# A New Approach to the Study of Fixed Point Theorems withw-Distances via R-Functions

Authors: Farzad Zarinfar; Farshid Khojasteh; Seyyed Mansour Vaezpour
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We introduce some new generalization of fixed point theorems in complete metric spaces endowed withw-distances via R-functions. Our results extend many of known fixed point theorems such as Reich type contraction, Geraghty contraction, Meir-Keeler contraction, and Z-contraction. In addition, the result and corollaries show that our approach has a constructive attitude and many known and unknown results can be constructed in such way.


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## Body

## 1. Introduction

Recently, extensive researches have been performed on various topics in different spaces equippedw-distances, for example, fuzzy spaces [1], polish spaces [2], quasimetric spaces [3], metric spaces endowed with a graph [4], TVS-Cone metric spaces [5–8], partially ordered metric spaces [9–15], and other related papers such as [16, 17].In 2014, Du and Khojasteh [18] introduced a very strong class of mappings calledmanageable function of which many of known contractions in the set of multivalued mappings have closely related to such class and obtain from them.Moreover, the notion ofsimulation function was introduced by Khojasteh et al. [19]. They introduced Z-contraction with respect to simulation functions. The Z-contraction generalizes the Banach contraction and unifies several known types of contractions in complete metric space (X,d) involving the combination of dTx,Ty and dx,y and satisfies some particular conditions in such spaces.Very recently, in order to extend the previous families of functions, Roldán López de Hierro and Shahzad [20] introduced a new generalized class of mappings called R-functions and a new contraction called R-contraction and show that every Geraghty contraction, Meir-Keeler contraction, Z-contraction, Man^(R)-contraction, and many of other known contractions are special case of R-contraction.In this work, we introduce some new generalization of fixed point theorems in complete metric spaces endowed withw-distances via R-functions and many of known fixed point theorems such as Reich type contraction, Geraghty contraction [9], Meir-Keeler contraction [21], and Z-contraction [19] which in such spaces are concluded as apparent corollaries.


## 2. Preliminaries

In the following, we assert some notions which are needed in continuance.Definition 1 (see [22]).
Let(X,d) be a metric space. The mapping p:X×X→0,1 is said to be w-distance on X, if it satisfies the following axioms: (1)p(x,z)≤p(x,y)+p(y,z), for any x,y,z∈X;(2) for anyx∈X, px,·:X→0,1 is lower semicontinuous;(3) for anyϵ>0, there exists δ>0 such that p(z,x)≤δ and p(z,y)≤δ implies d(x,y)<ϵ.Denoteranp=px,y:x,y∈X.Lemma 2 (see [22]).
Let(X,d) be a metric space and p be w-distance on X. Assume that xn, yn are sequences in X and αn, βn are sequences in 0,1 which are convergent to 0 and x,y∈X. Then the following holds: (i)
ifpxn,y≤αn and pxn,z≤βn, for any n∈N, then y=z;(ii)
ifpxn,yn≤αn and pxn,z≤βn, for any n∈N, then yn converges to z;(iii)
ifpxn,xm≤αn, for any m,n∈N with m>n, then xn is a Cauchy sequence;(iv)
ifpy,xn≤αn, for any n∈N, then xn is a Cauchy sequence.Definition 3 (see [23]).
A symmetricw-distance on a metric space (X,d) is w-distance p on (X,d) such that(SY):qx,y=qy,x for all x,y∈X.If p is w-distance satisfying (SY), we then say that it is a symmetric w-distance on X,d.Definition 4 (see [24]).
A functionϕ:0,∞→0,∞ is calledL-function if (a)
ϕ0=0,(b)
ϕt>0 for all t>0,(c)
for allε>0, there exists δ>0 such that ϕt≤ε, for all t∈ε,ε+δ. Moreover, for each ε>0 we have ϕ(ε)≤ε.Definition 5 (see [17, 21]).
A mappingT:X→X, from a metric space X,d into itself, is called Meir-Keeler mapping, whenever, for all ε>0, there exists δ>0 such that, for all x,y∈X, if ε≤dx,y<ε+δ, then dTx,Ty<ε.Meir and Keeler [21] proved that every Meir-Keeler mapping on a complete metric space has a unique fixed point.Definition 6 (see [19]).
