# Admissible Almost TypeZ-Contractions and Fixed Point Results

Authors: Said Melliani; Abdelhamid Moussaoui; L. S. Chadli
Journal: International Journal of Mathematics and Mathematical Sciences
 (2020)

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## Abstract

In this paper, we introduce a new concept ofα-admissible almost type Z-contraction and prove some fixed point results for this new class of contractions in the context of complete metric spaces. The presented results generalize and unify several existing results in the literature.


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## Body

## 1. Introduction and Preliminaries

Fixed point theory is one of the fundamental research areas in nonlinear functional analysis, and it plays a powerful role in the resolution of several mathematical problems with potential applicability in various fields such economics and game theory (see, e.g., [1, 2]) and computer sciences (see, e.g., [3, 4]).The well-known Banach contraction principle (BCP) is the most impressive theoretical development in the evolution of the metric fixed point theory, and this principle has been extended, improved, and generalized in different approaches (see [5–18]). In particular, in 2015, Khojasteh et al. [5] coined the concept of simulation functions and defined a new class of nonlinear contractions, namely, Z-contractions, which generalize the Banach contraction principle and unify several known types of contractions. For other results related to this interesting approach, refer [8, 17–21].In [6], Samet et al. introduced the notion of α-ψ-contractive-type mappings by defining the concept of α-admissibility and using Bianchini–Grandolfi gauge functions, and the authors inspected for the existence and the uniqueness of fixed points for such mappings. Later on, Karapinar and Samet [7] generalized the results derived in [6] by proposing the concept of generalized α-ψ-contractive type. In 2016, Karapinar [8] originated the concept of α-admissible Z-contraction by combining the ideas in [5, 6] to obtain some interesting fixed point results in the context of complete metric spaces. In fact, Karapinar [8] proved, among other results, that several existing fixed point theorems can be expressed easily from the main results of [8].In the present paper, we introduce a new concept ofα-admissible almost type Z-contraction with respect to a simulation function ζ, and we prove some results about existence and uniqueness of fixed points for such mappings; the presented results unify several well-known types of contraction and generalize several existing results in the literature.LetΨ be the family of nondecreasing functions ψ:ℝ+⟶ℝ+ satisfying the following condition.B There exist n0∈ℕ and α∈0,1 and a convergent series of nonnegative terms ∑n=1∞vn such that(1)ψn+1t≤αψnt+vn,forn≥n0and all at∈ℝ+.The class of such functions is called as Bianchini–Grandolfi gauge functions [10, 22] or c-comparison functions [11].Lemma 1 (see [11]).
Ifψ∈Ψ, then the following holds:(i)
ψntn∈ℕ converges to 0 as n⟶∞ for all t∈ℝ+.(ii)
ψt<t, for any t∈ℝ+.(iii)
ψ is continuous at 0.(iv)
The series∑k=1∞ψkt converges for all t∈ℝ+.In 2012, Karapinar and Samet [7] introduced a new class of contractive mappings via the following concept of α-admissibility presented in [6].Definition 1 (see [6]).
LetT:X⟶X be a self-mapping and α:X×X⟶0,∞ be a function. T is said to be α-admissible if(2)αx,y≥1⟹αTx,Ty≥1,for allx,y∈X.Example 1 (see [6]).
LetX=0,∞, and we define the mappings T:X⟶X and α:X×X⟶0,∞ as follows. Tx=x, for allx∈X, and(3)αx,y=ex−y,ifx≥y,0,ifx<y.
Then,T is α-admissible.Definition 2 (see [7].
LetX,d be a metric space and T:X⟶X be a given mapping. We say that T is a generalized α-ψ-contractive mapping if there exist two functions α:X×X⟶0,∞ and ψ∈Ψ such that(4)αx,ydTx,Ty≤ψMx,y,for allx,y∈X,where Mx,y=maxdx,y,dx,Tx+dy,Ty/2,dx,Ty+dy,Tx/2.
In 2014, Popescu [12] suggested the concept of triangular α-orbital admissible as an improvement of the triangular α-admissible notion proposed in [9].Definition 3 (see [12]).
LetT:X⟶X be a mapping and α:X×X⟶0,∞ be a function. Then, T is said to be α-orbital admissible if(5)αx,Tx≥1⟶αTx,T2x≥1.
Moreover,T is called a triangular α-orbital admissible if it satisfies the following conditions:(i)
T1: T is α-orbital admissible.(ii)
T2: αx,y≥1 and αy,Ty≥1⟹αx,Ty≥1.
On the contrary, Khojasteh et al. [5] defined a new family of contractions by using the following notion of simulation functions.Definition 4 (see [5]).
The functionζ:0,∞×0,∞⟶ℝ is said to be a simulation function, if it satisfies the following conditions:(i)
ζ1: ζ0,0=0.(ii)
ζ2: ζt,s<s−t for all t,s>0.(iii)
ζ3: if tn,sn are sequences in 0,∞ such that limn⟶∞tn=limn⟶∞sn>0, then limn⟶∞supζtn,sn<0.
The set of all simulation functions is denoted byZ.Example 2 (see [5]).
Letζi:0,∞×0,∞⟶ℝ,i=1,2,3 be defined as follows:(1)
ζ1t,s=ϕ˜s−ϕt for all t,s∈0,∞, where ϕ˜,ϕ:0,∞⟶0,∞ are two continuous functions such that ϕ˜t=ϕt=0 if and only if t=0 and ϕ˜t<t≤ϕt for all t>0.(2)
ζ2t,s=s−φs−t for all t,s∈0,∞, where φ:0,∞⟶0,∞ is a continuous function such that φt=0 if and only if t=0.(3)
ζ3t,s=s−ft,s/gt,st for all t,s∈0,∞, where f,g:0,∞⟶0,∞ are continuous functions with respect to each variable such that ft,s>gt,s for all t,s>0.(4)
ζi for i=1,2,3 are simulation functions. For other interesting examples of simulation functions, readers are referred to [5, 19, 20].Definition 5 (see [5]).
LetX,d be a metric space, T:X⟶X a self-mapping, and ζ∈Z. We say that T is a Z-contraction with respect to ζ, if the following condition is satisfied:(6)ζdTx,Ty,dx,y≥0,for allx,y∈X.
Now, we state the result proved in [5] as follows.Theorem 1 (see [5]).
LetX,d be a complete metric space and T:X⟶X be a Z-contraction with respect to a simulation function ζ∈Z. Then, T has unique fixed point in X, and for every x0∈X, the Picard sequence xn, where xn=Txn−1 for all n∈ℕ, converges to the fixed point of T.


