# Some Fixed-Point Results via Mix-Type Contractive Condition

Authors: Özlem Acar
Journal: Journal of Function Spaces
 (2021)

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## Abstract

We consider a fixed-point problem for mappings involving a rational type and almost type contraction on complete metric spaces. To do this, we are usingF-contraction and H,φ-contraction. We also present an example to illustrate our result.


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## Body

## 1. Introduction

The beginning of metrical fixed point theory is related to Banach’s Contraction Principle, presented in 1922 [1], which says that any contraction self-map on M has a unique fixed point whenever M,d is complete. Afterwards, the crucial role of the principle in existence and uniqueness problems arising in mathematics has been realized which fact directed the researchers to extend and generalize the principle in many ways (see [2–7]).In the studies of generalizations and modifications of contractions, an interesting generalization was given by Wardowski [8] using a new concept F-contraction. Then, many authors gave some results using this concept in different type metric spaces. One of them is given by Jleli et al. [9] by introducing a family H of functions H:0,∞3→0,∞ with the certain assumption. Also, you can find this type generalizations in [10–12].In this paper, we consider a fixed-point problem for mappings involving a rational type contraction and almost contraction. Firstly, we recall some basic on the notions ofF-contraction and H,ϕ-contraction.


## 2. Preliminaries

LetF be the family of all functions F:ℝ+=0,∞→ℝ satisfying the following conditions:(F1) F is nondecreasing;(F2) for every sequence αn of positive numbers limn→+∞αn=0 if and only if limn→+∞Fαn=−∞;(F3) there exists k∈0,1 such that limα→0+αkFα=0. ([8])Definition 1. (see [8]).
LetM,d be a metric space and Y:M→M be a mapping. Given F∈F, we say that Y is F-contraction, if there exists τ>0 such that
(1)μ,γ∈M,dYμ,Yγ>0⇒τ+FdYμ,Yγ≤Fdμ,γ.
Taking in (1) different functions F∈F, one gets a variety of F-contractions, and some of them being already known in the literature. You can see this contractions in [8]. In addition, Wardowski concluded that every F-contraction Y is a contractive mapping, i.e.,
(2)dYμ,Yγ<dμ,γ,forallμ,γ∈M,Yμ≠Yγ.
Thus, everyF-contraction is a continuous mapping.Theorem 2. (see [8]).
LetM,d be a complete metric space (C.M.S) and let Y:M→M be an F-contraction. Then, Y has a unique fixed point in M.
In [9], Jleli et al. introduced a family H of functions H:0,+∞3→0,+∞ satisfying the following conditions:
(H1) maxα,β≤Hα,β,γ for all α,β,γ∈0,+∞;
(H2) H0,0,0=0;
(H3) H is continuous.
Some examples of functions belonging toH are given as follows:
(i)
Hα,β,γ=α+β+γ for all α,β,γ∈0,+∞(ii)
Hα,β,γ=maxα,β+γ for all α,β,γ∈0,+∞(iii)
Hα,β,γ=α+β+αβ+γ for all α,β,γ∈0,+∞
Using a functionH∈H, the authors of [9] introduced the following notion of H,ϕ-contraction.Definition 3. (see [9]).
LetM,d be a metric space, ϕ:M→0,+∞ be a given function, and H∈H. Then, Y:M→M is called a H,ϕ-contraction with respect to the metric d if and only if
(3)HdYμ,Yγ,ϕYμ,ϕYγ≤kHdμ,γ,ϕμ,ϕγforallμ,γ∈M,for some constant k∈0,1.Now, we set(4)Zϕ≔μ∈M:ϕμ=0,FY≔μ∈M:Yμ=μ.Furthermore, we say thatY is a ϕ-Picard operator if and only if the following condition holds
(5)FY∩Zϕ=ςandYnμ→ς,asn→+∞,foreachμ∈M.Theorem 4. (see [9]).
LetM,d be a C.M.S, ϕ:M→0,+∞ be a given function and H∈H. Suppose that the following conditions hold
(A1) ϕ is lower semicontinuous (l.s.c.);
(A2) Y:M→M is a H,ϕ-contraction with respect to the metric d.
Then,(6)FY⊂Zϕ;(i)
Y is a ϕ-Picard operator(ii)
For allμ∈M and for all n∈ℕ, we have(7)dYnμ,ς≤kn1−kHdYμ,μ,ϕYμ,ϕμ,where ς=FY∩Zϕ=FY.
Recently, Vetro ([13]) generalized Theorem 4 by using F-H-contraction.Definition 5. (see [13]).
LetM,d be a metric space and let Y:M→M be a mapping. The mapping Y is called an F-H-contraction if there exists F∈F, H∈H, a real number, τ>0 and ϕ:M→0,+∞ s.t.
(8)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHdμ,γ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0.
We remark that everyF-contraction is an F-H-contraction such that H∈H defined by Hx,y,z=x+y+z for all x,y,z∈0,+∞ and ϕ:M→0,+∞ defined by ϕμ=0 for all μ∈M.Lemma 6. (see [13]).
LetM,d be a metric space and let Y:M→M be an F-H-contraction with respect to the functions F∈F, H∈H, ϕ:M→0,+∞, and the real number τ>0. If μn is a sequence of Picard starting at μ0∈M, then
(9)limn→+∞Hdμn−1,μn,ϕμn−1,ϕμn=0,and hence
(10)limn→+∞dμn−1,μn=0andlimn→+∞ϕμn=0.Theorem 7. (see [13]).
LetM,d be a C.M.S and Y:M→M be an F-H-contraction with respect to the functions F∈F, H∈H, the real number τ>0, and a l.s.c. function ϕ:M→0,+∞ such that (8) holds; that is,
(11)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHdμ,γ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0. Then, Y has a unique fixed point ς such that ϕς=0.Theorem 8. (see [13]).
LetM,d be a C.M.S and let Y:M→M be a mapping. Assume that there exists a continuous function F that satisfies the conditions (F1) and (F2), a function H∈H, a real number τ>0, and a l.s.c. function ϕ:M→0,+∞ such that (8) holds; that is,
(12)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHdμ,γ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0. Then, Y has a unique fixed point ς such that ϕς=0.


