# Existence and Uniqueness of Fixed Points of Generalized F-Contraction Mappings

Authors: A. P. Farajzadeh; M. Delfani; Y. H. Wang
Journal: Journal of Mathematics
 (2021)

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## Abstract

The newest generalization of the Banach contraction through the notions of the generalized F-contraction, simulation function, and admissible function is introduced. The existence and uniqueness of fixed points for a self-mapping on complete metric spaces by the new constructed contraction are investigated. The results of this article can be viewed as an improvement of the main results given in the references.


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## Body

## 1. Introduction and Preliminaries

In 1922, Banach proved the following famous and fundamental result in fixed-point theory [1]. Let X,d be a complete metric space. Let T be a contractive mapping on X; that is, there exists q∈0,1 satisfying(1)dTx,Ty≤q.dx,y,∀x,y∈X.Then, there exists a unique fixed pointx0∈X of T. This theorem, which is called the Banach contraction principle that is a forceful tool in nonlinear analysis [9–14] and fixed-point theory, is a fascinating subject, with an enormous number of algorithms and applications in various fields of mathematics, see, e.g., [15–18]. This principle has been generalized in different directions by various researchers. One of them is the following theorem that is presented by Bryant.Theorem 1.
(see [2]). If f is a mapping of a complete metric space into itself and if, for some positive integer k, fk is a contraction, then f has a unique fixed point.
It is obvious thatfk is continuous but there are examples that show it cannot imply the continuity of f and so Theorem 1 is a real extension of the Banach principle.
In 1969, Sehgal [19] proved the following interesting generalization of Theorem 1.Theorem 2.
(see [19]). Let X,d be a complete metric space, q∈0,1, and T:X⟶X be a continuous mapping. If for each x∈X there exists a positive integer k=kx such that(2)dTkxx,Tkxy≤qdx,y,for all y∈X, then T has a unique fixed point u∈X. Moreover, for any x∈X, u=limn⟶∞Tnx.
Several researchers are interested to generalize Banach contraction. Here, we state two of them. Wardowski [8] generalized the Banach contraction as follows.Definition 1.
(see [8]). Let X,d be a metric space. The mapping T:X⟶X is called an F-contraction, if there exist F∈ℱ and τ>0 such that, for all x,y∈X,(3)dTx,Ty>0⇒τ+FdTx,Ty≤Fdx,y,where F:0,∞⟶ℝ is strictly increasing limn⟶∞Fαn=−∞ iff limn⟶∞αn=0 and there exists a number k∈0,1 such that limα⟶0+αkFα=−∞.
Notation. The family of all functions F:0,+∞⟶ℝ is denoted by ℱ (see [8]) if F satisfies the following conditions:(F1)F is strictly increasing(F2) for every sequenceαn in 0,+∞, we have limn⟶∞Fαn=−∞ iff limn⟶∞αn=0(F3) there exists a numberk∈0,1 such that limα⟶0+αkFα=−∞
The collection of all functionsF:0,+∞⟶ℝ is denoted by G ( [20]) if F satisfies the following conditions:(G1)F is strictly increasing(G2) there exists a sequenceαn in 0,+∞ such that limn⟶∞Fαn=−∞, or infF=−∞(G3)F is a continuous mapping
Another way to generalize the Banach contraction is through the following notion.Definition 2.
(see [3, 21]). Let ζ:0,∞×0,∞⟶ℝ be a mapping, then ζ is called a simulation function if it satisfies the following conditions:ζ1ζ0,0=0ζ2ζt,s<s−t for all t,s>0ζ3 if tn, sn are sequences in 0,∞ such that limn⟶∞tn=limn⟶∞sn>0 and tn<sn for all n∈ℕ, then(4)limsupn⟶∞ζtn,sn<0
We denote the set of all simulation functions byZ.
Ozturk [4], by using the simulation function and Wardowski [8] idea, extended Theorem 2 as follows.Theorem 3.