## 2. Main Results

First, we present the following concept.Definition 6.
LetX,d be a metric space and ζ∈Z. We say that T:X⟶X is an α-admissible almost Z-contraction if there exists α:X×X⟶0,∞ and a constant L≥0 such that(7)ζαx,ydTx,Ty,Mx,y+LNx,y≥0,for allx,y∈X,where(8)Nx,y=mindx,Tx,dy,Ty,dx,Ty,dy,Tx,Mx,y=maxdx,y,dx,Tx+dy,Ty2,dx,Ty+dy,Tx2.Remark 1.
IfT is an α-admissible almost Z-contraction with respect to ζ∈Z, then(9)αx,ydTx,Ty<Mx,y+LNx,y,for allx,y∈X.
Our first result is the following theorem.Theorem 2.
LetX,d be a complete metric space, and let T:X⟶X be an α-admissible almost Z-contraction with respect to ζ∈Z and satisfying the following conditions:(i)
T is triangular α-orbital admissible.(ii)
There existsx0∈X such that αx0,Tx0≥1.(iii)
T is continuous.
Then, there existsz∈X such that Tz=z.Proof. Using condition (ii), there existsx0∈X such that αx0,Tx0≥1, and let xn be the iterative sequence in X defined by(10)xn+1=Txn,for alln∈ℕ.Ifxm+1=Txm for some m∈ℕ, then xm is a fixed point of T. Therefore, to continue our proof, we suppose that xn+1≠xn for all n∈ℕ. Since T is an α-admissible mapping, we have(11)αx0,x1=αx0,Tx0≥1,implies thatαTx0,Tx1=αx1,x2≥1.By induction, we obtain that(12)αxn,xn+1≥1,for alln∈ℕ.Applying the condition (7) with x=xn and y=xn−1 and using (12), we get(13)0≤ζαxn,xn−1dTxn,Txn−1,Mxn,xn−1+LNxn,xn−1=ζαxn,xn−1dxn+1,xn,Mxn,xn−1+LNxn,xn−1<Mxn,xn−1+LNxn,xn−1−αxn,xn−1dxn+1,xn,where(14)Nxn,xn−1=mindxn,Txn,dxn−1,Txn−1,dxn−1,Txn−1,dxn−1,Txn−1=mindxn,xn+1,dxn−1,Txn−1,dxn,xn,dxn−1,xn+1=0,(15)Mxn−1,xn=maxdxn,xn−1,dxn,Txn+dxn−1,Txn−12,dxn,xn−1+dxn−1,Txn2=maxdxn,xn−1,dxn,xn+1+dxn−1,xn2,dxn−1,xn+12≤maxdxn,xn−1,dxn,xn+1+dxn−1,xn2≤maxdxn,xn−1,dxn,xn+1.By (13) and taking in account (12), (14), and (15), we derive that(16)dxn,xn+1≤αxn,xn−1dxn,xn+1<maxdxn,xn−1,dxn,xn+1for all n≥1. Now, if maxdxn,xn−1,dxn,xn+1=dxn,xn+1 for some n≥1, then from the above inequality, we get(17)dxn,xn+1≤αxn,xn−1dxn,xn+1<dxn,xn+1,which is a contradiction. Therefore,(18)maxdxn−1,xn,dxn,xn+1=dxn,xn−1,for alln≥1.Hence,(19)dxn,xn+1≤αxn,xn−1dxn,xn+1<dxn,xn−1,for alln≥1.Consequently, we deduce that the sequencedxn,xn−1 is a decreasing positive real number. Thus, there exists r≥0 such that limn⟶∞dxn,xn−1=r≥0. We claim that(20)limn⟶∞dxn,xn−1=0.On the contrary, assume thatr>0. It follows from the inequality (19) that(21)limn⟶∞αxn,xn−1dxn,xn+1=r.Now, if we take the sequencesδn=αxn,xn−1dxn,xn+1 