Asimulation function is a mapping ζ:0,∞×0,∞→R which satisfies the following conditions: ζ1
ζt,s<s-t for all t,s>0,ζ2
iftn,sn are sequences in (0,∞) such that limn→∞⁡tn=limn→∞sn>0, then (1)lim⁡supn→∞⁡ζtn,sn<0.Many examples of simulation functions can be found in [18, 19, 25]. Very recently, Roldán López de Hierro and Shahzad [20] introduced the following category of mappings which are connected to extend the simulation functions.Definition 7.
LetA⊆R be a nonempty subset and let ϱ:A×A→R be a function. We say that ϱ is R-function if it satisfies the following two conditions:(ϱ1)
Ifan⊂0,∞∩A is a sequence such that ϱan+1,an>0, for all n∈N, then limn→∞an=0.(ϱ2)
Ifan,bn⊂0,∞∩A are two sequences which converge to the same limit L≥0 such that L<an and ϱan,bn>0, for all n∈N, then L=0. We denote by RA the family of all R-functions whose domain is A×A.In some cases, for a given functionϱ:A×A→R, we will also consider the following property:(ϱ3)
Ifan,bn⊂0,∞∩A are two sequences such that limn→∞bn=0 and ϱan,bn>0, for all n∈N, then limn→∞an=0.In the present research, we will introduce our main result and demonstrate that many of known results which have been invented in the set of metric spaces, with or withoutw-distances, can be deduced easily from our main result. We draw the respondents’ attention to Alegre et al. [3, Theorem 2, Corollary 1] about Meir-Keeler contraction in metric spaces with w-distances which can be acquired and generalized with Theorem 9 (for more detail see Corollary 13).


## 3. Main Result

The following definition plays the crucial role in our main result.Definition 8.
LetX,d be a metric space, let p be w-distance on X, and let T:X→X be a mapping. We say that T is Rp-contraction, if there exists R-function ϱ:A×A→R such that ranp⊆A and (2)ϱpTx,Ty,px,y>0,for all x,y∈X. In such case, we say that T is Rp-contraction with respect to ϱ. The family of all Rp-contractions from X,d into itself, with respect to ϱ, is denoted by RX,p,A,ϱ or, when no confusion is possible, by Rϱ,p.In what follows, we assume thatp is a symmetric w-distance, in the sense of Definition 3, unless the contrary is asserted. Now we are ready to give the main result.Theorem 9.
LetX,d be a complete metric space, let p be w-distance on X, and let T:X→X be Rp-contraction with respect to ϱ∈RA. Assume that, at least, one of the following conditions holds: (a)
T is p-continuous; that is, for any sequence xn, if pxn,z→0 then pTxn,Tz→0 as n→∞.(b)
The functionϱ satisfies ϱ3. Then T has a unique fixed point.Proof.
Letx0∈X be chosen arbitrary. If x0=Tx0, then we have nothing to prove. So let x0≠Tx0. Define x1=Tx0. Similarly one can define the Picard sequence xn+1=Txn, for all n∈N, such that xn+1≠xn. Let an=p(xn,xn+1)>0 and divide the proof into three steps.
Step  1. { x n } is p-asymptotically regular; that is, limn→∞an=limn→∞pxn,xn+1=0.
T is Rp-contraction with respect to ϱ∈RA. Thus (3)ϱan+1,an=ϱpxn+1,xn+2,pxn,xn+1=ϱpTxn,Txn+1,pxn,xn+1>0.By (ϱ1) of Definition 7 we have limn→∞an=0 and desired result is obtained.
Step  2. x n is a Cauchy sequence and so it converges to some z∈X.
On the contrary, assume thatxn failed to be a Cauchy sequence. By (iii) of Lemma 2, xn failed to be p-Cauchy sequence. Hence, there exists ϵ0>0 such that, for all k∈N, there exist subsequences xnk and xmk of xn and mk can be chosen as the smallest integer such that mk>nk≥k and for all s∈mk-1,…,nk,nk-1,…,1(4)pxnk,xs≤ϵ0<pxnk,xmk.Obviously, limk→∞pxnk,xmk=ϵ0, because (5)ϵ0<pxnk,xmk≤pxnk,xmk-1+pxmk-1,xmk≤pxmk-1,xmk+ϵ0.Also, applying the symmetry property we have (6)ϵ0<pxnk,xmk≤pxnk,xnk-1+pxnk-1,xmk-1+pxmk-1,xmk≤2pxnk,xnk-1+pxnk,xmk-1+pxmk-1,xmk≤2pxnk,xnk-1+ϵ0+pxmk-1,xmk.Therefore, limk→∞pxnk-1,xmk-1=ϵ0. Now, taking L=ϵ0, ak=pxnk,xmk, and bk=pxnk-1,xmk-1 and by the fact that L=ϵ0<pxnk,xmk=ak, we have (7)ϱak,bk=ϱpxnk,xmk,pxnk-1,xmk-1=ϱpTxnk-1,Txmk-1,pxnk-1,xmk-1>0.By (ϱ2) of Definition 7, we have L=ϵ0=0 and this is a contradiction. Therefore, xn is p-Cauchy sequence and by (iii) of Lemma 2 is Cauchy. Thus it converges to some z∈X.
