# Fixed Points Results forα-Admissible Mapping of Integral Type on Generalized Metric Spaces

Authors: Erdal Karapınar
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We introduce generalized(α,ψ)-contractive mappings of integral type in the context of generalized metric spaces. The results of this paper generalize and improve several results on the topic in literature.


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## Body

## 1. Introduction and Preliminaries

In fixed point theory, one of the interesting research trends is to investigate the existence and uniqueness of certain mappings in the various abstract spaces. As a result of this approach, the notion of metric has been extended in several ways to get distinct abstract spaces. Among all, we mention the concept of generalized metric space that was introduced by Branciari [1] in 2001. The notion of generalized metric can be considered as a natural extension of the concept of a metric since it is obtained by replacing the the triangle inequality condition by a weaker condition, namely, quadrilateral inequality. Branciari [1] proved Banach’s fixed point theorem in such a space. For more details, the reader can refer to [2–21].At this point, we emphasize why the generalized metric space is interesting. Although the definitions of metric and generalized metric are very close to each other, the topology of the corresponding spaces is very different. In particular, a generalized metric may or may not be continuous. Furthermore, a convergent sequence in generalized metric spaces need not be Cauchy. Besides them, we cannot guarantee that a generalized metric space is Hausdorff, and hence the uniqueness of limits cannot be provided easily.On the other hand, a notion ofα-admissible mappings was defined by Samet et al. [22]. By using this notion, the authors introduced α-ψ contractive mappings and investigated the existence and uniqueness of a fixed point of such mappings in the context of metric space. Their results have attracted several authors since they are very interesting and that several existing fixed point theorems listed as consequences of the main result of this paper [22]. The approaches used in this paper have been extended and improved by a number of authors to get similar results in different settings; see, for example, [13, 15, 23–26].The aim of this paper is to examine the existence and uniqueness of fixed points ofα-admissible mappings of integral type in the setting of generalized metric spaces. We also underline that the phrase “a generalized metric” has been used for distinct notions since all such concepts generalize the notion of metric. For this reason, when we mention a “generalized metric” we mean the distance function introduced by Branciari [1]. It is evident that any metric space is a generalized metric space but the converse is not true [1].For the sake of completeness, we recall some basic definitions and notations and fundamental results that will be used in the sequel.N and R+ denote the set of positive integers and the set of nonnegative reals, respectively. Let Ψ be the family of functions ψ:0,∞→[0,∞) satisfying the following conditions:(i)
ψ is upper semicontinuous;(ii)
ψntn∈N converges to 0 as n→∞ for all t>0;(iii)
ψt<t, for any t>0.In the following, we recall the notion of a generalized metric space.Definition 1 (see [1]).
LetX be a nonempty set and let as d:X×X→[0,∞] satisfy the following conditions for all x,y∈X and all distinct u,v∈X each of which is different from x and y. Consider(1)GMS1dx,y=0ifandonlyifx=yGMS2dx,y=dy,xGMS3dx,y≤dx,u+du,v+dv,y.
Then, the mapd is called a generalized metric and abbreviated as GMS. Here, the pair (X,d) is called a generalized metric space.In the above definition, ifd satisfies only (GMS1) and (GMS2), then it is called a semimetric (see, e.g., [27]).The concepts of convergence, Cauchy sequence, completeness, and continuity on a GMS are defined below.Definition 2.
(1)
A sequence{xn} in a GMS (X,d) is GMS convergent to a limit x if and only if d(xn,x)→0 as n→∞.
(2)
A sequence{xn} in a GMS (X,d) is GMS Cauchy if and only if for every ε>0 there exists positive integer N(ε) such that d(xn,xm)<ε for all n>m>N(ε).
(3)
A GMS(X,d) is called complete if every GMS Cauchy sequence in X is GMS convergent.
(4)
A mappingT:(X,d)→(X,d) is continuous if for any sequence {xn} in X for which limn→∞d(xn,x)=0, we have limn→∞d(Txn,Tx)=0.The following assumption was suggested by Wilson [27] to replace the triangle inequality with the weakened condition.(W):
for each pair of (distinct) pointsu,v, there is a number ru,v>0 such that for every z∈X(2)ru,v<du,z+dz,v.Proposition 3 (see [28]).
In a semimetric space, the assumption(W) is equivalent to the assertion that limits are unique.Proposition 4 (see [28]).
