DDNS Discrete Dynamics in Nature and Society 1607-887X 1026-0226 Hindawi 10.1155/2018/4745764 4745764 Research Article Nemytzki-Edelstein-Meir-Keeler Type Results in b-Metric Spaces http://orcid.org/0000-0003-3896-3809 Aydi Hassen 1 2 Banković Radoje 3 Mitrović Ivan 4 http://orcid.org/0000-0002-1274-1936 Nazam Muhammad 5 Candito Pasquale 1 Department of Mathematics College of Education of Jubail Imam Abdulrahman Bin Faisal University P.O. 12020 Industrial Jubail 31961 Saudi Arabia iau.edu.sa 2 Department of Medical Research China Medical University Hospital China Medical University Taichung Taiwan cmu.edu.tw 3 Vojnogeografski Institut Beograd Vojska Srbije Serbia vgi.mod.gov.rs 4 First Technical School 35 000 Jagodina Serbia 5 Department of Mathematics and Statistics International Islamic University Islamabad Pakistan iiu.edu.pk 2018 472018 2018 12 04 2018 05 06 2018 19 06 2018 472018 2018 Copyright © 2018 Hassen Aydi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider some Nemytzki-Edelstein-Meir-Keeler type results in the context of b-metric spaces. In some cases, we assume that the b-metric is continuous. Our results generalize several known ones in existing literature. We also present some examples to illustrate the usability of our results.

1. Definitions, Notations, and Preliminaries

Let X,d be a metric space and T:XX be a self-mapping. The following Meir-Keeler conditions are well known: For each ε>0, there exists δ=δε>0 such that(1)εdx,y<ε+δ  implies  dTx,Ty<ε,or(2)ε<dx,y<ε+δ  implies  dTx,Tyε,or T is contractive and(3)εdx,y<ε+δ  implies  dTx,Tyε.It is clear that (1) implies (2) and (2) implies (3), while the converse is not true. One says that the mapping T defined on the metric space X,d is contractive if dTx,Ty<dx,y holds, whenever xy. For more details, see  (pages 30-33 and 56-58) and . In 1969, Meir-Keeler  proved the following.

Theorem 1 ([<xref ref-type="bibr" rid="B31">2</xref>], Theorem).

Let X,d be a complete metric space and let T be a self-mapping on X satisfying (1). Then T     has a unique fixed point, say uX, and, for each xX, limnTnx=u.

For other fixed point results via generalized Meir-Keeler contractions, see . Inspired from Meir-Keeler theorem, Ćirić proved the next slightly more general result.

Theorem 2 ([<xref ref-type="bibr" rid="B16">1</xref>], Theorem 2.5).

Let X,d be a complete metric space and let T be a self-mapping on X satisfying (2). Then T     has a unique fixed point, say uX, and, for each xX, limnTnx=u.

The following example shows that Ćirić result is a proper generalization of Meir-Keeler theorem.

Example 3.

Let X=[0,1]2,3+1/3,,3n-1,3n+1/3n, be a subset of reals with the Euclidean metric and let T be a self-mapping on X defined by (4)Tx=0,if  0x1,x=2,,3n-1,(5)Tx=1,if  x=3+13,6+16,,3n+13n,Then one can verify that T satisfies (2), while it does not satisfy the Meir-Keeler condition (1). For all details, see .

Remark 4.

Theorems 1 and 2 are true if the self-mapping T:XX satisfies condition (3). For more details, see , pages 30-33.

On the other hand, Bakhtin  and Czerwik  introduced the concept of b-metric spaces (a generalization of metric spaces) and proved the Banach contraction principle. The definition of a b-metric space is the following.

Definition 5 (Bakhtin [<xref ref-type="bibr" rid="B11">6</xref>] and Czerwik [<xref ref-type="bibr" rid="B17">7</xref>]).

Let X be a nonempty set and let s1 be a given real number. A function d:X×X[0,) is said to be a b-metric if and only if, for all x,y,zX, the following conditions are satisfied:

dx,y=0 if and only if x=y.

dx,y=dy,x.

dx,zsdx,y+dy,z.

The triplet X,d,s1 is called a b-metric space with coefficient s.

In the last period, many authors obtained several fixed point results for single-valued or set-valued mappings in the context of b-metric spaces. For more details, see [5, 835]. It should be noted that the class of b-metric spaces is effectively larger than that of standard metric spaces, since a b-metric is a metric when s=1. The following example shows that, in general, a b-metric does not necessarily need to be a metric.

Example 6.