Step  3. z is a unique fixed point and pz,z=0.
If we consider a sequence{αn}, in a sense of Lemma 2, xn is p-Cauchy sequence; that is, for all m,n∈N, m>n, pxn,xm≤αn. Also xn→z as n→∞ and px,· is a lower semicontinuous, for all x∈X. Hence (8)pz,xn≤lim⁡infm→∞⁡pxm,xn≤αn.It means that p(z,xn)→0 as n→∞.
Now suppose that(a) holds. Since T is p-continuous mapping, limn→∞pTxn,Tz=0 and for all x∈X, specially x=z and x=Tz, we have (9)pz,z≤lim⁡infn→∞⁡pz,xn+1=0,and so (10)pz,Tz≤lim⁡infn→∞⁡pxn+1,Tz=lim⁡infn→∞⁡pTxn,Tz=0.Therefore, pz,Tz=pz,z=0. By (i) of Lemma 2, we have Tz=z and so z is the fixed point of T.
On the other hand, let(b) hold. For all n∈N, suppose that an=p(xn+1,Tz) and bn=p(xn,z). By (8), we have limn→∞bn=0. In addition, for all n∈N, (11)ϱan,bn=ϱpxn+1,Tz,pxn,z=ϱpTxn,Tz,pxn,z>0.Henceforth, if bn=0, then xn=z. So xn+1=Txn=Tz; thus an=0.
Now consider the following set:(12)Ω=n∈N:an=0=n∈N:pxn+1,Tz=0.One can consider two cases: (a1)
Ω is a finite set:  in this case, one can choose n0∈N, such that an>0, for all n>n0, and so bn>0, for all n>n0. Applying (11) concludes that limn→∞an=0. It means that limn→∞p(xn+1,Tz)=0 and so applying (i) of Lemma 2 yields Tz=z.(a2)
Ω is infinite set:  in this case, one can find a subsequence xnk of xn in Ω. We have (13)pxnk+1,Tz=0,for allk∈N. Analogous the argument in the first case and applying (i) of Lemma 2 results Tz=z again.
Moreover, ifx,y are two distinct fixed points for T, then by considering an=px,x≥0 we have(14)ϱan+1,an=ϱpx,x,px,x=ϱpTx,Tx,px,x>0.Applying ϱ1 deduces that limn→∞an=0 and so px,x=0. Applying the same argument on bn=px,y concludes that px,y=0 and (i) of Lemma 2 eventuates x=y.Example 10.
The following items are some examples ofR-functions: (a)
ϱt,s=sφs-t, where φ:0,+∞→0,1 is a mapping such that limsupt→s+φt<1, for all s∈0,+∞,(b)
ϱt,s=sφs-t, where φ:0,+∞→0,1 is a mapping such that, for all sequence tn⊂0,+∞, (15)limn→∞φtn=1implieslimn→∞tn=0,(c)
ϱt,s=φs-t, where φ:0,+∞→0,+∞ is L-function in the sense of Definition 4,(d)
every simulation functionζ:0,∞×0,∞→R,(e)
ϱφ,ψt,s=ψs-φs-ψt, where ψ,φ:0,∞→0,∞ are two functions such that ψ is nondecreasing and continuous from the right, φ is lower semicontinuous, and φ-10=0,(f)
ϱt,s=s/t+1-t,(g)
ϱt,s=s/et-t,(h)
ϱt,s=lns+1-t.Proof.