## 3. Main Results

We first introduce the rational typeF-H-contraction.Definition 9.
LetM,d be a metric space and Y:M→M be a mapping. Y is called a rational type F-H-contraction if there exists F∈F, H∈H, a real number τ>0, and ϕ:M→0,+∞s.t.(13)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHMμ,γ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0 where
(14)Mμ,γ=maxdμ,γ,dμ,Yμ1+dγ,Yγ1+dYμ,Yγ.Lemma 10.
LetM,d be a metric space and Y:M→M be a rational type F-H-contraction with respect to the functions F∈F, H∈H, ϕ:M→0,+∞, and the real number τ>0. If μn is a sequence of Picard starting at μ0∈M, then
(15)limn→+∞Hdμn−1,μn,ϕμn−1,ϕμn=0,and hence
(16)limn→+∞dμn−1,μn=0andlimn→+∞ϕμn=0.Proof.
By replacing the contradiction in [[13], (29)] with contradiction (13) and following the proof of [[13], Lemma 1], we immediately have the desired result.Theorem 11.
LetM,d be a C.M.S and let Y:M→M be an rational type F-H-contraction with respect to the functions F∈F, H∈H, the real number τ>0, and a l.s.c. function ϕ:M→0,+∞ such that (13) holds for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0. Then, Y has a unique fixed point ς such that ϕς=0.Proof.
First, we shall proof the uniqueness. Arguing by contradiction, we assume that there existς,w∈M such that ς=Yς, w=Yw, and ς≠w. The hypothesis ς≠w ensures, by the property (H1) of the function H, that
(17)HdYς,Yw,ϕYς,ϕYw≥dYς,Yw=dς,w>0.
Using (13) with μ=ς and γ=w, we obtain
(18)τ+FHdYς,Yw,ϕYς,ϕYw=τ+FHdς,w,ϕς,ϕw≤FHMς,w,ϕς,ϕw≤FHmaxdς,w,dς,Yς1+dw,Yw1+dYς,Yw,ϕς,ϕw≤FHmaxdς,w,dς,ς1+dw,w1+dς,w,ϕς,ϕw≤FHdς,w,ϕς,ϕw,which is a contradiction. So, we have w=ς, and the fixed point is unique.
Now, we can show the existence of a fixed point. Take a pointμ0∈M and create the μn sequence starting at μ0. We emphasize that if μk−1=μk for some k∈ℕ, then ς=μk−1=μk=Yμk−1=Yς; that is, ς is a fixed point of Y such that ϕς=0. In fact, by Lemma 10, Hdμk−1,μk,ϕμk−1,ϕμk=0 and by the property (H1) of the function H, we have ϕς=0. So, we can suppose that μn−1≠μn for every n∈ℕ.
In this step, we show thatμn is a Cauchy. By Lemma 10, we say that
(19)0<hn−1=Hdμn−1,μn,ϕμn−1,ϕμn→0asn→+∞.
There existsk∈0,1 such that hnkFhn→0 as n→+∞ by he property (F3) of F. Using (13) with μ=μn−1 and γ=μn, we get
(20)FHdμn,μn+1,ϕμn,ϕμn+1≤FHMμn−1,μn,ϕμn−1,ϕμn−τ≤FHmaxdμn−1,μn,dμn−1,Yμn−11+dμn,Yμn1+dYμn−1,Yμn,ϕμn−1,ϕμn−τ≤FHdμn−1,μn,ϕμn−1,ϕμn−τ≤FHdμ0,μ1,ϕμ0,ϕμ1−nτ,for all n∈ℕ; that is,
(21)Fhn≤Fhn−1−τ≤⋯≤Fh0−nτforalln∈ℕ.