Let X,ρ be a metric space and dx,y=ρx,yp where p>1 is a real number. Then d is a b-metric with s=2p-1, but d is not a metric on X.

The concepts of b-convergence, b-completeness, b-Cauchy, and b-closed set in b-metric spaces have been initiated in [6, 7].

The following two lemmas are very significant in the class of b-metric spaces.

Lemma 7 ([<xref ref-type="bibr" rid="B22">21</xref>], Lemma 3.1).

Let yn be a sequence in a b-metric space X,d,s1 such that(6)dyn,yn+1λdyn-1,ynfor some λ[0,1/s), and each n=1,2,. Then yn is a b-Cauchy sequence in a b-metric space X,d.

Lemma 8 ([<xref ref-type="bibr" rid="B32">30</xref>], Lemma 2.2).

Let yn be a sequence in a b-metric space X,d,s1 such that(7)dyn,yn+1λdyn-1,ynfor some λ[0,1), and each n=1,2,. Then yn is a b-Cauchy sequence in a b-metric space X,d.

Since in general a b-metric is not continuous, we need the following two lemmas.

Lemma 9 ([<xref ref-type="bibr" rid="B4">36</xref>], Lemma 2.1).

Let X,d,s1 be a b-metric space with s1. Suppose that xn and yn are b-convergent to x and y, respectively. Then (8)1s2dx,yliminfndxn,ynlimsupndxn,yns2dx,y.In particular, if x=y, then we have limndxn,yn=0. Moreover, for each zX, we have (9)1sdx,zliminfndxn,zlimsupndxn,zs2dx,z.

Lemma 10 (see [<xref ref-type="bibr" rid="B18">37</xref>]).

Let X,d,s1 be a b-metric space and xn be a sequence in X such that (10)limndxn,xn+1=0.If xn is not b-Cauchy, then there exist ε>0 and two sequences mk and nk of positive integers such that, for the following four sequences (11)dxmk,xnk,dxmk,xnk+1,dxmk+1,xnk,dxmk+1,xnk+1,we have (12)εliminfkdxmk,xnklimsupkdxmk,xnkεs,εsliminfkdxmk,xnk+1limsupkdxmk,xnk+1εs2,εsliminfkdxmk+1,xnklimsupkdxmk+1,xnkεs2,εs2liminfkdxmk+1,xnk+1limsupkdxmk+1,xnk+1εs3.

Essential to the proofs of fixed point theorems for the most contractive conditions in the context of b-metric spaces are the above two lemmas (see, for example, [3, 5, 9, 13, 17, 23, 25, 28]). However, it is not hard to show that the proofs of the most fixed point theorems in the context of b-metric spaces become simpler and shorter if they are based on Lemma 8.

2. Main Result

To our knowledge, it is not known whether Meir-Keeler and Ćirić theorems hold in the context of b-metric spaces. Also, it is not known that if there are examples such that condition (1) or (2) or (3) holds in the context of b-metric spaces, but T has no fixed point.

Our first result generalizes Lemma 1 of . For some results also see recent paper .

Lemma 11.

Let X,d,s>1 be a b-complete b-metric space and T:XX such that condition (1) holds. If Tnx is a b-Cauchy sequence for each xX, then T has a unique fixed point, say uX and Tnxu.

Proof.

Since X,d,s>1 is b-complete, each {Tnx} has a limit point, say ηx. Since condition (1) implies the continuity of T, we have(13)Tηx=TlimnTnx=limnTn+1x=ηx.Thus, ηx is a fixed point, and therefore all ηx are equal.

Remark 12.

If condition (1) holds on b-metric spaces X,d,s>1, we do not know whether every sequence Tnx is b-Cauchy.

However, with a stronger condition than (1), we have a positive response. It will be the subject of Theorem 13.

Now, we announce a Meir-Keeler type result in the context of b-metric spaces.

Theorem 13.

Let X,d be a complete b-metric space and let T be a self-mapping on X satisfying the following condition.

Given ε>0 there exists δ>0 such that (14)εdx,y<ε+δ  implies  sadTx,Ty<ε,where a>0 is given.

Then T has a unique fixed point, say uX, and, for each xX,limnTnx=u.

Proof.

It is clear that, for all x,yX with dx,y>0, we obtain(15)dTx,Tyλdx,y,where λ=1/sa[0,1).

Let x0X be an arbitrary point. Define the sequence xn by xn+1=Txn for all n0. If, for some n, xn=xn+1, then xn is a fixed point of T. From now on, suppose that xnxn+1 for all n0. From condition (15), we obtain(16)dxn,xn+1λdxn-1,xn.Further, according to (, Lemma 2.2.), the sequence xn is b-Cauchy in the b-metric space X,d. By b-completeness of X,d, there exists uX such that(17)limnxn=u.Finally, (15) and (17) imply that Tu=u; that is, u is the unique fixed point of T in X.