It is sufficient to show thatϱ1 and ϱ2 of Definition 7 and ϱ3 hold for all (a)–(h).(a)
ϱ1
Letan⊂0,+∞ be a sequence and ϱan+1,an>0. Then (16)0<ϱan+1,an=anφan-an+1.It means that(17)an+1an<φan,and so taking limit on both sides of (17), we have (18)lim⁡supn→∞⁡an+1an<lim⁡supn→∞⁡φan<1.By ratio test,∑n=1∞‍an<∞ and then concludes that limn→∞an=0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, L<an, and ϱan,bn>0 for all n∈N. Suppose that L>0; then bnφbn-an>0 and so (19)anbn<φbn.By taking limit on both sides of (19), we have (20)1=lim⁡supn→∞⁡anbn≤lim⁡supn→∞⁡φbn<1,which is a contradiction. Thus,L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ϱan,bn>0. Therefore, (21)0<ϱan,bn=bnφbn-an≤bn-an,so0<an<bn which implies that limn→∞an=0.(b)
ϱ1
Letan⊂0,+∞ be a sequence such that ϱan+1,an>0. Therefore, (22)0<ϱan+1,an=φanan-an+1.Sincean>0 and φan<1, one can see easily (23)an+1<φanan<an.Hence,an is a strictly decreasing sequence of nonnegative real numbers and so converges to L≥0. On the contrary, assume that L>0. Therefore, (24)0<L<an+1<φanan<an.Taking limit on both sides of (24) deduces that limn→∞φan=1. Since φ is a Geraghty function, so L=limn→∞an=0 which contradicts L>0. So L=0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, L<an hold, and ϱan,bn>0 hold. On the contrary, assume that L>0. Therefore, (25)0<ϱan,bn=φbnbn-an,and so(26)L<an<φbnbn<bn.Taking limit on both sides of (26) deduces that limn→∞φbn=1. Since φ is a Geraghty function, it concludes that L=limn→∞bn=0 which contradicts L>0. So L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ϱ(an,bn)>0. Therefore, (27)0<ϱan,bn=φbnbn-an≤bn-an,so0<an<bn which implies that limn→∞an=0.(c)
ϱ1
Letan⊂(0,+∞) be a sequence and let ϱan+1,an>0. Then (28)0<ϱan+1,an=φan-an+1.It means that(29)an+1<φan≤an.Therefore,an is a nonincreasing sequence; so is converges to L≥0. On the contrary, suppose that L>0. By considering ϵ=L, there exists δ>0 such that, for all t∈L,L+δ, φ(t)≤ϵ. Since limn→∞an→L, there exists n0>0 such that, for all n≥n0, we have L<an<L+δ. Therefore, an0+1<φan0≤L and this is a contradiction to L<an<L+δ. So L=0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, L<an hold, and ϱan,bn>0 hold. On the contrary, assume L>0. Therefore, (30)0<ϱan,bn=φbn-an.Also(31)L<an<φbn≤bn.Sinceφ is L-function, taking ϵ=L>0, there exists δ>0 such that (32)φt≤ϵ=L∀t∈L,L+δ.Sincelimn→∞bn=L, there exists n0∈N such that L<bn0<L+δ. Applying (32), one deduces that φ(bn0)≤L which is a contradiction, because φ(bn0)≤L<an0<φ(bn0). So L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ϱan,bn>0. Therefore, (33)0<ϱan,bn=φbn-an≤bn-an,so0<an<bn which implies that limn→∞an=0.(d)
Suppose thatζ is a simulation function.
ϱ1
Letan⊂0,+∞ be a sequence and let ζan+1,an>0. Therefore 0<ζan+1,an<an-an+1. Hence an is a strictly decreasing sequence so it converges to L≥0. On the contrary, suppose that L>0. Let tn=an+1 and sn=an. Then tn and sn are convergent to L and tn<sn. By condition ζ3 we have (34)0≤lim⁡supn→∞⁡ζan+1,an=lim⁡supn→∞⁡ζtn,sn<0,and this is a contradiction. SoL=0.ϱ2
Let{an},{bn}⊂(0,+∞) be two sequences which converge to the same limit L≥0, L<an, and ζ(an,bn)>0. In order to prove that L=0, assume that L>0. On the other hand, (ζ2) shows that 0<ζ(an,bn)=bn-an so L<an<bn. Applying (ζ3), one can yield that (35)0≤lim⁡supn→∞⁡ζan,bn<0,which is a contradiction. SoL=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ζ(an,bn)>0. Since ζ is a simulation function 0<ζ(an,bn)<bn-an. Hence, 0<an<bn, which means that limn→∞an=0.(e)
First of all, we show that, fort,s∈0,∞, (36)s>0,ϱψ,φ≥0⟹t<s.Letϱψ,φt,s≥0 and t≥s; we are going to show that s=0. As ψ is nondecreasing, (37)ψs≤ψt≤ψs-φs≤ψs,which implies thatφs=0. Hence s=0 and this is a contradiction. So we have t<s.