Suppose that{xn} is a Cauchy sequence in a GMS (X,d) with limn→∞⁡d(xn,u)=0, where u∈X. Then limn→∞⁡d(xn,z)=d(u,z) for all z∈X. In particular, the sequence {xn} does not converge to z if z≠u.The following concepts were defined by Samet et al. [22].Definition 5 (see [22]).
For a nonempty setX, let T:X→X and α:X×X→[0,∞) be mappings. We say that T is α-admissible if for all x,y∈X, one has (3)αx,y≥1⟹αTx,Ty≥1.In what follows we recall the notion of aα-ψ contractive mapping.Definition 6 (see [22]).
Let(X,d) be a metric space and let T:X→X be a given mapping. One says that T is a α-ψ contractive mapping if there exist two functions α:X×X→[0,∞) and a certain ψ such that (4)αx,ydTx,Ty≤ψdx,y,∀x,y∈X.Notice that any contractive mapping, that is a mapping satisfying the Banach contraction, is aα-ψ contractive mapping with α(x,y)=1 for all x,y∈X and ψ(t)=kt, k∈(0,1).Inspired by the results of Samet et al. [22], Karapınar [13] gave the analog of the notion of a α-ψ contractive mapping in the context of generalized metric spaces as follows.Definition 7.
Let(X,d) be a generalized metric space and let T:X→X be a given mapping. One says that T is a α-ψ contractive mapping if there exist two functions α:X×X→[0,∞) and a certain ψ such that (5)αx,ydTx,Ty≤ψdx,y,∀x,y∈X.Let(X,d) be a generalized metric space. A sequence {xn} is called regular if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞; then α(xn,x)≥1 for all n.Karapınar [13] also stated the following fixed point theorems.Theorem 8.
Let(X,d) be a complete generalized metric space and let T:X→X be a α-ψ contractive mapping. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
eitherT is continuous or {xn} is regular.
Then there exists au∈X such that Tu=u.For the uniqueness, an additional condition was considered.(U):
for allx,y∈Fix(T), one has α(x,y)≥1, where Fix(T) denotes the set of fixed points of T.Theorem 9.
Adding condition(U) to the hypotheses of Theorem 8, one obtains that u is the unique fixed point of T.As an alternative condition for the uniqueness of a fixed point of aα-ψ contractive mapping, one will consider the following hypothesis.(H):
for allx,y∈Fix(T), there exists z∈X such that α(x,z)≥1 and α(y,z)≥1.Theorem 10.
Adding conditions(H) and (W) to the hypotheses of Theorem 8, one obtains that u is the unique fixed point of T.Corollary 11.
Adding condition(H) to the hypotheses of Theorem 8 and assuming that (X,d) is Hausdorff, one obtains that u is the unique fixed point of T.


## 2. Main Results

In this section, we will present our main results. For this purpose, we first define the following class of functions:Φ={φ:φ:R+→R} such that φ is nonnegative, Lebesgue integrable and satisfies (6)∫0ϵφtdt>0foreachϵ>0.Definition 12 (see [29]).
One says thatϕ∈Φ is an integral subadditive if for each a,b>0, one has (7)∫0a+bϕtdt≤∫0aϕtdt+∫0bϕtdt.One denotes byΦs the class of all integral subadditive functions ϕ∈Φ.Example 13 (see [29]).
Letϕ1(t)=(1/2)(t+1)-1/2 for all t≥0, ϕ2(t)=(2/3)(t+1)-1/3 for all t≥0, and ϕ3(t)=e-t for all t≥0. Then ϕi∈Φs, where i=1,2,3.In what follows we introduce notions of generalizedα-ψ-contractive type mappings of integral type I and type II.Definition 14.
Let(X,d) be a generalized metric space and let T:X→X be a given mapping. One says that T is generalized α-ψ-contractive type mappings of integral type I if there exist two functions α:X×X→[0,+∞) and ψ∈Ψ such that for each x,y∈X(8)αx,y∫0dTx,Tyφtdt≤ψ∫0Mx,yφtdt,where φ∈Φs and (9)Mx,y=max⁡dx,y,dx,Tx,dy,Ty.Definition 15.