From(22)0=limn→+∞hnkFhn≤limn→+∞hnkFh0−nτ≤0,we deduce that
(23)limn→+∞hnkn=0.
This provides that∑n=1+∞hn is convergent. By the property (H1) of the function H, also, the series ∑n=1+∞dμn,μn+1 is convergent and hence μn is a Cauchy sequence. Now, since M,d is complete, there exists ς∈M such that
(24)limn→+∞μn=ς.
By (13), taking into account that ϕ is a l.s.c. function, we have
(25)0≤ϕς≤liminfn→+∞ϕμn=0;that is, ϕς=0. Now, show that ς is a fixed point. If there exists a subsequence μnk of μn such that μnk=ς or Yμnk=Yς, for all k∈ℕ, then ς is a fixed point. Otherwise, we can assume that μn≠ς and Yμn≠Yς for all n∈ℕ. So, using (13) with μ=μn and γ=ς, we deduce that
(26)τ+FHdYμn,Yς,ϕYμn,ϕYς≤FHMμn,ς,ϕμn,ϕς.
Sinceτ>0, we obtain
(27)HdYμn,Yς,ϕYμn,ϕYς<HMμn,ς,ϕμn,ϕςforalln∈ℕ,and so
(28)dς,Yς≤dς,μn+1+dYμn,Yς≤dς,μn+1+HdYμn,Yς,ϕYμn,ϕYς<dς,μn+1+HMμn,ς,ϕμn,ϕς<dς,μn+1+Hmaxdμn,ς,dμn,Yμn1+dς,Yς1+dYμn,Yς,ϕμn,ϕς≤dς,μn+1+Hmaxdμn,ς,dμn,μn+11+dς,Yς1+dμn+1,Yς,ϕμn,ϕς,for all n∈ℕ.
Finally, lettingn→+∞ in the above calculations and using that H is continuous in 0,0,0, we deduce that dς,Yς≤H0,0,0=0; that is, ς=Yς.Imposing thatF is a continuous function and relaxing the hypothesis F3, we can give t Theorem 12.Theorem 12.
LetM,d be a C.M.S and Y:M→M be a mapping. Assume that there exists a continuous function F that satisfies the conditions (F1) and (F2), a function H∈H, a real number τ>0, and a l.s.c. function ϕ:M→0,+∞ s.t.
(29)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHMμ,γ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0. Then, Y has a unique fixed point ς such that ϕς=0.Proof.
Following the similar arguments as in the proof of Theorem11, we obtain easily the uniqueness of the fixed point. The existence of a fixed point, we take a point μ0∈M and create the μn sequence starting at μ0. Clearly, if μk−1=μk for some k∈ℕ, then ς=μk−1=μk=Yμk−1=Yς; that is, ς is a fixed point of Y such that ϕς=0 (see the proof of Theorem 11), and so we have already done.
So, we can suppose thatμn−1≠μn for every n∈ℕ. Now, showing that μn is a Cauchy. Let us admit the opposite. Then, there exists a positive real number ε and two sequences mk and nk such that
(30)nk>mk≥kanddμmk,μnk≥ε>dμmk,μnk−1forallk∈ℕ.
By Lemma10, we say that dμn−1,μn→0, ϕμn→0, as n→+∞. This implies
(31)limk→+∞dμmk,μnk=limk→+∞dμmk−1,μnk−1=ε.
Now, the hypothesis thatdμmk,μnk>ε ensures that
(32)Hdμmk,μnk,ϕμmk,ϕμnk>0forallk∈ℕ.
Using the continuity ofH, we have