(see [4]). Let X,d be a complete metric space and T:X⟶X a mapping which satisfies the condition: If there exist f∈ℱ and τ>0 such that for each x∈X there is a positive integer nx such that for all y∈X,(5)dTnxx,Tnxy>0⇒ζFdx,y,τ+FdTnxx,Tnxy≥0,then T has a unique fixed point z∈X and Tnx0⟶z for each x0∈X, as n⟶∞.The first aim of this paper is to generalize Theorem2 by introducing a more general contraction type mapping through the notions of the generalized F-contraction, simulation function, and admissible function. Then, by the new constructed contraction and suitable conditions, the existence and uniqueness of fixed points are investigated.The following definitions and preliminary results are needed in the next section.Definition 3.
(see [6, 22]). Let α:X×X⟶0,+∞ be a given mapping. The mapping T:X⟶X is said to be an α-admissible, whenever αTx,Ty≥1 provided αx,y≥1 and x,y∈X.Definition 4.
(see [23]). An α-admissible map T is said to have the K-property, while for each sequence xn⊆X with αxn,xn+1≥1 for all n∈ℕ0, the nonnegative integer numbers, there exists a positive integer number k such that αTxn,Txm≥1, for all m>n≥k.Lemma 1.
(see [5]). Let F:0,+∞⟶ℝ be an increasing function and αn be a sequence of positive real numbers. Then, the following holds:(a)
iflimn⟶∞Fαn=−∞, then limn⟶∞αn=0(b)
ifinfF=−∞ and limn⟶∞αn=0, then limn⟶∞Fαn=−∞Lemma 2.
(see [24]). Let X,d be a metric space and xn be a sequence in X such that limn⟶∞dxn,xn+1=0. If xn is not a Cauchy sequence, then there exist ε>0 and two sequences of positive integers nk and mk with nk>mk>k such that dxmk,xnk>ε, dxmk,xnk−1<ε, and(1)
limk⟶∞dxmk,xnk=ε(2)
limk⟶∞dxmk−1,xnk=ε(3)
limk⟶∞dxmk,xnk+1=ε(4)
limk⟶∞dxmk−1,xnk+1=ε


## 2. Main Results

In this section, the main achievements of this article are presented. The existence and uniqueness of fixed points of the self-mappings on complete metric spaces satisfying the generalized F-contraction (relation (6) of the following theorem) with suitable assumptions are established by the first theorem. The second theorem can be viewed as a generalized version of Suzuki’s theorem given in [21]. Of course it ensures existence of fixed points for self-mappings under suitable hypothesis.Theorem 4.
LetX,d be a complete metric space and α:X×X⟶0,+∞ be a symmetric function, where αx,y≥1 and T:X⟶X be a continuous mapping which satisfies the condition: if there exist F∈ℱ, τ>0, L≥0, and simulation function ζ such that for all x∈X there is a positive integer nx such that for all y∈X and dTnxx,Tnxy>0,(6)ζτ+αx,yFdTnxx,Tnxy,Fmx,y+LN1x,y≥0,where(7)mx,y=maxdx,y,dx,Tnxx,dy,Tnxy,dx,Tnxy+dy,Tnxx2,N1x,y=mindx,Tnxx,dx,Tnxy,dy,Tnxx,then T has a unique fixed point.Proof.
We shall built a recursive sequencexk as follows: for the chosen arbitrary point x0∈X with n0=nx0, we set x1=Tn0x0 and inductively we get xi+1=Tnixi with ni=nxi.
We assert thatxi≠xi+1 for all i∈ℕ0. Suppose, on the contrary, there exists i0∈N0 such that xi0=xi0+1=Tni0xi0. Then, xi0 turns to be a fixed point of Tni0. On the other hand,(8)Txi0=TTni0xi0=Tni0Txi0.
Thus,Txi0 form a fixed point of Tni0. If Txi0=xi0, then we conclude that T has a fixed point and that terminate the proof. Suppose, on the contrary, that Txi0≠xi0 and hence dTni0Txi0,Tni0xi0>0. Then, by (6), we have(9)0≤τ+αxi0,Txi0FdTni0xi0,Tni0Txi0,Fmxi0,Txi0+LN1xi0,Txi0,≤Fmxi0,Txi0+LN1xi0,Txi0−τ+αxi0,Txi0FdTni0xi0,Tni0Txi0.