ϱ1
Let{an}⊂(0,+∞) be a sequence and let ϱψ,φan+1,an>0. By (36), an+1<an. Hence an is a strictly decreasing sequence so it converges to L≥0 where L=infn∈N⁡an; hence L<an, for all n∈N. On the contrary, suppose that L>0. Therefore, (38)0<ϱψ,φan+1,an=ψan-φan-ψan+1,and so0≤φan<ψan-ψan+1. Since ψ is continuous from the right and limn→∞an=L, we deduce that limn→∞φan=0. Since φ is lower semicontinuous, (39)0≤φL≤lim⁡infr→L⁡φr=limn→∞φan=0.Henceφ(L)=0, so L=0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, verifying L<an, and ϱψ,φan,bn>0. By (36) one can see L<an<bn. In order to prove that L=0, assume that L>0. Therefore, (40)0<ϱψ,φan,bn=ψbn-φbn-ψan,and so0≤φbn<ψbn-ψan. As ψ is continuous from the right and limn→∞bn=L, we deduce that limn→∞φ(bn)=0. Since φ is lower semicontinuous, (41)0≤φL≤lim⁡infr→L⁡φr=limn→∞φbn=0.Hence,φL=0, so L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ϱφan,bn>0. By (36), one can see 0<an<bn and this implies that limn→∞an=0.(f)
ϱ1
Letan⊂0,+∞ be a sequence and let ϱan+1,an>0. It means that (42)0<ϱan+1,an=anan+1+1-an+1.Hence,(43)0<an+1<anan+1+1.Therefore,(44)0<an+1an<1an+1+1<1.Suppose thatlimn→∞an=r. If r>0, then by letting n→∞ we deduce that (45)0<limn→∞an+1an=1≤1r+1<1,which is a contradiction. Solimn→∞an=0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, L<an, and ϱan,bn>0, for all n∈N. We have (46)0<ϱan,bn=bnan+1-an.Therefore,(47)0<an<bnan+1.Hence,(48)0≤L<an<bnan+1.Lettingn tend to infinity, we deduce 0≤L≤L/L+1, so 0≤LL+1≤L and hence 0≤L2≤0. It means that L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that bn→0 and ϱan,bn>0. Therefore, (49)0<ϱan,bn=bnan+1-an,so0<an<bn/an+1, which implies that an→0.(g)
ϱ1
Letan⊂0,+∞ be a sequence and let ϱan+1,an>0. It means that (50)0<ϱan+1,an=anean+1-an+1.Hence,(51)0<an+1<anean+1.Therefore,(52)0<an+1an<1ean+1<1.Suppose thatan→r. If r>0, then taking limit on both sides of (52), we deduce that (53)0<limn→∞an+1an=1≤1er<1,which is a contradiction. Soan→0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, L<an, and ϱan,bn>0, for all n∈N. On the contrary, suppose that L>0. We have (54)0<ϱan,bn=bnean-an.Therefore,(55)0<an<bnean.Hence,(56)0<L<an<bnean.Lettingn→∞, we deduce that 0<L≤L/eL, so 0<LeL≤L and hence 0<eL≤1 and this is a contradiction. So L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ϱan,bn>0. Therefore, (57)0<ϱan,bn=bnean-an,so0<an<bn/ean which implies that an→0.(h)
ϱ1
Letan⊂0,+∞ be a sequence and let ϱan+1,an>0. It means that (58)0<ϱan+1,an=ln⁡an+1-an+1.Thus,(59)0<an+1<ln⁡an+1.Therefore,(60)0<an+1an<ln⁡an+1an<1.Suppose thatlimn→∞an=r. If r>0, then by letting n→∞, we deduce that (61)0<lim⁡n→∞an+1an=1≤ln⁡r+1r<1,which is a contradiction. Solimn→∞an=0.ϱ2
Letan,bn⊂0,+∞ be two sequences which converge to the same limit L≥0, L<an, and ϱan,bn>0, for all n∈N. We have (62)0<ϱan,bn=ln⁡bn+1-an.Therefore,(63)0<an<ln⁡bn+1.Hence,(64)0≤L<an<ln⁡bn+1.Lettingn→∞, we deduce that 0≤L≤lnL+1 and so L=0.ϱ3
Letan,bn⊂0,+∞ be two sequences such that limn→∞bn=0 and ϱan,bn>0. Therefore, (65)0<ϱan,bn=ln⁡bn+1-an,so0<an<lnbn+1 which implies that limn→∞an=0.The following corollary is the generalization of Reich’s theorem (see [21, 24]).Corollary 11.
