes most fixed point theorems dealing with F-contractions, linear contractions, and many others. Several examples are presented to illustrate this fact.
Throughout this paper, we denote by $N$ the set of natural numbers, that is, $N = { 0 , 1 , 2 , ⋯ }$. We denote by $N *$ the set $N ∖ { 0 }$. Let $T : X → X$ be a certain self-mapping on X. For $n ∈ N$, we denote by $T n$ the nth-iterate of T (we suppose that $T 0$ is the identity mapping on X).

## 2. The Class of Generalized Ćirić-Contractions

Let $Ψ$ be the set of functions $ψ : [ 0 , + ∞ ) → ( − ∞ , 0 )$ such that $ψ$ is upper semi-continuous from the right. We denote by $Φ$ the set of functions $φ : ( 0 , + ∞ ) → R$ such that
($Φ 1$)
$φ$ is non-decreasing, i.e., $0 < t < s ⇒ φ ( t ) ≤ φ ( s )$.
($Φ 2$)
For every sequence ${ t n } ⊂ ( 0 , + ∞ )$,
$lim n → + ∞ φ ( t n ) = − ∞$
if and only if
$lim n → + ∞ t n = 0 .$
($Φ 3$)
There exists $k ∈ ( 0 , 1 )$ such that $lim t → 0 + t k φ ( t ) = 0$.
Let $( X , d )$ be a metric space. For a given mapping $T : X → X$, let
$M T ( x , y ) = min { d ( T x , T y ) , d ( x , T x ) , d ( y , T y ) } − min { d ( x , T y ) , d ( y , T x ) } , ( x , y ) ∈ X × X .$
Definition 1.
A mapping $T : X → X$ is said to be a generalized Ćirić-contraction, if there exists $( φ , ψ ) ∈ Φ × Ψ$ such that
$φ ( M T ( x , y ) ) ≤ φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $M T ( x , y ) > 0$.
We have the following fixed point result.
Theorem 3.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be a continuous mapping. If T is a generalized Ćirić-contraction for some $( φ , ψ ) ∈ Φ × Ψ$, then for any $x ∈ X$, the Picard sequence ${ T n x }$ converges to a fixed point of T.
Proof.
Let $x ∈ X$ be fixed, and let ${ x n } ⊂ X$ be the sequence defined by
$x n = T n x , n ∈ N .$
If $x p + 1 = x p$ for some $p ∈ N$, then $x p$ will be a fixed point of T. Therefore, we may assume that
$d ( x n , x n + 1 ) > 0 , n ∈ N .$
On the other hand, for every $n ∈ N$, we have
$M T ( x n , x n + 1 ) = M T ( T n x , T n + 1 x ) = min { d ( T n + 1 x , T n + 2 x ) , d ( T n x , T n + 1 x ) , d ( T n + 1 x , T n + 2 x ) } − min { d ( T n x , T n + 2 x ) , d ( T n + 1 x , T n + 1 x ) } = min { d ( x n + 1 , x n + 2 ) , d ( x n , x n + 1 ) } .$
Therefore, from (4), we have
$M T ( x n , x n + 1 ) > 0 , n ∈ N .$
From (3), we obtain
$φ ( M T ( x n , x n + 1 ) ) ≤ φ ( d ( x n , x n + 1 ) ) + ψ ( d ( x n , x n + 1 ) ) , n ∈ N .$
If for some $n ∈ N$, we have $M T ( x n , x n + 1 ) = d ( x n , x n + 1 )$, then we obtain
$φ ( d ( x n , x n + 1 ) ) ≤ φ ( d ( x n , x n + 1 ) ) + ψ ( d ( x n , x n + 1 ) ) ,$
that is,
$0 ≤ ψ ( d ( x n , x n + 1 ) ) ,$
which is a contradiction with the fact that $ψ ( t ) < 0$, for all $t > 0$. As a consequence, we have
$M T ( x n , x n + 1 ) = d ( x n + 1 , x n + 2 ) , n ∈ N .$
Hence, we find
$φ ( d ( x n + 1 , x n + 2 ) ) ≤ φ ( d ( x n , x n + 1 ) ) + ψ ( d ( x n , x n + 1 ) ) , n ∈ N .$
Taking $n = 0$ in (5), we obtain
$φ ( d ( x 1 , x 2 ) ) ≤ φ ( d ( x 0 , x 1 ) ) + ψ ( d ( x 0 , x 1 ) ) .$
Taking $n = 1$ in (5) and using the above inequality, we obtain
$φ ( d ( x 2 , x 3 ) ) ≤ φ ( d ( x 1 , x 2 ) ) + ψ ( d ( x 1 , x 2 ) ) ≤ φ ( d ( x 0 , x 1 ) ) + ψ ( d ( x 0 , x 1 ) ) + ψ ( d ( x 1 , x 2 ) ) .$
Continuing this process, by induction we have
$φ ( d ( x n , x n + 1 ) ) ≤ φ ( d ( x 0 , x 1 ) ) + ∑ i = 0 n − 1 ψ ( d ( x i , x i + 1 ) ) , n ∈ N * .$
Next, let us denote by ${ u n }$ the real sequence defined by