and τn=dxn,xn−1 and considering (21), then limn⟶∞δn=limn⟶∞τn=r; therefore, by ζ3, we get that(22)0≤limn⟶∞supζαxn,xn−1dxn,xn+1,dxn,xn−1<0is a contradiction, we deduce that r=0, and equation (20) holds.Next, we show thatxn is Cauchy sequence in X, reasoning by the method of reductio ad absurdum. Suppose to the contrary that xn is not a Cauchy sequence. So, there exists ε>0; for every N∈ℕ, there exist n,m∈ℕ such that N<m<n and dxm,xn>ε. In account of (20), there exists n0∈ℕ such that(23)dxn,xn+1<ε,for alln>n0.We can find two subsequencesxnk and xmk of xn such that mk>nk≥n0 and(24)dxmk,xnk>ε,for allk,where mk is the smallest index satisfying (24). Then,(25)dxmk−1,xnk≤ε,for allk.Now, using (24), (25), and the triangular inequality, we obtain(26)ε<dxmk,xnk≤dxmk,xmk−1+dxmk−1,xnk≤dxmk,xmk−1+ε.Lettingk⟶∞ and using equation (20), we derive that(27)limn⟶∞dxmk,xnk=ε.Again, using the triangular inequality, we get(28)dxmk,xnk≤dxmk,xmk+1+dxmk+1,xnk+1+dxnk+1,xnk,for allk.Also, we have(29)dxmk+1,xnk+1≤dxmk+1,xmk+dxmk,xnk+dxnk,xnk+1,for allk.By taking the limit ask⟶∞ on both sides of (28), (29), and using (20), we deduce that(30)limn⟶∞dxmk+1,xnk+1=ε.By the same reasoning as above, we get that(31)limn⟶∞dxmk,xnk+1=limn⟶∞dxmk+1,xnk=ε.AsT is triangular α-orbital admissible, we have(32)αxmk,xnk≥1.Moreover, sinceT is an α-admissible almost Z-contraction with respect to ζ∈Z, we obtain(33)0≤ζαxmk,xnkdTxmk,Txnk,Mxmk,xnk+LNxmk,xnk=ζαxmk,xnkdxmk+1,xnk+1,Mxmk,xnk+LNxmk,xnk<Mxmk,xnk+LNxmk,xnk−αxmk,xnkdxmk+1,xnk+1.Hence,(34)0<dxmk+1,xnk+1≤αxmk,xnkdxmk+1,xnk+1<Mxmk,xnk+LNxmk,xnk,for all k≥n1, where(35)Mxmk,xnk=maxdxmk,xnk,dxmk,Txmk+dxnk,Txnk2,dxmk,Txnk+dxnk,Txmk2=maxdxmk,xnk,dxmk,xmk+1+dxnk,xnk+12,dxmk,xnk+1+dxnk,xmk+12,(36)Nxmk,xnk=mindxmk,Txmk,dxnk,Txnk,dxmk,Txnk,dxnk,Txmk=mindxmk,xmk+1,dxnk,xnk+1,dxmk,xnk+1,dxnk,xmk+1.Taking the limit ask⟶∞ in (35) and (36), using (20), (27), (30), and (31), we get(37)limk⟶∞Mxmk,xnk=ε,(38)limk⟶∞Nxmk,xnk=0.From (34), (37), and (38), we derive that μn=αxmk,xnkdxmk+1,xnk+1⟶ε as νn=Mxmk,xnk+LNxmk,xnk⟶ε; therefore by ζ3, we get(39)limk⟶∞supζαxmk,xnkdxmk+1,xnk+1,Mxmk,xnk+LNxmk,xnk<0,which is a contradiction. It follows that xn is a Cauchy sequence in the complete metric space X,d. Therefore, there exists z∈X such that(40)limn⟶∞dxn,z=0.Furthermore, by the continuity ofT, we obtain that(41)limn⟶∞dxn+1,Tz=limn⟶∞dTxn,Tz=0.Taking into account (40), (41), and the uniqueness of the limit, we deduce that z is a fixed point of T and Tz=z.Theorem 3.