LetX,d be a complete metric space, let p be w-distance on X, and let T:X→X be a mapping such that (66)pTx,Ty<φpx,ypx,y,for all x,y∈X, where φ:0,+∞→0,1 is a mapping such that limsupt→s+φt<1, for all s∈0,+∞. Then, T has a unique fixed point.Proof.
Definingϱt,s=sφs-t, applying (a) of example, and using Theorem 9, one can conclude desired result.The following corollary is the generalization of Geraghty fixed point theorem (see [9, Theorem 1.3]).Corollary 12.
Let(X,d) be a complete metric space, let p be w-distance on X, and let T:X→X be a mapping such that (67)pTx,Ty<φpx,ypx,y,for all x,y∈X, where φ:0,+∞→0,1 is a mapping such that, for all sequence tn⊂0,+∞, (68)limn→∞φtn=1implieslimn→∞tn=0.Then, T has a unique fixed point.Proof.
Definingϱt,s=sφs-t, applying (b) of example, and using Theorem 9, one can conclude desired result.The following corollary is the generalization of Meir-Keeler fixed point theorem (also see [24, Theorem 1]).Corollary 13.
Let(X,d) be a complete metric space, let p be w-distance on X, and let T:X→X be a mapping such that (69)pTx,Ty<φpx,y,for all x,y∈X, where φ:0,+∞→0,+∞ is L-function in a sense of Definition 4. Then T has a unique fixed point.Proof.
Definingϱt,s=φs-t, applying (c) of example, and using Theorem 9, one can conclude desired result.The following corollary is the generalization of Khojasteh’s result [19, Theorem 2.8] (see also [25]).Corollary 14.
Let(X,d) be a complete metric space, let p be w-distance on X, and let T:X→X be a mapping such that (70)ζpTx,Ty,px,y≥0,for all x,y∈X, where ζ is a simulation function. Then T has a unique fixed point.Proof.
Applyingϱt,s=ζt,s, (d) of example and considering Theorem 9, one can conclude desired result.Corollary 15.
Let(X,d) be a complete metric space, let p be w-distance on X, and let T:X→X be a mapping such that (71)ψpTx,Ty<ψpx,y-φpx,y,for all x,y∈X, where ψ,φ:0,∞→0,∞ are two functions such that ψ is nondecreasing and continuous from the right, φ is lower semicontinuous, and φ-10=0. Then, T has a unique fixed point.Proof.
Consideringϱφ,ψt,s=ψs-φs-ψt, (e) of example, and applying Theorem 9, one can conclude desired result.In the following, we are going to present some new corollaries to show that our results are constructive and new results can be built byR-contractions.Corollary 16.
Let(X,d) be a metric space, let p be w-distance on X, and let T:X→X be a self-mapping. Let, for every x,y∈X, with x≠y, the following hold: (72)pTx,Ty<px,ypTx,Ty+1.Then, T is Rp-contraction.Proof.
Definingϱ(t,s)=s/t+1-t, applying (f) of example, and using Theorem 9, one can conclude desired result.Corollary 17.
Let(X,d) be a metric space, let p be w-distance on X, and let T:X→X be a self-mapping. Let, for every x,y∈X, with x≠y, the following hold: (73)pTx,Ty<px,yepTx,Ty.Then, T is Rp-contraction.Proof.
Consideringϱ(t,s)=s/et-t, applying (g) of example, and using Theorem 9, one can conclude desired result.Corollary 18.
LetX,d be a metric space, let p be w-distance on X, and let T:X→X be a self-mapping. Let for every x,y∈X, with x≠y, (74)pTx,Ty<ln⁡px,y+1.Then, T is Rp-contraction.Proof.
Definingϱt,s=lns+1-t, applying (h) of example, and using Theorem 9, one can conclude desired result.



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*Source: 6978439-2016-11-01.xml*