$u n = d ( x n , x n + 1 ) , n ∈ N .$
Observe that from (5), and using ($Φ 1$) and the fact that $ψ ( t ) < 0$ for all $t > 0$, we deduce that ${ u n }$ is a decreasing sequence. Therefore, there exists some $r ≥ 0$ such that
$u n ↓ r as n → + ∞ .$
Since $ψ$ is upper semi-continuous from the right, there exists some $N ∈ N$ such that
$ψ ( u p ) < ψ ( r ) − ψ ( r ) 2 = ψ ( r ) 2 , p ≥ N .$
Further, using (6) and the fact that $ψ ( t ) < 0$ for all $t > 0$, we obtain
$φ ( u n ) ≤ φ ( u 0 ) + ∑ i = N n − 1 ψ ( u i ) , n ≥ N + 1 .$
Therefore, from (7) we deduce that
$φ ( u n ) ≤ φ ( u 0 ) + ( n − N ) 2 ψ ( r ) , n ≥ N + 1 .$
Let $n → + ∞$ in (8) and we obtain
$lim n → + ∞ φ ( u n ) = − ∞ ,$
which implies from ($Φ 2$) that
$lim n → + ∞ u n = 0 = r .$
Next, we prove that ${ x n }$ is a Cauchy sequence. From ($Φ 3$) and (9), there exists some $k ∈ ( 0 , 1 )$ such that
$lim n → + ∞ u n k φ ( u n ) = 0 .$
Using (8), we obtain
$u n k φ ( u n ) − u n k φ ( u 0 ) ≤ ( n − N ) 2 ψ ( r ) u n k ≤ 0 , n ≥ N + 1 .$
Let $n → + ∞$, and using (9) and (10), we deduce that
$lim n → + ∞ n u n k = 0 .$
Then there exists some $q ∈ N$ such that
$u n < 1 n 1 / k , n ≥ q .$
Using (11) and the triangle inequality, for $n ≥ q$ and $m ∈ N *$, we have
$d ( x n , x n + m ) ≤ ∑ i = n n + m − 1 u i ≤ ∑ i = n + ∞ 1 i 1 / k .$
The convergence of the Riemann series $∑ n 1 n 1 / k$ (since $0 < k < 1$) yields ${ x n }$ is a Cauchy sequence. Since $( X , d )$ is complete, there exists some $ω ∈ X$ such that
$lim n → + ∞ d ( T n x , ω ) = lim n → + ∞ d ( x n , ω ) = 0 .$
The continuity of T yields
$lim n → + ∞ d ( T n + 1 x , T ω ) = 0 .$
Finally, the uniqueness of the limit implies that $ω = T ω$, i.e., $ω$ is a fixed point of T. □
Let us give some examples to illustrate the result given by Theorem 3.
Example 1.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be a continuous mapping. Let $F : ( 0 , + ∞ ) → R$ be a function that belongs to Φ. Suppose that there exists a constant $τ > 0$ such that
$τ + F ( M T ( x , y ) ) ≤ F ( d ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $M T ( x , y ) > 0$. Then for any $x ∈ X$, the Picard sequence ${ T n x }$ converges to a fixed point of T. In order to prove this result, we apply Theorem 3 with $( φ , ψ ) = ( F , − τ )$.
Example 2.
Suppose that all the assumptions of Theorem 2 are satisfied. Then T satisfies (3) with $φ ( t ) = ln t$, $t > 0$, and $ψ ≡ ln q$. Therefore, the result of Theorem 2 follows from Theorem 3.
Example 3.
Let
$X = x n = n ( n + 1 ) 2 : n ∈ N * .$
We endow X with the metric
$d ( x , y ) = | x − y | , ( x , y ) ∈ X × X .$
Then $( X , d )$ is a complete metric space. Consider the mapping $T : X → X$ defined by
$T x 1 = x 1 a n d T x n + 1 = x n , n ∈ N * .$
One observes easily that
${ ( x , y ) ∈ X × X : M T ( x , y ) > 0 } = { ( x n , x n + 1 ) : n ∈ N * } .$
Furthermore, for all $n ∈ N *$, one has
$M T ( x n , x n + 1 ) d ( x n , x n + 1 ) = n n + 1 → 1 a s n → ∞ ,$
which shows that (2) is not satisfied. Hence Theorem 2 cannot be applied in this case. On the other hand, taking $τ = 1$ and
$F ( t ) = t + ln t , t > 0 ,$
one obtains
$τ + F ( M T ( x n , x n + 1 ) ) = 1 + F ( n ) = 1 + n + ln n ≤ 1 + n + ln ( n + 1 ) = F ( d ( x n , x n + 1 ) ) ,$
for all $n ∈ N *$. Hence (12) is satisfied for all $( x , y ) ∈ X × X$ with $M T ( x , y ) > 0$. Therefore, by Example 1, one deduces that T has a fixed point $x * ∈ X$. In this case, one observes that $x * = x 1 = 1$.