Hence,(10)τ+Fdxi0,Txi0=τ+FdTni0xi0,Tni0Txi0≤τ+αxi0,Txi0FdTni0xi0,Tni0Txi0≤Fmxi0,Txi0+LN1xi0,Txi0.
However,(11)mxi0,Txi0=maxdxi0,Txi0,dxi0,Tni0xi0,dTxi0,Tni0Txi0,dxi0,Tni0Txi0+dTxi0,Tni0xi02=dxi0,Txi0,N1=mindxi0,Tni0xi0,dxi0,Tni0Txi0,dTxi0,Tni0xi0=0.
Therefore,(12)τ+Fdxi0,Txi0≤Fdxi0,Txi0.
So,τ≤0, which is a contradiction. Consequently, we deduce that for all i∈ℕ0, xi≠xi+1. Then, dxi+1,xi>0, by (6),(13)τ+Fdxi+1,xi+2=τ+FdTnixi,Tnixi+1≤τ+αxi,xi+1FdTnixi,Tnixi+1≤Fmxi,xi+1+LN1xi,xi+1≤Fmxi,xi+1+Ldxi+1,xi+1=Fmxi,xi+1.
Then,(14)τ+Fdxi+1,xi+2≤Fmxi,xi+1.
However,(15)mxi,xi+1=maxdxi,xi+1,dxi,Tnixi,dxi+1,Tnixi+1,dxi,Tnixi+1+dxi+1,Tnixi2=maxdxi,xi+1,dxi+1,xi+2,dxi,xi+22≤maxdxi,xi+1,dxi+1,xi+2,dxi,xi+1+dxi+1,xi+22≤maxdxi,xi+1,dxi+1,xi+2.
Ifdxi0+1,xi0+2≥dxi0,xi0+1 for some i0∈ℕ0, then(16)mxi0,xi0+1≤dxi0+1,xi0+2,and since F is strictly increasing,(17)Fmxi0,xi0+1≤Fdxi0+1,xi0+2,so, it follows from (14) that(18)τ+Fdxi0+1,xi0+2≤Fdxi0+1,xi0+2.
So,τ≤0, which is a contradiction. Consequently,(19)dxi+1,xi+2<dxi,xi+1,∀i∈ℕ0.
Hence, from (14) and (19), we have(20)τ+Fdxi+1,xi+2≤Fdxi,xi+1or(21)Fdxi+1,xi+2≤Fdxi,xi+1−τ.
In general, one can get(22)Fdxi+1,xi+2≤Fdx0,x1−iτ.
Hence,(23)limi⟶∞Fdxi,xi+1=−∞.
So, fromF2, we have(24)limi⟶∞dxi,xi+1=0.
Therefore, with notice toF3, there exists k∈0,1 such that(25)limi⟶∞dxi,xi+1kFdxi,xi+1=0.
Now, (22) implies that(26)dxi,xi+1kFdxi,xi+1≤dxi,xi+1kFdx0,x1−iτ.Then, it can be easily seen that(27)limi⟶∞idxi,xi+1k=0.
So, there existsi0∈ℕ0 such that(28)dxi,xi+1≤1i1/k,∀i≥i0.
Consequently, ifm>n>n0, then(29)dxn,xm≤∑j=nmdxj,xj+1≤∑j=nm1j1/k≤∑j=n0∞1j1/k.
Sincek∈0,1, the series ∑j=n0∞1/j1/k is convergent. Therefore, xi is a Cauchy sequence, and since X is complete, there exists u∈X such that xi⟶u as i⟶∞.As a next step, we show that u is a fixed point of Tnu. Indeed, due to the continuity of T, we have(30)dTu,u=limi⟶∞dTxi,xi=limi⟶∞dxi+1,xi=0,and so u is a fixed point of T. For proving the uniqueness of the fixed point, let us consider u and v be two distinct fixed points and n=nu. So, we have du,v>0, and hence, we get that dTu,Tv>0; then, by (6) and ζ2,(31)0≤ζτ+αu,vFdTu,Tv,Fmu,v+LN1u,v≤Fmu,v+LN1u,v−τ+αu,vFdTu,Tv.