Let(X,d) be a generalized metric space and let T:X→X be a given mapping. One says that T is generalized α-ψ-contractive type mappings of integral type II if there exist two functions α:X×X→[0,+∞) and ψ∈Ψ such that for each x,y∈X(10)αx,y∫0dTx,Tyφtdt≤ψ∫0Nx,yφtdt,where φ∈Φs and (11)Nx,y=max⁡dx,y,dx,Tx+dy,Ty2.Now, we state our first fixed point result.Theorem 16.
Let(X,d) be a complete generalized metric space and let T:X→X be a generalized α-ψ-contractive type mappings of integral type I. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
T is continuous.
Then there exists au∈X such that Tu=u.Proof.
Regarding assumption (ii), we guarantee that there exists a pointx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1. Starting this initial value x0∈X, we define an iterative sequence {xn} in X as follows: (12)xn+1=Txn=Tn+1x0∀n≥0.
Notice that ifxn0=xn0+1 for some n0, then the proof is completed in this case. Indeed, we have u=xn0=xn0+1=Txn0=Tu. As a consequence of this observation, throughout the proof, we assume that (13)xn≠xn+1∀n.
It is evident that(14)αx0,x1=αx0,Tx0≥1⟹αTx0,Tx1=αx1,x2≥1,
sinceT is α-admissible. Recursively, we find that (15)αxn,xn+1≥1,∀n=0,1,….
By repeating the same arguments, used above, we also derive that(16)αx0,x2=αx0,T2x0≥1⟹αTx0,Tx2=αx1,x3≥1.
From the previous inequalities, we conclude that(17)αxn,xn+2≥1,∀n=0,1,….
We divide the proofs into 4 steps.
Step  1. We show that (18)limn→∞⁡dxn,xn+1=0.
By taking (8) and (15) into account, we obtain that (19)∫0dxn,xn+1φtdt‍=∫0dTxn-1,Txnφtdt‍≤αxn-1,xn∫0dTxn-1,Txnφtdt≤ψ∫0Mxn-1,xnφtdt,for all n≥1, where (20)Mxn-1,xn=max⁡dxn-1,xn,dxn-1,Txn-1,dxn,Txn=max⁡dxn-1,xn,dxn-1,xn,dxn,xn+1=max⁡dxn-1,xn,dxn,xn+1.
If we haveM(xn-1,xn)=d(xn,xn+1) for some n∈N, then inequality (19) turns into (21)∫0dxn,xn+1φtdt≤ψ∫0Mxn-1,xnφtdt=ψ∫0dxn,xn+1φtdt<∫0dxn,xn+1φtdt,
by regarding the property (iii) of the auxiliary functionψ. This is a contradiction. Consequently, we have M(xn-1,xn)=d(xn-1,xn) for all n∈N and (19) becomes (22)∫0dxn,xn+1‍φtdt≤ψ∫0dxn-1,xn‍φtdt∀n∈N.
This yields that(23)∫0dxn,xn+1‍φtdt<∫0dxn-1,xn‍φtdt∀n∈N,
by recalling the property (iii) of the auxiliary functionψ. Due to (22), we find that (24)∫0dxn,xn+1‍φtdt≤ψn∫0dx0,x1‍φtdt,∀n∈N.
By property ofψ again, we deduce that (25)limn→∞∫0dxn,xn+1‍φtdt=0,and hence (26)limn→∞dxn,xn+1=0.
Step  2. We show that (27)limn→∞dxn,xn+2=0.
Combining (8) and (17), we conclude that (28)∫0dxn,xn+2φtdt=∫0dTxn-1,Txn+1φtdt≤αxn-1,xn+1∫0dTxn-1,Txn+1φtdt≤ψ∫0Mxn-1,xn+1φtdt,
for alln≥1, where (29)Mxn-1,xn+1=max⁡dxn-1,xn+1,dxn-1,Txn-1,dxn+1,Txn+1=max⁡dxn-1,xn+1,dxn-1,xn,dxn+1,xn+2.
By (23), we have (30)Mxn-1,xn+1=max⁡dxn-1,xn+1,dxn-1,xn=maxen,dn,where en=d(xn,xn+2) and dn=d(xn,xn+1). Thus, inequality (28) can be considered as (31)∫0enφtdt=∫0dxn,xn+2φtdt≤ψ∫0Mxn-1,xn+1‍φtdt=ψ∫0maxen-1,dn-1φtdt∀n∈N.