## 3. A Larger Class of Mappings

In this part, we discuss the existence of fixed points for a larger class of mappings than the one studied in the previous section. First, let us recall some concepts introduced recently by Samet in [29] (see also [28]).
Let $( X , d )$ be a metric space, and let $α : X × X → R$ be a given function.
Definition 2.
Let ${ x n } ⊂ X$ be a given sequence. We say that ${ x n }$ is α-regular if
$α ( x n , x n + 1 ) ≥ 1 , n ∈ N .$
Definition 3.
We say that $T : X → X$ is α-admissible if
$( x , y ) ∈ X × X , α ( x , y ) ≥ 1 ⇒ α ( T x , T y ) ≥ 1 .$
Definition 4.
We say that $T : X → X$ is α-continuous if for every α-regular sequence ${ x n } ⊂ X$ and $u ∈ X$,
$lim n → + ∞ d ( x n , u ) = 0$
implies that there exists a sub-sequence ${ x n k }$ of ${ x n }$ such that
$lim k → + ∞ d ( T x n k , T u ) = 0 .$
Definition 5.
Let ${ x n } ⊂ X$ be a given sequence. We say that ${ x n }$ is α-Cauchy if
(i)
${ x n }$ is α-regular.
(ii)
${ x n }$ is a Cauchy sequence.
Definition 6.
We say that $( X , d )$ is α-complete if every α-Cauchy sequence is convergent.
Next, we introduce the following class of mappings.
Let $T α$ be the class of mappings $T : X → X$ satisfying the following conditions:
($T 1$)
T is $α$-continuous.
($T 2$)
There exists $( φ , ψ ) ∈ Φ × Ψ$ such that for all $( x , y ) ∈ X × X$ with $d ( T x , T y ) > 0$,
$α ( x , y ) exp φ ( d ( T x , T y ) ) ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
We now give some examples of mappings $T : X → X$ that belong to the set $T α$, for some $α : X × X → R$. Let $( X , d )$ be a metric space.
Proposition 1 (The class of generalized Ćirić-contractions).
Let $T : X → X$ be a continuous mapping. If T is a generalized Ćirić-contraction, then there exists a function $α : X × X → R$ such that $T ∈ T α$.
Proof.
Let us consider the function $α : X × X → R$ defined by
$α ( x , y ) = 1 if y = T x , 0 if y ≠ T x .$
Let $( x , y ) ∈ X × X$ be such that $d ( T x , T y ) > 0$. We discuss two possible cases.
Case 1: $y ≠ T x$. In this case,
$α ( x , y ) exp φ ( d ( T x , T y ) ) = 0 ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
Case 2: $y = T x$. In this case, we have
$M T ( x , y ) = M T ( x , T x ) = min { d ( T x , T 2 x ) , d ( x , T x ) } .$
Since $d ( T x , T 2 x ) = d ( T x , T y ) > 0$, we have $d ( x , T x ) > 0$. Therefore, $M T ( x , y ) > 0$. Using the fact that T is a generalized Ćirić-contraction, we deduce that
$φ ( M T ( x , T x ) ) ≤ φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) ,$
that is,
$φ ( min { d ( T x , T 2 x ) , d ( x , T x ) } ) ≤ φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) ,$
which yields (since $ψ ( t ) < 0$, for all $t > 0$)
$φ ( d ( T x , T 2 x ) ) ≤ φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) .$
Hence, we obtain
$α ( x , T x ) exp φ ( d ( T x , T 2 x ) ) ≤ exp φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) .$
Therefore, T satisfies ($T 2$) with $α$ given by (13). Obviously, since T is continuous, then T is $α$-continuous. Then T satisfies ($T 1$). As a consequence, we have $T ∈ T α$. □
Proposition 2 (The class of F-contractions).