(33)limk→+∞Hdμmk−1,μnk−1,ϕμmk−1,ϕμnk−1=limk→+∞Hdμmk,μnk,ϕμmk,ϕμnk=Hε,0,0>0.
Using again (29), with μ=μmk−1 and γ=μnk−1, we get
(34)τ+FHdμmk,μnk,ϕμmk,ϕμnk≤FHMμmk−1,μnk−1,ϕμmk−1,ϕμnk−1≤FHmaxdμmk−1,μnk−1,dμmk−1,Yμmk−11+dμnk−1,Yμnk−11+dYμmk−1,Yμnk−1,ϕμmk−1,ϕμnk−1≤FHmaxdμmk−1,μnk−1,dμmk−1,μmk1+dμnk−1,μnk1+dμmk,μnk,ϕμmk−1,ϕμnk−1,for all k∈ℕ. Letting k→+∞ in the previous inequality, since the function F is continuous, we get
(35)τ+FHε,0,0≤FHε,0,0,which leads to contradiction. It follows that μn is a Cauchy sequence.
Now, sinceM,d is complete, there exists some ς∈M such that
(36)limn→+∞μn=ς.
By (29), using lower semicontinuity of ϕ, we get
(37)0≤ϕς≤liminfn→+∞ϕμn=0;that is, ϕς=0. Now, show that ς is a fixed point of Y. Clearly, ς is a fixed point of Y if there exists a subsequence μnk of μn such that μnk=ς or Yμnk=Yς, for all k∈ℕ. Otherwise, we can assume that μn≠ς and Yμn≠Yς for all n∈ℕ. Then, the property (H1) of the function H ensures that HdYμn,Yς,ϕYμn,ϕYς>0 for all n∈ℕ. So, using (29) with μ=μn and γ=ς, we deduce that
(38)τ+FHdYμn,Yς,ϕYμn,ϕYς≤FHMμn,ς,ϕμn,ϕς≤FHmaxdμn,ς,dμn,Yμn1+dς,Yς1+dYμn,Yς,ϕμn,ϕςforalln∈ℕ.
Sinceτ>0, we conclude that
(39)HdYμn,Yς,ϕYμn,ϕYς<FHmaxdμn,ς,dμn,Yμn1+dς,Yς1+dYμn,Yς,ϕμn,ϕςforalln∈ℕ,and so
(40)dς,Yς≤dς,μn+1+dYμn,Yς≤dς,μn+1+HdYμn,Yς,ϕYμn,ϕYς<dς,μn+1+Hmaxdμn,ς,dμn,Yμn1+dς,Yς1+dYμn,Yς,ϕμn,ϕς=dς,μn+1+Hmaxdμn,ς,dμn,μn+11+dς,Yς1+dμn+1,Yς,ϕμn,ϕς,for all n∈ℕ. Finally, letting n→+∞ and using that H is continuous in 0,0,0, we deduce that dς,Yς≤H0,0,0=0; that is, ς=Yς.Definition 13.
LetM,d be a metric space and let Y:M→M be a mapping. The mapping Y is called almost F-H-contraction if there exists a function F∈F, H∈H, a real number τ>0, and L≥0 and a l.s.c. function ϕ:M→0,+∞ such that
(41)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHdμ,γ+Ldγ,Yμ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0.Theorem 14.
LetM,d be a C.M.S and let Y:M→M be an almost F-H-contraction with respect to the functions F∈F, H∈H, the real number τ>0, and L≥0 and a l.s.c. function ϕ:M→0,+∞ s.t.
(42)τ+FHdYμ,Yγ,ϕYμ,ϕYγ≤FHdμ,γ+Ldγ,Yμ,ϕμ,ϕγ,for all μ,γ∈M with HdYμ,Yγ,ϕYμ,ϕYγ>0. Then, Y has a fixed point ς such that ϕς=0.Proof.
The existence of a fixed point we take a pointμ0∈M and create the μn sequence starting at μ0. We stress that if μk−1=μk for some k∈ℕ, then ς=μk−1=μk=Yμk−1=Yς; that is, ς is a fixed point of Y such that ϕς=0. In fact, by Lemma 10, Hdμk−1,μk,ϕμk−1,ϕμk=0 and by the property (H1) of the function H, we have ϕς=0. So, we can suppose that μn−1≠μn for every n∈ℕ.