Therefore,(32)τ+αu,vFdTu,Tv≤Fmu,v+LN1u,v.
Hence, (32) implies that(33)τ+Fdu,v=τ+FdTu,Tv≤τ+αu,vFdTu,Tv≤Fmu,v+LN1u,v≤Fmu,v+Ldu,Tu=Fmu,v+0=Fmu,v,where(34)mu,v=maxdu,v,du,Tu,dv,Tv,du,Tv+dv,Tu2=maxdu,v,0,0,du,v+dv,u2=du,v.
So, we have(35)τ+Fdu,v≤Fdu,v,which is a contradiction, as τ>0. So, u=v.Corollary 1.
Theorem 3.3 of [7] of Theorem 4 by taking nx=1. Because in this case,(36)ζτ+αx,yFdTx,Ty,Fmx,y+LN1x,y≥0.
Now, byζ2, we have(37)0≤ζτ+αx,yFdTx,Ty,Fmx,y+LN1x,y≤Fmx,y+LN1x,y−τ+αx,yFdTx,Ty.
Therefore,(38)τ+αx,yFdTx,Ty≤Fmx,y+LN1x,y.Corollary 2.
Theorem3 is contained in Theorem 4 by taking mx,y=dx,y, αx,y=1, and L=0. Also, Theorem 4 is reduced to theorem [8] by setting nx=1.The following example shows that if the mappingT satisfies the condition of Theorem 4, it cannot guarantee in general the continuity of the mapping T.Example 1.
LetX=ℝ denote the real numbers with the usual metric d. Define function T:X⟶X by(39)Tx=1,x∈ℚ,0,x∈ℚc.
Then,T discontinues at each point of X, and T2=1. If α is an arbitrary element of 0,1, then(40)∀x∈X,∃nx=2;∀y∈X:dTnxx,Tnxy=0≤αdx,y.
Now, it is obvious that the functionζt,s=αs−t of condition (6) of Theorem 4 on 0,∞×0,∞ is a simulation function and T satisfies following condition:(41)ζdTnx,Tny,dx,y≥0,but T discontinues at each point of X. Moreover, T satisfies all the assumptions of Theorem 4, when L=0 and the unique fixed point of T is x=1 and Picard’s iteration of T; that is, if y∈X is an arbitrary point of X, then Tny is convergent to the fixed point.Theorem 5.
LetX,d be a complete metric space and α:X×X⟶0,+∞ a symmetric function, where αx,y≥1. Assume that T:X⟶X is a mapping in which there exist F∈G, τ>0, and the simulation function ζ such that for all x,y∈X with Tnxx≠Tnxy, where nx is a positive integer and 1/2dx,Tnxx≤dx,y implies(42)ζτ+αx,yFdTnxx,Tnxy,Fmx,y≥0,where mx,y is defined as in Theorem 4, satisfying the following conditions:(i)
T is α-admissible,(ii)
there existsx0∈X such that αx0,Tx0≥1,(iii)
ifxn is a sequence in X such that xn⟶x as n⟶∞ and αxn,xn+1≥1 for all n∈ℕ0, then αxn,x≥1 for all n∈ℕ0, and(iv)
T has the K-property,then T has a fixed point in X.Proof.
Letx0∈X be an arbitrary point. The recursive sequence xk is inductively constructed as follows: n0=nx0, and we set x1=Tn0x0 and inductively get xi+1=Tnixi with ni=nxi.
We assert thatxi≠xi+1 for all i∈ℕ0. Suppose, on the contrary, that there exists i0∈N0 such that xi0=xi0+1=Tni0xi0. Then, xi0 turns to be a fixed point of Tni0. On the other hand,(43)Txi0=TTni0xi0=Tni0Txi0.