Let $T : X → X$ be an F-contraction, for some $F ∈ Φ$, that is, there exists a constant $τ > 0$ such that
$τ + F ( d ( T x , T y ) ) ≤ F ( d ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $d ( T x , T y ) > 0$. Then there exists a function $α : X × X → R$ such that $T ∈ T α$.
Proof.
Let
$α ( x , y ) = 1 , ( x , y ) ∈ X × X .$
Let $φ = F$ and $ψ ≡ − τ$. Then $( φ , ψ ) ∈ Φ × Ψ$. Let $( x , y ) ∈ X × X$ be such that $d ( T x , T y ) > 0$. Then
$φ ( d ( T x , T y ) ) ≤ φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) ,$
which yields
$α ( x , y ) exp φ ( d ( T x , T y ) ) ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
Then T satisfies $T 2$ with $α$ given by (14). On the other hand, it can be easily seen that any F-contraction is continuous, so it is $α$-continuous. Then T satisfies also $T 1$. As a consequence, we have $T ∈ T α$. □
Proposition 3.
Let $T : X → X$ be an orbitally continuous mapping, that is, for every $x ∈ X$, if
$lim n → + ∞ d ( T n x , u ) = 0 , u ∈ X ,$
then
$lim n → + ∞ d ( T T n x , T u ) = 0 .$
Suppose that there exist $F ∈ Φ$ and a constant $τ > 0$ such that
$τ + F ( d ( T x , T y ) ) ≤ F ( N T ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $d ( T x , T y ) > 0$, where
$N T ( x , y ) = max d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) + d ( y , T x ) 2 .$
Then there exists a function $α : X × X → R$ such that $T ∈ T α$.
Proof.
Let $α : X × X → R$ be the function defined by (13). Let $φ = F$ and $ψ ≡ − τ$. Then $( φ , ψ ) ∈ Φ × Ψ$. Let $( x , y ) ∈ X × X$ be such that $d ( T x , T y ) > 0$. We discuss two possible cases.
Case 1. $y ≠ T x$. In this case,
$α ( x , y ) exp φ ( d ( T x , T y ) ) = 0 ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
Case 2. $y = T x$. In this case,
$N T ( x , y ) = max d ( x , T x ) , d ( T x , T 2 x ) , d ( x , T 2 x ) 2 .$
On the other hand, by the triangle inequality, we have
$d ( x , T 2 x ) 2 ≤ d ( x , T x ) + d ( T x , T 2 x ) 2 ≤ max { d ( x , T x ) , d ( T x , T 2 x ) } .$
Therefore,
$N T ( x , y ) = max d ( x , T x ) , d ( T x , T 2 x ) .$
Suppose that $N T ( x , y ) = d ( T x , T 2 x )$. Then by (15), we have
$τ + φ ( d ( T x , T 2 x ) ) ≤ φ ( d ( T x , T 2 x ) ) ,$
which yields $τ ≤ 0$, which is a contradiction. Then we have $N T ( x , y ) = d ( x , T x )$. Again, by (15), we deduce that
$φ ( d ( T x , T 2 x ) ) ≤ φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) ,$
which yields
$α ( x , T x ) exp φ ( d ( T x , T 2 x ) ) ≤ exp φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) .$
Then T satisfies $T 2$ with $α$ given by (13). Next, we prove that T is $α$-continuous. Let ${ x n } ⊂ X$ be an $α$-regular sequence. By the definition of $α$, this means that
$x n + 1 = T x n , n ∈ N ,$
that is,
$x n = T n x 0 , n ∈ N .$
Suppose that there exists $u ∈ X$ such that
$lim n → + ∞ d ( x n , u ) = lim n → + ∞ d ( T n x 0 , u ) = 0 .$
Since T is orbitally continuous, we obtain
$lim n → + ∞ d ( T x n , T u ) = 0 .$
Then T is $α$-continuous, and it satisfies ($T 1$). As a consequence, we have $T ∈ T α$. □
Remark 1.
Let $T : X → X$ be a given mapping. Suppose that there exists a constant $0 < q < 1$ such that