LetX,d be a complete metric space and T:X⟶X be an α-admissible almost Z-contraction with respect to ζ∈Z satisfying the following conditions:(i)
T is triangular α-orbital admissible.(ii)
There existsx0∈X such that αx0,Tx0≥1.(iii)
Ifxn is a sequence in X such that αxn,xn+1≥1 for all n and xn⟶x∈X as n⟶∞, then there exists a subsequence xnk of xn such that αxnk,x≥1 for all k.
Then, there existsz∈X such that Tz=z.Proof. Following the lines of the proof of Theorem2, we obtain that the sequence xn defined by xn+1=Txn for all n≥0 is a Cauchy sequence in X. Since X,d is complete, there exists z∈X such that xn⟶z as n⟶∞. By (12) and the condition (iii), there exists a subsequence xnk of xn such that αxnk,x≥1 for all k∈ℕ. Using (7), we obtain that(42)0≤ζαxnk,zdTxnk,Tz,Mxnk,z+LNxnk,z=ζαxnk,zdxnk+1,Tz,Mxnk,z+LNxnk,z<Mxnk,z+LNxnk,z−αxnk,zdxnk+1,Tz.Hence,(43)dxnk+1,Tz≤αxnk,zdxnk+1,Tz<Mxnk,z+LNxnk,z,where(44)Mxnk,z=maxdxnk,z,dxnk,xnk+1+dz,Tz2,dxnk,Tz+dz,xnk+12,Nxnk,z=mindxnk,xnk+1,dz,Tz,dxnk,Tz,dz,xnk+1.Lettingk⟶∞ in the above equalities, we get that(45)limk⟶∞Mxnk,z=dz,Tz2,(46)limk⟶∞Nxnk,z=0.Suppose thatdz,Tz>0. From (43), we derive that(47)dxnk+1,Tz<Mxnk,z+LNxnk,z.Now, lettingk⟶∞ in the above inequality, taking into account (45) and (46), we obtain that(48)dz,Tz≤dz,Tz2is a contradiction, and hence, dz,Tz=0, that is, z=Tz.To ensure the uniqueness of a fixed point of aα-admissible almost Z-contraction with respect to ζ∈Z, we shall consider the following condition: C For all x,y∈ Fix (T), we have αx,y≥1, where Fix (T) denotes the set of fixed points of T.Theorem 4.
Adding conditionC to the hypotheses of Theorem 2 (resp., Theorem 3), we obtain the uniqueness of the fixed point of T.Proof. We argue by contradiction, suppose that there existz,z∗∈X such that z=Tz and z∗=Tz∗ with z≠z∗. From assumption C, we have(49)αz,z∗≥1.Therefore, it follows from equation (7) and ζ2 that(50)0≤ζαz,z∗dTz,Tz∗,Mz,z∗+LNz,z∗=ζαz,z∗dz,z∗,Mz,z∗+LNz,z∗<Mz,z∗+LNz,z∗−αz,z∗dz,z∗,where(51)Mz,z∗=maxdz,z∗,dz,Tz+dz∗,Tz∗2,dz,Tz∗+dz∗,Tz2=maxdz,z∗,dz,z+dz∗,z∗2,dz,z∗+dz∗,z2=dz,z∗,(52)Nz,z∗=mindz,Tz,dz∗,Tz∗,dz,Tz∗,dz∗,Tz=mindz,z,dz∗,z∗,dz,z∗,dz∗,z=0.From (50), together with (51) and (52), we deduce that(53)0<dz,z∗−αz,z∗dz,z∗.Using (49), it follows that(54)dz,z∗≤αz,z∗dz,z∗<dz,z∗,which is a contradiction. Hence, z=z∗.