Thus,Txi0 form a fixed point of Tni0. If Txi0=xi0, then we conclude that T has a fixed point and that terminate the proof. Suppose, on the contrary, that Txi0≠xi0 and hence dTni0Txi0,Tni0xi0>0. Then, by (42), we have(44)0≤τ+αxi0,Txi0FdTni0xi0,Tni0Txi0,Fmxi0,Txi0,≤Fmxi0,Txi0−τ+αxi0,Txi0FdTni0xi0,Tni0Txi0.
Hence,(45)τ+Fdxi0,Txi0=τ+FdTni0xi0,Tni0Txi0≤τ+αxi0,Txi0FdTni0xi0,Tni0Txi0≤Fmxi0,Txi0.
However,(46)mxi0,Txi0=maxdxi0,Txi0,dxi0,Tni0xi0,dTxi0,Tni0Txi0,dxi0,Tni0Txi0+dTxi0,Tni0xi02=dxi0,Txi0.
Therefore,(47)τ+Fdxi0,Txi0≤Fdxi0,Txi0.
So,τ≤0, which is a contradiction. Consequently, we deduce that, for all i∈ℕ0, xi≠xi+1. Then, dxi+1,xi>0, and so(48)12dxi,Tnixi=12dxi,xi+1≤dxi,xi+1.
Now, by (42),(49)τ+Fdxi+1,xi+2=τ+FdTnixi,Tnixi+1≤τ+αxi,xi+1FdTnixi,Tnixi+1≤Fmxi,xi+1.
Hence,(50)τ+Fdxi+1,xi+1≤Fmxi,xi+1.
However,(51)mxi,xi+1=maxdxi,xi+1,dxi,Tnixi,dxi+1,Tnixi+1,dxi,Tnixi+1+dxi+1,Tnixi2=maxdxi,xi+1,dxi+1,xi+2,dxi,xi+22≤maxdxi,xi+1,dxi+1,xi+2,dxi,xi+1+dxi+1,xi+22≤maxdxi,xi+1,dxi+1,xi+2.
Ifdxi0+1,xi0+2≥dxi0,xi0+1 for some i0∈ℕ0, then(52)mxi0,xi0+1≤dxi0+1,xi0+2,and since F is strictly increasing,(53)Fmxi0,xi0+1≤Fdxi0+1,xi0+2,so, it follows from (50) that(54)τ+Fdxi0+1,xi0+2≤Fdxi0+1,xi0+2.
Hence,τ≤0, which is a contradiction. Therefore,(55)dxi+1,xi+2<dxi,xi+1,∀i∈ℕ0.
Hence, from (50) and (55), we have(56)τ+Fdxi+1,xi+2≤Fdxi,xi+1or(57)Fdxi+1,xi+2≤Fdxi,xi+1−τ.
Consequently,(58)Fdxi+1,xi+2≤Fdx0,x1−iτ.
Hence,(59)limi⟶∞Fdxi,xi+1=−∞.
So, fromG2, we have(60)limi⟶∞dxi,xi+1=0.
Now, we claim thatxi is a Cauchy sequence. If it is not true, then by Lemma 2, there exist ε0>0 and two sequences of positive integers nk and mk with nk>mk>k such that dxmk,xnk>ϵ0, dxmk,xnk−1<ε0, andL1limk⟶∞dxnk,xmk=ε0L2limk⟶∞dxnk,xmk−1=ε0L3limk⟶∞dxnk+1,xmk=ε0L4limk⟶∞dxnk+1,xmk−1=ε0
Therefore, the definition ofmx,y implies(61)limk⟶∞mxnk,xmk−1=limk⟶∞maxdxnk,xmk−1,dxnk,Tnxxnk,dxmk−1,Tnxxmk−1,dxnk,Tnxxmk−1+dxmk−1,Tnxxnk2=limk⟶∞maxdxnk,xmk−1,dxnk,xnk+1,dxmk−1,xmk,dxnk,xmk+dxmk−1,xnk+12=maxε0,0,0,ε0+ε02=ε0.
So,(62)limk⟶∞mxnk,xmk−1=ε0.
On the other hand, sincelimk⟶∞dxnk,xmk−1=ε0>0 and limk⟶∞dxnk,xnk+1=0, with considering a subsequence if it is needed, one can suppose that there exists k1∈ℕ such that for any k>k1 and nk>mk>k,(63)dxnk,xnk+1≤dxnk,xmk−1.