$d ( T x , T y ) ≤ q N T ( x , y ) , ( x , y ) ∈ X × X .$
It can be easily seen that T is orbitally continuous mapping, and it satisfies (15) with $τ = − ln q$ and $F ( t ) = ln t$, $t > 0$. Therefore, $T ∈ T α$, where α is given by (13) and $( ϕ , ψ ) = ( F , − ln q )$.
Proposition 4.
Let $T : X → X$ be an orbitally continuous mapping. Suppose that there exist $F ∈ Φ$ and a constant $τ > 0$ such that
$τ + F ( d ( T x , T y ) ) ≤ F ( μ T ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $d ( T x , T y ) > 0$, where
$μ T ( x , y ) = max d ( x , y ) , d ( y , T y ) 1 + d ( x , T x ) 1 + d ( x , y ) .$
Then there exists a function $α : X × X → R$ such that $T ∈ T α$.
Proof.
Let $α : X × X → R$ be the function defined by (13). Let $φ = F$ and $ψ ≡ − τ$. Then $( φ , ψ ) ∈ Φ × Ψ$. Let $( x , y ) ∈ X × X$ be such that $d ( T x , T y ) > 0$. We discuss two possible cases.
Case 1. $y ≠ T x$. In this case, we have
$α ( x , y ) exp φ ( d ( T x , T y ) ) = 0 ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
Case 2. $y = T x$. In this case,
$μ T ( x , y ) = max d ( x , T x ) , d ( T x , T 2 x ) .$
If $μ T ( x , y ) = d ( T x , T 2 x )$, then by (16), we have
$τ + F ( d ( T x , T 2 x ) ) ≤ F ( d ( T x , T 2 x ) ) ,$
that is
$τ ≤ 0 ,$
which is a contradiction. Therefore, $μ T ( x , y ) = d ( x , T x )$. Again, by (16), we deduce that
$φ ( d ( T x , T 2 x ) ) ≤ φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) ,$
which yields
$α ( x , T x ) exp φ ( d ( T x , T 2 x ) ) ≤ exp φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) .$
Then T satisfies $T 2$ with $α$ given by (13). Since T is orbitally continuous, from the proof of Proposition 3, T is $α$-continuous, and it satisfies $T 1$. As a consequence, we have $T ∈ T α$. □
Remark 2.
Let $T : X → X$ be a given mapping. Suppose that there exists a constant $0 < q < 1$ such that
$d ( T x , T y ) ≤ q μ T ( x , y ) , ( x , y ) ∈ X × X .$
It can be easily seen that T is orbitally continuous mapping, and it satisfies (16) with $τ = − ln q$ and $F ( t ) = ln t$, $t > 0$. Therefore, $T ∈ T α$, where α is given by (13) and $( ϕ , ψ ) = ( F , − ln q )$.
Proposition 5 (The class of almost F-contractions).
Let $T : X → X$ be an almost F-contraction (see [22]), that is, there exist $F ∈ Φ$, $τ > 0$ and $L ≥ 0$ such that
$τ + F ( d ( T x , T y ) ) ≤ F ( d ( x , y ) + L d ( y , T x ) ) ,$
for all $( x , y ) ∈ X × X$ with $d ( T x , T y ) > 0$. Then there exists a function $α : X × X → R$ such that $T ∈ T α$.
Proof.
Let $α : X × X → R$ be the function defined by (13). Let $φ = F$ and $ψ ≡ − τ$. Then $( φ , ψ ) ∈ Φ × Ψ$. Let $( x , y ) ∈ X × X$ be such that $d ( T x , T y ) > 0$. We discuss two possible cases.
Case 1. $y ≠ T x$. In this case, we have
$α ( x , y ) exp φ ( d ( T x , T y ) ) = 0 ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
Case 2. $y = T x$. In this case, from (17), we have
$φ ( d ( T x , T 2 x ) ) ≤ φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) ,$
which yields
$α ( x , T x ) exp φ ( d ( T x , T 2 x ) ) ≤ exp φ ( d ( x , T x ) ) + ψ ( d ( x , T x ) ) .$