Now, showing thatμn is a Cauchy. By Lemma 10, we say that
(43)0<hn−1=Hdμn−1,μn,ϕμn−1,ϕμn→0asn→+∞.
The property (F3) of the function F ensures that there exists k∈0,1 such that hnkFhn→0 as n→+∞. Using (42), with μ=μn−1 and γ=μn, we get
(44)FHdμn,μn+1,ϕμn,ϕμn+1≤FHdμn−1,μn+Ldμn,Yμn−1,ϕμn−1,ϕμn−τ≤FHdμn−1,μn,ϕμ0,ϕμ1−τ≤FHdμ0,μ1,ϕμ0,ϕμ1−nτ,for ∀n∈ℕ; that is,
(45)Fhn≤Fhn−1−τ≤⋯≤Fh0−nτforalln∈ℕ.
From(46)0=limn→+∞hnkFhn≤limn→+∞hnkFh0−nτ≤0,we deduce that
(47)limn→+∞hnkn=0.
This ensures that the series∑n=1+∞hn is convergent. By the property (H1) of the function H, also, the series ∑n=1+∞dμn,μn+1 is convergent, and hence μn is a Cauchy sequence. Now, since M,d is complete, there exists some ς∈M such that
(48)limn→+∞μn=ς.
By (42), using lower semicontinuity of ϕ, we get
(49)0≤ϕς≤liminfn→+∞ϕμn=0;that is, ϕς=0. We assert that ς is a fixed point of Y. Clearly, ς is a fixed point of Y if there exists a subsequence μnk of μn such that μnk=ς or Yμnk=Yς, for all k∈ℕ. Otherwise, we can assume that μn≠ς and Yμn≠Yς for all n∈ℕ. So, using (42) with μ=μn and γ=ς, we deduce that
(50)τ+FHdYμn,Yς,ϕYμn,ϕYς≤FHdμn,ς+Ldς,Yμn,ϕμn,ϕς.
Sinceτ>0, this inequality leads to
(51)HdYμn,Yς,ϕYμn,ϕYς<Hdμn,ς+Ldς,Yμn,ϕμn,ϕςforalln∈ℕ,and so
(52)dς,Yς≤dς,μn+1+dYμn,Yς≤dς,μn+1+HdYμn,Yς,ϕYμn,ϕYς<dς,μn+1+Hdμn,ς+Ldς,Yμn,ϕμn,ϕς,for all n∈ℕ.
Finally, lettingn→+∞ in the above calculations and using that H is continuous in 0,0,0, we deduce that dς,Yς≤H0,0,0=0; that is, ς=Yς.Example 15.
LetM=0,1 endowed with the standart metric dμ,γ=μ−γ for all μ,γ∈M. Consider the mapping Y:M→M defined by
(53)Yμ=μ/2;μ∈0,11;μ=1.
Clearly,Y is not a F−contraction but Y is an almost F-H-contraction with respect to the functions F∈F defined by Fα=lnα for all α>0,H∈H defined by Ha,b,c=maxa,b+c for all a,b,c∈0,+∞, the real number τ=ln2 and L=4, and a l.s.c. function ϕ:M→0,+∞,ϕt=t for all t∈M, indeed.Case 1.
μ=0,γ=1, we have
(54)τ+FHdYμ,Yγ,ϕYμ,ϕYγ=τ+FHdY0,Y1,ϕY0,ϕY1=τ+FHd0,1,ϕ0,ϕ1=τ+FH1,0,1=ln4≤ln6=FHd0,1+4d1,Y0,ϕ0,ϕ1=FHdμ,γ+Ldγ,Yμ,ϕμ,ϕγ.Case 2.
μ=1,γ=0, we have
(55)τ+FHdYμ,Yγ,ϕYμ,ϕYγ=τ+FHdY1,Y0,ϕY1,ϕY0=τ+FHd1,0,ϕ1,ϕ0=τ+FH1,1,0=ln2≤ln5=FHd1,0+4d0,Y1,ϕ1,ϕ0=FHdμ,γ+Ldγ,Yμ,ϕμ,ϕγ.Case 3.
μ,γ∈0,1 with μ>γ, we have
(56)τ+FHdYμ,Yγ,ϕYμ,ϕYγ=τ+FHdμ2,γ2,ϕμ2,ϕγ2=τ+FHμ−γ2,μ2,γ2=τ+Fμ+γ2=lnμ+γ≤maxlnμ+γ,ln4γ−μ=FHdμ,γ+Ldγ,Yμ,ϕμ,ϕγ.



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