Example 14.

Let X=0,1,2 and define d:X×X[0,+) as follows: dx,x=0, dx,y=dy,x for all x,yX, d0,1=1, d0,2=2.2, and d1,2=1.1. Then X,d,22/21>1 is a b-complete b-metric space, but it is not a metric space. Let T:XX be defined by (18)Tx=0,if  x2,1,if  x=2.We shall check that, for all x,yX, the contractive condition (15) holds. For this, we distinguish three cases.

(a) x=0,y=1dT0,T1=d0,0=0. Obviously, condition (15) holds.

(b) x=0,y=2dT0,T2=d0,1. Since 22/21ad0,1d0,2, i.e., 22/21a·12.2,a>0, which is true, hence, again (15) holds.

(c) x=1,y=2dT1,T2=d0,1=1. Now, we have 22/21a·11.1, i.e., 22/21a1.1, which is also true because a>0.

Therefore, condition (15) holds for each a>0. However, condition (14) is not true for a=1. Indeed, for x=0 and y=2, it becomes (19)εd0,2<ε+δ  implies  2221<ε,or equivalently (20)ε2.2<ε+δ  implies  2221<ε.Take ε=1/2. Then there exists δ=δ1/2>0 such that 1/22.2<1/2+δ (for example, any δ>17/10). But 22/21<1/2 is false.

Now, we give an example supporting Theorem 13.

Example 15.

Let X=0,1,dx,y=x-y2. Then X,d,2 is a b-complete b-metric space. Let T:XX be defined as Tx=1/4x2. Taking δ=ε, we get, for x and y satisfying εdx,y<ε+δ=2ε, (21)s·dTx,Ty=2·dTx,Ty=2·116x-y2x+y212x-y2<ε.Hence, all the conditions of Theorem 13 are satisfied. The mapping T has a unique fixed point, which is u=0.

Let Gs be the class of all mappings g:[0,)[0,1/s) which satisfy the condition: gtn1/s, whenever tn0. Note that Gs. As an example, consider the mapping g:[0,)[0,1/s) given by gt=1/se-t for t>0 and g0[0,1/s).

The following is Geraghty type result in the context of b-metric spaces (see, for instance, , where authors use Lemma 1.4.).

Theorem 16.

Let X,d,s>1 be a complete b-metric space. Suppose that the mapping T:XX satisfies the condition (22)dTx,Tygdx,ydx,yfor all x,yX and some gGs. Then T has a unique fixed point uX and for each xX the Picard sequence Tnx converges to u in X,d,s>1.

Proof.

Since g:[0,)[0,1/s), we get(23)dTx,Ty1sdx,y=λdx,y.In view of λ0,1, the result follows according to Lemma 8. and condition (15).

It is well known that, in compact metric spaces, fixed point results can be obtained under the strict contractive condition (dTx,Ty<dx,y whenever xy). In the case of b-metric spaces with a continuous b-metric, the following results of Nemytzki and Edelstein can be obtained in the same way as in the metric case (see , pages 56-58).

Theorem 17.

Let X,d,s>1 be a compact b-metric space with continuous b-metric d and let T:XX be a self-mapping. Suppose that the following condition holds: (24)dTx,Ty<dx,yfor  xy.Then T has a unique fixed point, say uX, and, for each xX,limnTnx=u.

Proof.

Define a function f:X[0,+) by(25)fx=dx,Tx.Since T is continuous, f is also continuous. So, as X,d,s>1 is a compact b-metric space, there exists a point uX such that(26)fu=du,Tu=minxXdx,Tx.If we assume that uTu, then as T is contractive (dTx,Ty<dx,y whether xy), one writes(27)fTu=dTu,TTu<du,Tu=fu,which is a contradiction. Therefore, u is a fixed point of T. The uniqueness is obvious.

Theorem 18.

Let X,d,s>1 be a b-metric space with a continuous b-metric d and let T:XX be a self-map. Suppose that the following condition holds: (28)dTx,Ty<dx,yfor  xy.If there exists a point x0X such that the sequence Tnx0 contains a convergent subsequence Tnix0 to u, then u is the unique fixed point of T.

Proof.