On the other hand, by (23) (32)∫0dn‍φtdt≤∫0dn-1‍φtdt≤∫0max⁡en-1,dn-1‍φtdt.
Therefore,(33)∫0max⁡en,dn‍φtdt≤∫0max⁡en-1,dn-1‍φtdt∀n∈N.
Then, the sequence{∫0max{en,dn}‍φ(t)dt} is monotone nonincreasing, so it converges to some t≥0. Assume that L>0. Now, by (18) (34)limsupn→∞∫0enφtdt=limsupn→∞∫0max⁡en,dnφtdt=limn→∞⁡∫0max⁡en,dnφtdt=L.
Takingn→∞ in (31) (35)L=limsupn→∞∫0enφtdt≤limsupn→∞ψ∫0max⁡en-1,dn-1φtdt≤ψlimn→∞⁡∫0max⁡en-1,dn-1φtdt=ψL<L,which is a contradiction; that is, (27) is proved.
Step  3. We will prove that (36)xn≠xm∀n≠m.
We argue by contradiction. Suppose thatxn=xm for some m,n∈N with m≠n. Since d(xp,xp+1)>0 for each p∈N, so without loss of generality, assume that m>n+1. Consider now (37)∫0dxn,xn+1φtdt=∫0dxn,Txnφtdt=∫0dxm,Txmφtdt=∫0dTxm-1,Txmφtdt≤αxm-1,xm∫0dTxm-1,Txmφtdt≤ψ∫0Mxm-1,xmφtdt,where (38)Mxm-1,xm=max⁡dxm-1,xm,dxm-1,Txm-1,dxm,Txm=max⁡dxm-1,xm,dxm-1,xm,dxm,xm+1=max⁡dxm-1,xm,dxm,xm+1.
IfM(xm-1,xm)=d(xm-1,xm), then from (37) we get that(39)∫0dxn,xn+1‍φtdt=∫0dxn,Txn‍φtdt=∫0dxm,Txm‍φtdt=∫0dxm,xm+1‍φtdt≤αxm,xm+1∫0dTxm-1,Txm‍φtdt≤ψ∫0Mxm-1,xm‍φtdt=ψ∫0dxm-1,xm‍φtdt≤ψm-n∫0dxn,xn+1‍φtdt.
IfM(xm-1,xm)=d(xm,xm+1), inequality (37) becomes (40)∫0dxn,xn+1φtdt=∫0dxn,Txnφtdt=∫0dxm,Txmφtdt=∫0dTxm-1,Txmφtdt≤αxm-1,xm∫0dTxm-1,Txmφtdt≤ψ∫0Mxm-1,xmφtdt=ψ∫0dxm,xm+1φtdt≤ψm-n+1∫0dxn,xn+1φtdt.
Due to a property ofψ, inequalities (39) and (40) together yield that (41)∫0dxn,xn+1φtdt≤ψm-n∫0dxn,xn+1φtdt<∫0dxn,xn+1φtdt,(42)∫0dxn,xn+1φtdt≤ψm-n+1∫0dxn,xn+1φtdt<∫0dxn,xn+1φtdt,
respectively. In each case, there is a contradiction.
Step  4. We will prove that {xn} is a Cauchy sequence; that is, (43)Limn→∞∫0dxn,xn+k‍φtdt=0∀k∈N.
The casesk=1 and k=2 are proved, respectively, by (18) and (27). Now, take k≥3 arbitrary. It is sufficient to examine two cases.
Case (I). Suppose that k=2m+1 where m≥1. Then, by using step 3 and the quadrilateral inequality together with (24), we find (44)∫0dxn,xn+kφtdt=∫0dxn,xn+2m+1φtdt≤∫0dxn,xn+1+dxn+1,xn+2+⋯+dxn+2m,xn+2m+1φtdt≤∫0dxn,xn+1φt+∫0dxn+1,xn+2φt+⋯+∫0dxn+2m,xn+2m+1φtdt≤∑p=nn+2mψp∫0dx0,x1φtdt≤∑p=n+∞ψp∫0dx0,x1φtdt⟶0asn⟶∞.