## 3. Consequences

In this section, we will show that several known fixed point results can be easily concluded from our obtained results.Corollary 1 (see Karapinar and Samet [7]).
LetX,d be a complete metric space. Suppose that T:X⟶X is a generalized α-ψ-contractive mapping and satisfies the following conditions:(i)
T is α-admissible.(ii)
There existsx0∈X such that αx0,Tx0≥1.(iii)
EitherT is continuous or(iv)
Ifxn is a sequence in X such that αxn,xn+1≥1 for all n and xn⟶x∈X as n⟶∞, then there exists a subsequence xnk of xn such that αxnk,x≥1 for all k.
Then, there existsu∈X such that Tu=u.Proof. TakingL=0 and a simulation function ζEB:0,∞×0,∞⟶ℝ defined by ζEBt,s=ψs−t for all s,t∈0,∞ where ψ∈Ψ, in Theorem 4, we obtain that(55)αx,ydTx,Ty≤ψMx,y,for allx,y∈X.The mappingT is an α-admissible almost Z-contraction with respect to ζEB∈Z, and the conclusion follows.Corollary 2 (see [7]).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a function ψ∈Ψ such that(56)dTx,Ty≤ψMx,y,for allx,y∈X.
Then,T has a unique fixed point.
Proof. It suffices to choose the mappingα:X×X⟶0,∞ such that αx,y=1, for all x,y∈X and L=0 with ζ=ζEB in Theorem 4.Corollary 3 (see Berinde [23]).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a function ψ∈Ψ such that(57)dTx,Ty≤ψdx,y,for allx,y∈X.Then,T has a unique fixed point.Corollary 4 (see Ćirić [13]).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a constant k∈0,1 such that(58)dTx,Ty≤kmaxdx,y,dx,Tx+dy,Ty2,dx,Ty+dy,Tx2,for all x,y∈X. Then, T has a unique fixed point.Proof.
It is enough to choose the mappingα:X×X⟶0,∞such that αx,y=1, for all x,y∈X and L=0 with ζCt,s=ks−t in Theorem 4.Corollary 5 (see Hardy and Rogers [14]).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a constant β,γ,η≥0 with β+2γ+2η∈0,1 such that(59)dTx,Ty≤βdx,y+γdx,Tx+dy,Ty+ηdx,Ty+dy,Tx,for all x,y∈X. Then, T has a unique fixed point.Corollary 6 (see Banach contraction principle).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a constant k∈0,1 such that(60)dTx,Ty≤kdx,y,for allx,y∈X.
Then,T has a unique fixed point.Corollary 7 (see Kannan [15]).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a constant k∈0,1/2 such that(61)dTx,Ty≤kdx,Tx+dy,Ty,for allx,y∈X.Then,T has a unique fixed point.Corollary 8 (see Chatterjea [16]).
LetX,d be a complete metric space and T:X⟶X be a given mapping. Suppose that there exists a constant k∈0,1/2 such that(62)dTx,Ty≤kdx,Ty+dy,Tx,for allx,y∈X.Then,T has a unique fixed point.



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