Then T satisfies $T 2$ with $α$ given by (13). Next, we shall prove that T is $α$-continuous. Let ${ x n } ⊂ X$ be an $α$-regular sequence, i.e.,
$x n + 1 = T x n , n ∈ N .$
Suppose that there exists $u ∈ X$ such that
$lim n → + ∞ d ( x n , u ) = 0 .$
Let us define the set
$I = { n ∈ N : d ( x n , T u ) = 0 } .$
If $| I | < + ∞$, then there exists some $N ∈ N$ such that
$d ( x n + 1 , T u ) > 0 , n ≥ N .$
From (17) and ($Φ 1$), we have
$d ( x n + 1 , T u ) ≤ d ( x n , u ) + L d ( u , x n + 1 ) , n ≥ N .$
Let $n → + ∞$ and we obtain
$lim n → + ∞ d ( x n + 1 , T u ) = 0 .$
If $| I | = + ∞$, then there exists a sub-sequence ${ x n k }$ of ${ x n }$ such that
$d ( x n k , T u ) = 0 , k ∈ N .$
Therefore, we have
$lim k → + ∞ d ( T x n k , T u ) = lim k → + ∞ d ( x n k + 1 , T u ) = 0 .$
Then T is $α$-continuous, and it satisfies $T 1$. As a consequence, we have $T ∈ T α$. □
Remark 3.
Let $T : X → X$ be a mapping that belongs to the class of Berinde mappings (see [2]), that is, there exist $0 < q < 1$ and $ℓ ≥ 0$ such that
$d ( T x , T y ) ≤ q d ( x , y ) + ℓ d ( y , T x ) , ( x , y ) ∈ X × X .$
It can be easily seen that T is an almost F-contraction with $F ( t ) = ln t$, $t > 0$, and $( τ , L ) = ( − ln q , ℓ / q )$. Therefore, $T ∈ T α$, where α is given by (13) and $( ϕ , ψ ) = ( F , − ln q )$.
Now, we state and prove the main result of this section.
Theorem 4.
Let $( X , d )$ be a metric space, and let $T : X → X$ be a given mapping. Suppose that
(i)
There exists $α : X × X → R$ such that $( X , d )$ is α-complete.
(ii)
There exists $( φ , ψ ) ∈ Φ × Ψ$ such that $T ∈ T α$.
(iii)
(iv)
There exists some $x 0 ∈ X$ such that $α ( x 0 , T x 0 ) ≥ 1$.
Then there exists a sub-sequence ${ T n k x 0 }$ of ${ T n x 0 }$ that converges to a fixed point of T.
Proof.
Let ${ x n }$ be the Picard sequence defined by
$x n = T n x 0 , n ∈ N .$
Without loss of generality, we may suppose that
$d ( x n , x n + 1 ) > 0 , n ∈ N .$
From ($T 2$), we have
$α ( x n − 1 , x n ) exp φ ( d ( x n , x n + 1 ) ) ≤ exp φ ( d ( x n − 1 , x n ) ) + ψ ( d ( x n − 1 , x n ) ) , n ∈ N * .$
On the other hand, from $( i i i )$ and $( i v )$, we have
$α ( x n − 1 , x n ) ≥ 1 , n ∈ N * .$
Therefore, we obtain
$exp φ ( d ( x n , x n + 1 ) ) ≤ exp φ ( d ( x n − 1 , x n ) ) + ψ ( d ( x n − 1 , x n ) ) , n ∈ N * ,$
which yields
$φ ( d ( x n , x n + 1 ) ) ≤ φ ( d ( x n − 1 , x n ) ) + ψ ( d ( x n − 1 , x n ) ) , n ∈ N * .$
Next, following the same argument as in the proof of Theorem 3, we can prove that ${ x n }$ is a Cauchy sequence. Moreover, from (18), ${ x n }$ is $α$-Cauchy. Since $( X , d )$ is $α$-complete, there exists some $ω ∈ X$ such that
$lim n → + ∞ d ( x n , ω ) = 0 .$
From ($T 1$), there exists a sub-sequence ${ x n k }$ of ${ x n }$ such that
$lim k → + ∞ d ( x n k + 1 , T ω ) = 0 .$
The uniqueness of the limit yields $T ω = ω$, i.e., $ω$ is a fixed point of T. □
Remark 4.
From the proof of Theorem 4, it can be easily seen that if we replace ($T 1$) by the continuity of T, then the Picard sequence ${ T n x 0 }$ converges to a fixed point of T.
Next, we will show that most fixed point results from the literature involving F-contraction mappings follow easily from Theorem 4.