Case (II). Suppose that k=2m where m≥2. Again, by applying the quadrilateral inequality and step 3 together with (24), we find (45)∫0dxn,xn+kφtdt=∫0dxn,xn+2mφtdt≤∫0dxn,xn+2+dxn+2,xn+3+⋯+dxn+2m-1,xn+2mφtdt≤∫0dxn,xn+2φtdt+∫0dxn+2,xn+3φtdt+⋯+∫0dxn+2m-1,xn+2mφtdt≤∫0dxn,xn+2φtdt+∑p=n+2n+2m-1ψp∫0dx0,x1φtdt≤∫0dxn,xn+2φtdt+∑p=n+∞ψp∫0dx0,x1φtdt⟶0asn⟶∞.
By combining expressions (44) and (45), we have (46)limn→∞∫0dxn,xn+k‍φtdt=0∀k≥3.
Hence, we have(47)limn→∞⁡dxn,xn+k=0∀k≥3.
We conclude that{xn} is a Cauchy sequence in (X,d). Since (X,d) is complete, there exists u∈X such that (48)limn→∞⁡dxn,u=0.
SinceT is continuous, we obtain from (48) that (49)limn→∞⁡dxn+1,Tu=limn→∞⁡dTxn,Tu=0;that is, limn→∞⁡xn+1=Tu. Taking Proposition 4 into account, we conclude that Tu=u; that is, u is a fixed point of T.The following result is deduced from Theorem16 due to the obvious inequality N(x,y)≤M(x,y).Theorem 17.
Let(X,d) be a complete generalized metric space and let T:X→X be generalized α-ψ-contractive type mappings of integral type II. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
T is continuous.
Then there exists au∈X such that Tu=u.Theorem16 remains true if we replace the continuity hypothesis by the following property.If{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k.This statement is given as follows.Theorem 18.
Let(X,d) be a complete generalized metric space and let T:X→X be generalized α-ψ-contractive type mappings of integral type I. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k.
Then, there existsu∈X such that Tu=u.Proof.
Following the lines in the proof of Theorem8, we deduce that the sequence {xn} defined by xn+1=Txn for all n≥0 is Cauchy and converges to some u∈X. In view of Proposition 4, (50)limk→∞⁡dxnk+1,Tu=du,Tu.
By using the method ofreductio ad absurdum, we will show that Tu=u. Suppose, on the contrary, that Tu≠u; that is, d(Tu,u)>0. From (15) and condition (iii), there exists a subsequence {xn(k)} of {xn} such that α(xn(k),u)≥1 for all k.
By applying (8), we find that (51)∫0dxnk+1,Tuφtdt≤αxnk,u∫0dTxnk,Tuφtdt≤ψ∫0Mxnk,uφtdt,where (52)Mxnk,u=max⁡dxnk,u,dxnk,Txnk,du,Tu=max⁡dxnk,u,dxnk,xnk+1,du,Tu.
By (18) and (50), we obtain (53)Limk→∞∫0Mxnk,uφtdt=∫0du,Tuφtdt.
Sinceψ is upper semicontinuous, by letting k→∞ in (51) we derive that (54)∫0du,Tuφtdt≤ψ∫0du,Tuφtdt<∫0du,Tuφtdt.
This is a contradiction. Hence, we obtain thatu is a fixed point of T; that is, Tu=u.In the following, the hypothesis of upper semicontinuity ofψ is not required. Similar to Theorem 18, for the generalized α-ψ contractive mappings of type II, we have the following.Theorem 19.
Let(X,d) be a complete generalized metric space and let T:X→X be generalized α-ψ-contractive type mappings of integral type II. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
if{xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k.
Then, there existsu∈X such that Tu=u.Proof.
Following the proof of Theorem17 (which is the same as Theorem 16), we know that the sequence {xn} defined by xn+1=Txn for all n≥0 is Cauchy and converges to some u∈X. Similarly, in view of Proposition 4, (55)limk→∞⁡dxnk+1,Tu=du,Tu.
We will show thatTu=u. Suppose, on the contrary, that Tu≠u. From (15) and condition (iii), there exists a subsequence {xn(k)} of {xn} such that α(xn(k),u)≥1 for all k. By applying (10), for all k, we get that (56)∫0dxnk+1,Tuφtdt≤αxnk,u∫0dTxnk,Tuφtdt≤ψ∫0Nxnk,uφtdt,where (57)Nxnk,u=max⁡dxnk,u,dxnk,Txnk+du,Tu2.
Lettingk→∞ in (56), we have (58)limk→∞⁡∫0Nxnk,uφtdt=∫0du,Tu/2φtdt.