Consider the real sequence dTnx0,Tn+1x0. If Tk+1x0=Tkx0 for some kN, then Tnx0 for nk is a stationary sequence and so Tkx0=u. Thus, Tk+1x0=Tkx0 implies Tu=u. Assume now that Tn+1x0Tnx0 for all nN. Then as T is contractive, dTnx0,Tn+1x0 is a strictly decreasing sequence of positive reals. Therefore, it converges. Since Tnix0u as i and T is continuous, we have(29)limiTni+1x0=limiTTnix0=Tu;limiTni+2x0=T2u.Thus (30)limidTnix0,Tni+1x0=du,Tu,limidTni+1x0,Tni+2x0=dTu,T2u.Since {dTnix0,Tni+1x0} and {dTni+1x0,Tni+2x0} are subsequences of the convergent sequence dTnx0,Tn+1x0, they have the same limit. Therefore,(31)du,Tu=dTu,T2u.Hence Tu=u. If not, as T is contractive, we have(32)dTu,T2u<du,Tu,which is a contradiction.

Remark 19.

The two previous theorems are known in literature as Nemytzki and Edelstein theorems, respectively. It is clear that Edelstein theorem extends the result of Nemytzki.

In the sequel, we consider ε-contractive mappings in the context of b-metric spaces. Namely, we first introduce the following.

Definition 20.

A mapping T of a b-metric space X,d,s1 into itself is said to be ε-contractive, if and only if there exists ε>0 such that(33)0<dx,y<ε  implies  dTx,Ty<dx,y.The following results extend ones from standard metric spaces to b-metric spaces, with a continuous b-metric d.

Theorem 21.

Let X,d,s1 be a b-metric space with continuous b-metric d and T an ε-contractive self-mapping on X. If, for some x0X, the sequence Tnx0 has a convergent subsequence {Tnix0} to uX, then u is a periodic point; that is, there exists a positive integer k such that Tku=u.

Proof.

Since Tnx0 has a cluster point, there exist positive integers p and q with p<q such that dTpx0,Tqx0<ε; that is, dTpx0,TkTpx0<ε, where k=q-p. Then the sequence dTnx0,Tn+kx0n=p+ is nonincreasing due to the fact that T is ε-contractive. Thus, this sequence converges and so(34)limndTnx0,Tn+kx0=limidTnix0,Tni+kx0=limidTni+1x0,Tni+k+1x0.Since T and the b-metric d are both continuous, we have(35)du,Tku=limidTnix0,Tni+kx0dTpx0,TkTpx0<ε,and(36)dTu,TTku=limidTni+1x0,Tni+k+1x0.Thus,(37)dTu,TTku=du,Tku<ε.Hence, Tku=u. Otherwise, as T is ε-contractive, we would have dTu,TTku<du,Tku.

Now, consider a class of mappings T of a b-metric space X,d,s1 into itself which satisfy the following condition.

For every x,yX there exists a positive integer kx,y such that(38)0<dx,y  implies  dTnx,Tny<dx,ynkx,y.A mapping T satisfying (38) is called an eventually contractive mapping.

It is obvious that any contractive mapping is eventually contractive (it satisfies (38) with kx,y=1), but the implication is not reverse.

Contractive and ε-contractive mappings are continuous. However, eventually contractive mappings need not be continuous nor orbitally continuous. Recall that a mapping T is said to be orbitally continuous if, for each xX, TnixuX implies TTnixTu.

Now, we announce the next result.

Theorem 22.

Let X,d,s1 be a b-metric space with continuous b-metric d and T:XX be an eventually contractive and orbitally continuous mapping. If, for some x0X, the sequence of iterates Tnx0 has a subsequence {Tnix0} converging to uX, then u is the unique fixed point of T and limnTnx0=u.

Proof.

If dTn0x0,Tn0+1x0=0 for some n0N, then Tnx=Tn0x0 for all nn0. Thus, Tn0x0=u and so Tn0+1x0=Tn0x0 implies that Tu=u.