The following lemma will be used later.
Lemma 1.
Let $T : X → X$ be a given mapping. Let $α : X × X → R$ be the function defined by (13). Then T is α-admissible.
Proof.
Let $( x , y ) ∈ X × X$ be such that $α ( x , y ) ≥ 1$. By the definition of $α$, this means that $y = T x$. Then $T y = T 2 x$, which yields $α ( T x , T y ) = 1$. This proves that T is $α$-admissible. □
Corollary 1.
Theorem 4 ⇒ Theorem 3.
Proof.
Suppose that all the assumptions of Theorem 3 are satisfied. By Proposition 1, we know that $T ∈ T α$, where $α : X × X → R$ is given by (13). Since $( X , d )$ is complete, then it is $α$-complete. From Lemma 1, T is $α$-admissible. From the definition of $α$, we have $α ( x , T x ) = 1$, for all $x ∈ X$. Therefore, all the assumptions of Theorem 4 are satisfied. In particular $( i v )$ is satisfied for every $x ∈ X$. Taking in consideration Remark 4, we obtain that for any $x ∈ X$, the Picard sequence ${ T n x }$ converges to a fixed point of T. □
Corollary 2.
Theorem 4 ⇒ Theorem 1.
Proof.
It follows from Proposition 2, Lemma 1 and Remark 4. □
Corollary 3.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be an orbitally continuous mapping. Suppose that there exist $F ∈ Φ$ and a constant $τ > 0$ such that (15) is satisfied. Then, for any $x ∈ X$, there exists a sub-sequence ${ T n k x }$ of ${ T n x }$ such that ${ T n x }$ converges to a fixed point of T.
Proof.
It follows from Proposition 3, Lemma 1, and Theorem 4. □
Remark 5.
By Remark 4, if we replace the assumption T is orbitally continuous with T is continuous, then for any $x ∈ X$, the Picard sequence ${ T n x }$ converges to a fixed point of T. Such a result was established by Wardowski and Van Dung in [27].
Corollary 4.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be an orbitally continuous mapping. Suppose that there exist $F ∈ Φ$ and a constant $τ > 0$ such that (16) is satisfied. Then, for any $x ∈ X$, there exists a sub-sequence ${ T n k x }$ of ${ T n x }$ such that ${ T n x }$ converges to a fixed point of T.
Proof.
It follows from Proposition 4, Lemma 1, and Theorem 4. □
The next result was established by Minak et al. [22].
Corollary 5.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be an almost F-contraction, that is, there exit $F ∈ Φ$, $τ > 0$ and $L ≥ 0$ such that (17) is satisfied. Then, for any $x ∈ X$, there exists a sub-sequence ${ T n k x }$ of ${ T n x }$ such that ${ T n x }$ converges to a fixed point of T.
Proof.
It follows from Proposition 5, Lemma 1, and Theorem 4. □
Next, we will show that we can deduce easily from Theorem 4 several fixed point results in partially ordered metric spaces.
Corollary 6.
Let $( X , d )$ be a complete metric space, and let $T : X → X$ be continuous mapping. Suppose that X is partially ordered by a certain binary relation . Suppose that
(i)
T is non-decreasing with respect to , i.e.,
$T x ⪯ T y ,$
for all $( x , y ) ∈ X × X$ with $x ⪯ y$.
(ii)
There exists $x 0 ∈ X$ such that $x 0 ⪯ T x 0$.
(iii)
There exist $F ∈ Φ$ and $τ > 0$ such that
$τ + F ( d ( T x , T y ) ) ≤ F ( d ( x , y ) ) ,$
for all $( x , y ) ∈ X × X$ with $x ⪯ y$ and $d ( T x , T y ) > 0$.