From (58), for k large enough, we have N(xn(k),u)>0, which implies that (59)ψ∫0Nxnk,uφtdt<∫0Nxnk,uφtdt.
Thus, from (56) and (58), we have (60)∫0du,Tuφtdt≤∫0du,Tu/2φtdt,which is a contradiction. Hence, we obtain that u is a fixed point of T; that is, Tu=u.Theorem 20.
Adding condition(U) to the hypotheses of Theorem 16 (resp., Theorem 18), one obtains that u is the unique fixed point of T.Proof.
By using the method ofreductio ad absurdum, we will show that u is the unique fixed point of T. Let v be another fixed point of T with v≠u. By hypothesis (U), (61)1≤αu,v=αTu,Tv.
Now, due to (8), we have (62)∫0du,v‍φtdt≤αu,v∫0du,vφtdt=αTu,Tv∫0dTu,Tvφtdt≤ψ∫0Mu,vφtdt=ψ∫0maxdu,v,du,Tu,dv,Tvφtdt=ψ∫0du,vφtdt<∫0du,vφtdtwhich is a contradiction. Hence, u=v.Theorem 21.
Adding condition(U) to the hypotheses of Theorem 17 (resp., Theorem 19), one obtains that u is the unique fixed point of T.Proof.
As in Theorem20, we use the method ofreductio ad absurdum to show that u is the unique fixed point of T. Suppose, on the contrary, that v is another fixed point of T with v≠u. It is evident that 1≤α(u,v)=α(Tu,Tv).
Now, due to (10), we have (63)∫0du,vφtdt≤αu,v∫0du,vφtdt=αTu,Tv∫0dTu,Tvφtdt≤ψ∫0Nu,vφtdt=ψ∫0max⁡du,v,du,Tu+dv,Tv/2φtdt=ψ∫0du,vφtdt<∫0du,vφtdtwhich is a contradiction. Hence, u=v.For the uniqueness of a fixed point of a generalizedα-ψ contractive mapping, we will consider the following hypotheses suggested in [11].(H1):
for allx,y∈Fix(T), there exists z in X such that α(x,z)≥1 and α(y,z)≥1.(H2):
letx,y∈Fix(T). If there exists {zn} in X such that α(x,zn)≥1 and α(y,zn)≥1, then (64)dzn,zn+1≤inf⁡dx,zn,dy,zn∀n∈N.Theorem 22.
Adding conditions(H1), (H2), and (W) to the hypotheses of Theorem 16 (resp., Theorem 18), one obtains that u is the unique fixed point of T.Proof.
We will show thatu is the unique fixed point of T, by using the method ofreductio ad absurdum. Let v be another fixed point of T with v≠u; that is, d(u,v)>0. Due to (H1), there exists z∈X such that (65)αu,z≥1,αv,z≥1.
SinceT is α-admissible, from (65), we have (66)αu,Tnz≥1,αv,Tnz≥1,∀n.
Define the sequence{zn} in X by zn+1=Tzn for all n≥0 and z0=z. From (66), for all n, we have (67)∫0du,zn+1φtdt=∫0dTu,Tznφtdt≤αu,zn∫0dTu,Tznφtdt≤ψ∫0Mu,znφtdt,where (68)Mu,zn=max⁡du,zn,du,Tu,dzn,Tzn=max⁡du,zn,dzn,zn+1.
By(H2), we get (69)Mu,zn=du,zn∀n.
Iteratively, by using inequality (67), we get that (70)∫0du,znφtdt≤ψn∫0du,z0φtdt,for all n. Letting n→∞ in the above inequality, we obtain (71)limn→∞∫0dzn,uφtdt=0,and hence (72)limn→∞⁡dzn,u=0.
Similarly, one can show that(73)limn→∞⁡dzn,v=0.
Regarding(W), there exists ru,v>0 such that for all n(74)ru,v<du,zn+dv,zn,and hence (75)∫0ru,vφtdt<∫0du,zn+dv,znφtdt.
From (71) and (73), by passing n→∞, it follows that ru,v=0, which is a contradiction. Thus, we proved that u is the unique fixed point of T.Theorem 23.
Adding conditions(H1), (H2), and (W) to the hypotheses of Theorem 17 (resp., Theorem 19), one obtains that u is the unique fixed point of T.Proof.
Suppose thatv is another fixed point of T and u≠v. From (H1), there exists z∈X such that (76)αu,z≥1,αv,z≥1.