Assume now that dTn0x0,Tn0+1x0>0 for all nN. Consider u=limiTnix0. Since T is orbitally continuous, for any fixed positive integer p, we have(39)limidTni+px0,Tni+p+1x0=dTpu,Tp+1u.Assume that Tuu. As T is eventually contractive, there exists rN such that(40)dTru,TrTu<du,Tu.Hence ε0=1/2du,Tu-dTru,TrTu>0. Since(41)limidTni+rx0,Tni+r+1x0=dTru,TTu,for arbitrary ε>0, there exists a sufficiently large ni=q such that(42)dTq+rx0,Tq+r+1x0<dTru.TrTu+ε.For ε=ε0, we have(43)dTq+rx0,Tq+r+1x0<12du,Tu+dTru,TrTu.Since dTn0x0,Tn0+1x0>0 for all nN and T is eventually contractive, there exists some positive integer k=kTq+rx0,Tq+r+1x0 such that(44)dTnTq+rx0,TnTq+r+1x0<dTq+rx0,Tq+r+1x0nk;that is, dTnx0,Tn+1x0<dTq+rx0,Tq+r+1x0 for all nk+q+r. So, we obtain that(45)dTnx0,Tn+1x0<12du,Tu+dTru,TrTunk+q+r.Hence(46)dTnix0,Tni+1x0<12du,Tu+dTru,TrTunik+q+r.Thus we get(47)du,Tu=limidTnix0,Tni+1x012du,Tu+dTru,TrTu.Hence(48)du,TudTru,TrTu,which contradicts the choice of r. Therefore, du,Tu=0; that is, Tu=u.

Now, we show that limnTnx0=u. Let ε>0 be arbitrary. Since Tnix0u as i, there exists a sufficiently large ni=l such that du,Tlx0<ε. If Tnix0u, then Tl+1x0=Tu=u and hence Tnx0=u for all nl. Assume that du,Tlx0>0. As T is eventually ε-contractive and du,Tlx0<ε, there is k=ku,Tlx0N such that(49)dTnu,TnTlx0<du,Tlx0nk.Since Tu=u, we have(50)du,Tnx0=dTn-lu,Tn-lTlx0<du,Tlu<εnk+l.Therefore, limnTnx0=u.

Corollary 23.

If X,d,s1 is a compact b-metric space with a continuous b-metric d and T is a continuous and eventually contractive self-mapping on X,d, then T has a unique fixed point, say uX. Also, for every xX,limnTnx=u.

A mapping T of a b-metric space X,d,s1 into itself is said to be an eventually ε-contractive mapping if there exists ε>0 such that, for every x,yX, there is a positive integer kx,y such that (51)0<dx,y<ε  implies  dTnx,Tny<dx,ynkx,y.For eventually ε-contractive mappings in the context of b-metric spaces, we announce the next result.

Theorem 24.

Let X,d,s1 be a b-metric space with a continuous b-metric d and T:XX be an eventually ε-contractive and orbitally continuous self-map on X,d. If, for some x0X, the sequence Tnx0 has a convergent subsequence {Tnix0} to uX, then u is a periodic point of T.

Proof.

The proof is very similar to the ones in the previous theorem. Therefore, it is omitted.

Corollary 25.

If X,d,s1 is a compact b-metric space with a continuous b-metric d and T:XX is an eventually ε-contractive and continuous self-map on X,d, then the set of periodic points of T is not empty.

The following two results extend ones from standard metric spaces to b-metric spaces (see ).

Theorem 26.

Let X,d,s1 be a compact b-metric space with continuous b-metric d and T a continuous self-map on X such that for every x,yX there exists a positive integer nx,y such that (52)0<dx,y  implies  dTnx,yx,Tnx,yy<dx,y.Then T has a unique fixed point.

Proof.

Define f:X[0,+) by fx=dx,Tx. Since d and T are continuous, f is also continuous. Therefore, f attains its minimum on X in some point, say uX. Assume that fu=du,Tu>0. Then, by hypothesis, there exists a positive integer nu,Tu such that(53)dTnu,Tuu,Tnu,TuTu<du,Tu;that is, fTnu,Tuu<fu, which contradicts the choice of u. Therefore, Tu=u.

Theorem 27.

Let X,d,s1 be a compact b-metric space with a continuous b-metric d and T be a continuous self-map on X such that, for every x,yX, (54)0<dx,y<ε  implies  dTnx,yx,Tnx,yy<dx,y;nx,yN.Then the set of periodic points of T is not empty.

Proof.

Let x0X. Since X,d is compact, Tnx0 has a cluster point. Therefore, there exist positive integers p and k such that(55)dTpx0,TkTpx0=dTpx0,Tk+px0<ε.Define fx0=dx0,Tkx0. Since T is continuous, Tk is continuous and so f is continuous. Therefore, there is some uX such that fu=minfx:xX. Then(56)fu=du,Tku=mindx,Tkx:xXdTpx0,TkTpx0<ε.Assume that Tkuu. Then, as du,Tku<ε, by hypothesis there exists n=nu,TkuN such that(57)dTnu,TnTku<du,Tku.Thus, fTnu<fu, which contradicts the choice of u. Therefore, Tku=u.

Data Availability

No data is used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest regarding the publication of this paper.