Then ${ T n x 0 }$ converges to a fixed point of T.
Proof.
Let $α : X × X → R$ be the function defined by
$α ( x , y ) = 1 if x ⪯ y , 0 if x ⋠ y .$
From $( i )$ and the definition of $α$, it can be easily seen that T is $α$-admissible. Since T is continuous, it is $α$-continuous. Since $( X , d )$ is complete, it is $α$-complete. On the other hand, from $( i i i )$, we have
$exp F ( d ( T x , T y ) ) ≤ exp F ( d ( x , y ) ) − τ ,$
for all $( x , y ) ∈ X × X$ with $x ⪯ y$ and $d ( T x , T y ) > 0$. Let $( φ , ψ ) = ( F , − τ )$. Then $( φ , ψ ) ∈ Φ × Ψ$. Further, by the definition of $α$, for all $( x , y ) ∈ X × X$ with $( d ( T x , T y ) > 0$, we have
$α ( x , y ) exp φ ( d ( T x , T y ) ) ≤ exp φ ( d ( x , y ) ) + ψ ( d ( x , y ) ) .$
Therefore, $T ∈ T α$, where $α$ is given by (19). Note that by (ii), we have $α ( x 0 , T x 0 ) = 1$. Applying Theorem 4 and taking in consideration Remark 4, we obtain the desired result. □
Corollary 7
(Ran–Reurings fixed point theorem [13]). Let $( X , d )$ be a complete metric space, and let $T : X → X$ be continuous mapping. Suppose that X is partially ordered by a certain binary relation . Suppose that
(i)
T is non-decreasing with respect to .
(ii)
There exists $x 0 ∈ X$ such that $x 0 ⪯ T x 0$.
(iii)
There exists $0 < q < 1$ such that for all $( x , y ) ∈ X × X$ with $x ⪯ y$,
$d ( T x , T y ) ≤ q d ( x , y ) .$
Then ${ T n x 0 }$ converges to a fixed point of T.
Proof.
We have observe that T satisfies the condition $( i i i )$ of Corollary 6 with $F ( t ) = ln t$, $t > 0$, and $τ = − ln q$. Therefore, the result follows immediately from Corollary 6. □
Remark 6.
Note that several other fixed point results can be deduced from Theorem 4. For example, we mention the Banach fixed point theorem, the Berinde fixed point theorem [2], the Dass–Gupta fixed point theorem [7], the Chatterjea fixed point theorem [4], the Kannan fixed point theorem [11], the Reich fixed point theorem [14], the Hardy–Rogers fixed point theorem [8], etc.

## Author Contributions

All authors contributed equally to the writing of this paper.

## Funding

This work was supported by the International Scientific Partnership Program (ISPP) at King Saud University, under Grant ISPP No. 0027.

## Acknowledgments

The second and third authors extend their appreciation to the International Scientific Partnership Program ISPP at King Saud University for funding this research work through ISPP No. 0027.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Darwish, M.A.; Jleli, M.; O’Regan, D.; Samet, B. On the Study of Fixed Points for a New Class of α-Admissible Mappings. Mathematics 2019, 7, 1240. https://doi.org/10.3390/math7121240

AMA Style

Darwish MA, Jleli M, O’Regan D, Samet B. On the Study of Fixed Points for a New Class of α-Admissible Mappings. Mathematics. 2019; 7(12):1240. https://doi.org/10.3390/math7121240

Chicago/Turabian Style

Darwish, Mohamed Abdalla, Mohamed Jleli, Donal O’Regan, and Bessem Samet. 2019. "On the Study of Fixed Points for a New Class of α-Admissible Mappings" Mathematics 7, no. 12: 1240. https://doi.org/10.3390/math7121240

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