SinceT is α-admissible, from (76), we have (77)αu,Tnz≥1,αv,Tnz≥1,∀n.
Define the sequence{zn} in X by zn+1=Tzn for all n≥0 and z0=z. From (77), for all n, we have (78)∫0du,zn+1φtdt=∫0dTu,Tznφtdt≤αu,zn∫0dTu,Tznφtdt≤ψ∫0Nu,znφtdt,where (79)Nu,zn=max⁡du,zn,du,Tu+dzn,Tzn2=max⁡du,zn,dzn,zn+12.
By(H2), we get (80)Nu,zn=du,zn∀n.
Iteratively, by using inequality (78), we get that (81)∫0du,znφtdt≤ψn∫0du,z0φtdt,for all n. Letting n→∞ in the above inequality, we obtain (82)limn→∞⁡∫0dzn,uφtdt=0,and hence (83)limn→∞⁡dzn,u=0.
Analogously, one can show that(84)limn→∞⁡dzn,v=0.
Similarly, regarding(W) together with (83) and (84), it follows that u=v. Thus we proved that u is the unique fixed point of T.It is known that Hausdorffness property implies the uniqueness of the limit, so the(W) condition in Theorem 22 (resp., Theorem 23) can be replaced by Hausdorff property. Then, the proof of the following result is clear and hence it is omitted.Corollary 24.
Adding conditions(H1) and (H2) to the hypotheses of Theorem 16 (resp., Theorems 18, 17, and 19) and assuming that (X,d) is Hausdorff, one obtains that u is the unique fixed point of T.


## 3. Consequences

In what follows we introduce the notion ofα-ψ-contractive type mappings of integral type.Definition 25 (Karapınar, [14]).
Let(X,d) be a generalized metric space and let T:X→X be a given mapping. One says that T is an α-ψ-contractive mapping of integral type if there exist two functions α:X×X→[0,+∞) and ψ∈Ψ such that for each x,y∈X(85)αx,y∫0dTx,Tyφtdt≤ψ∫0dx,yφtdt,where φ∈Φs.Now, we state the following fixed point theorem.Theorem 26 (Karapınar, [14]).
Let(X,d) be a complete generalized metric space and let T:X→X be an α-ψ contractive mapping of integral type. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
eitherT is continuous or {xn} is regular.
Then there exists au∈X such that Tu=u.Proof.
The proof is verbatim of the proofs of Theorems16 and 18.Theorem 27 (Karapınar, [14]).
Adding condition(U) to the hypotheses of Theorem 26, one obtains that u is the unique fixed point of T.Proof.
The proof is verbatim of the proofs of Theorem20.Remark 28.
The uniqueness condition(U) in Theorem 27 can be replaced with alternative criteria (H1), (H2), and (W) as in Theorems 22 and 23.Corollary 29.
Let(X,d) be a complete generalized metric space and let T:X→X be a continuous mapping. Suppose that there exists a function ψ∈Ψ such that (86)∫0dTx,Tyφtdt≤ψ∫0Mx,yφtdt,for all x,y∈X, where φ∈Φs and (87)Mx,y=max⁡dx,y,dx,Tx,dy,Ty.
ThenT has a unique fixed point.Proof.
Letα:X×X→[0,∞) be the mapping defined by α(x,y)=1, for all x,y∈X. Then T is an α-ψ-contraction mapping of integral type I. It is clear that all conditions of Theorem 20 are satisfied. Hence, T has a unique fixed point.Corollary 30.
Let(X,d) be a complete generalized metric space and let T:X→X be a continuous mapping. Suppose that there exists a function ψ∈Ψ such that (88)∫0dTx,Tyφtdt≤ψ∫0Nx,yφtdt,for all x,y∈X, where φ∈Φs and (89)Nx,y=max⁡dx,y,dx,Tx+dy,Ty2.
ThenT has a unique fixed point.Proof.
As in the corollary, it is sufficient to defineα:X×X→[0,∞) such that α(x,y)=1, for all x,y∈X. Then, evidently, T is an α-ψ-contraction mapping of integral type II. Hence, all conditions of Theorem 21 are fulfilled. So, T has a unique fixed point.The following fixed point theorems follow immediately from Corollary29 by taking ψ(t)=λt, where λ∈(0,1).Corollary 31.