So, it is obvious that, for allk>k1 and nk>mk>k,(64)12dxnk,Tnxxnk=12dxnk,xnk+1<dxnk,xmk−1.
Also, using the K-property, there existsk2∈ℕ such that(65)αxnk,xmk−1≥1,∀k>k2.
Ifk≥maxk1,k2, then it follows from (65) that(66)τ+FdTnkxnk,xmk−1≤τ+αxnk,xmk−1FdTnxxnk,Tnxxmk−1≤Fmxnk,xmk−1.
Lettingn⟶∞, the continuity of F through (L1) and (62) implies(67)τ+Fε0≤Fε0,which is contradicted by τ>0. Consequently, xi is a Cauchy sequence in the complete metric space X. Hence, there exists u∈X such that xi⟶u, as n⟶∞. To complete the proof, we show that u is a fixed point of T. We first claim, for all n≥0, that(68)12dxi,xi+1≤dxi,u,or12dxi+1,xi+2≤dxi+1,u.
In fact, if we assume that, for somei0≥0, both of them are false, then(69)12dxi0,xi0+1>dxi0,u,and12dxi0+1,xi0+2>dxi0+1,u.
Hence, (55) implies(70)dxi0,xi0+1≤dxi0,u+du,xi0+1<12dxi0,xi0+1+12dxi0+1,xi0+2≤12dxi0,xi0+1+12dxi0,xi0+1=dxi0,xi0+1,which is a contradiction and the claim is proved.
Now, let us begin with the first part of (68); that is, suppose that(71)12dxi,xi+1≤dxi,u,and on the contrary, assume that Tu≠u. Without loss of generality, one can imagine that Txi≠Tu, for all i∈ℕ0 (because if xi+1=Txi=Tu for infinite values of i, then uniqueness of the limit concludes that Tu=u). Then, from (45) and (iii), we get(72)τ+Fdxi+1,Tu=τ+FdTxi,Tu≤τ+αxi,uFdTxi,Tu≤Fmxi,u.
And sinceF is continuous on 0,+∞, and du,Tu>0, as i⟶∞, we get(73)τ+Fdu,Tu≤Flimi⟶∞mxi,u.
However,(74)mxi,u=maxdxi,u,dxi,xi+1,du,Tu,dxi,Tu+du,xi+12.
So, we have(75)limi⟶∞mxi,u=max0,0,du,Tu,du,Tu+02=du,Tu.
Therefore, ifdu,Tu≠0, then from (73), we have(76)τ+Fdu,Tu≤Fdu,Tu,which is contradicted, as τ>0. So, du,Tu=0, i.e., Tu=u. Finally, if we assume that the second part of (68) is true, i.e.,(77)12dxi+1,xi+2≤dxi+1,u,then by using the same manner, we can prove that du,Tu=0, i.e., Tu=u.
Suppose thatu and v are two fixed points of T. If u≠v, then dTu,Tv>0. Furthermore, αu,v≥1, because u,v∈FixT. It is also clear that 1/2du,Tu=0<du,v. Hence, (45) implies(78)τ+Fdu,v=τ+FdTu,Tv≤τ+αu,vFdTu,Tv≤Fmu,v,where(79)mu,v=maxdu,v,du,Tu,dv,Tv,du,Tv+dv,Tu2=maxdu,v,0,0,du,v+dv,u2=du,v.
So, we get(80)τ+Fdu,v≤Fdu,v,which is a contradicted by τ>0 and so u=v. This completes the proof.Corollary 3.
If in Theorem5, we put nx=1, then(81)ζτ+αx,yFdTx,Ty,Fmx,y≥0.
Now, byζ2, we have(82)0≤ζτ+αx,yFdTx,Ty,Fmx,y≤Fmx,y−τ+αx,yFdTx,Ty.
Therefore,(83)τ+αx,yFdTx,Ty≤Fmx,y.
Hence, we get Theorem 3.3 of [7].



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