Let(X,d) be a complete generalized metric space and let T:X→X be a continuous mapping. Suppose that there exists a constant λ∈(0,1) such that(90)∫0dTx,Tyφtdt≤λ∫0Mx,y‍φtdt,for all x,y∈X, where φ∈Φs and (91)Mx,y=max⁡dx,y,dx,Tx,dy,Ty.
ThenT has a unique fixed point.By takingψ(t)=λt, where λ∈(0,1), in Corollary 30, we derive the following result.Corollary 32.
Let(X,d) be a complete generalized metric space and let T:X→X be a continuous mapping. Suppose that there exists a constant λ∈(0,1) such that (92)∫0dTx,Ty‍φtdt≤λ∫0Nx,y‍φtdt,for all x,y∈X, where φ∈Φs and (93)Nx,y=max⁡dx,y,dx,Tx+dy,Ty2.
ThenT has a unique fixed point.Corollary 33 (cf. [11]).
Let(X,d) be a complete generalized metric space and let T:X→X be a continuous mapping. Suppose that there exists a function ψ∈Ψ such that (94)dTx,Ty≤ψMx,y,for all x,y∈X. Then T has a unique fixed point.Proof.
Letα:X×X→[0,∞) be the mapping defined by α(x,y)=1, for all x,y∈X. Then T is an α-ψ-contraction mapping. It is evident that all conditions of Theorem 8 are satisfied. Hence, T has a unique fixed point.The following fixed point theorems follow immediately from Corollary33 by taking ψ(t)=λt, where λ∈(0,1).Corollary 34 (see e.g. [11]).
Let(X,d) be a complete generalized metric space and let T:X→X be a continuous mapping. Suppose that there exists a constant λ∈(0,1) such that (95)dTx,Ty≤λNx,y,for all x,y∈X. Then T has a unique fixed point.Now, we will show that many existing results in the literature can be deduced easily from our obtained results. The following theorems are the main results of Aydi et al. [11].Theorem 35 (Aydi et al. [11]).
Let(X,d) be a complete generalized metric space and let T:X→X be a generalized α-ψ contractive mapping of type I. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
eitherT is continuous or if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k.
Then there exists au∈X such that Tu=u.Proof.
It is sufficient to takeφ(t)=1 in Theorems 16 and 18.Theorem 36 (Aydi et al. [11]).
Let(X,d) be a complete generalized metric space and let T:X→X be a generalized α-ψ contractive mapping of type II. Suppose that (i)
T is α-admissible;(ii)
there existsx0∈X such that α(x0,Tx0)≥1 and α(x0,T2x0)≥1;(iii)
eitherT is continuous or if {xn} is a sequence in X such that α(xn,xn+1)≥1 for all n and xn→x∈X as n→∞, then there exists a subsequence {xn(k)} of {xn} such that α(xn(k),x)≥1 for all k.
Then there exists au∈X such that Tu=u.Proof.
If we takeφ(t)=1 in Theorems 17 and 19, then the proof follows immediately.Theorem 37 (Aydi et al. [11]).
Adding condition(U) to the hypotheses of Theorem 35 (resp., Theorem 36), one obtains that u is the unique fixed point of T.Proof.
Letφ(t)=1 in Theorems 20 and 21.Remark 38.
Notice that all consequences and corollaries of Aydi et al. [11] can be added here since their main results are corollaries of the main results of this paper. To avoid the repetition, we do not want to state them here but we underline this fact.Example 39.
LetX=[0,1] and A={1/n:n∈N}. We define the distance function d:X×X→[0,∞) as follows: (96)dx,y=0ifx=y,dx,y=dy,x∀x,y,d12,13=d14,15=15,d12,15=d13,14=25,d12,14=d13,15=1,dx,y=x-yotherwise.
It is clear that(X,d) is a generalized metric space. Notice also that d is not a metric since (97)1=d12,14>d12,13+d13,14=35.
We defineT:X→X as Tx=1-x. Furthermore, let ψ:[0,∞)→[0,∞) be defined as ψ(t)=t/3 and φ(t)=1. Now, we define αX×X→[0,∞) as follows: (98)αx,y=1ifx=y,5xifx,y∈12,13,14,15withx≠y,0otherwise.
Hence, all conditions of Theorem20 are satisfied and x=1/2 is a unique fixed point of T.



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*Source: 141409-2015-07-27.xml*