# Nemytzki-Edelstein-Meir-Keeler Type Results inb-Metric Spaces

Authors: Hassen Aydi; Radoje Banković; Ivan Mitrović; Muhammad Nazam
Journal: Discrete Dynamics in Nature and Society
 (2018)

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## Abstract

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.


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## Body

## 1. Definitions, Notations, and Preliminaries

LetX,d be a metric space and T:X→X be a self-mapping. The following Meir-Keeler conditions are well known: For each ε>0, there exists δ=δε>0 such that(1)ε≤dx,y<ε+δimpliesdTx,Ty<ε,or(2)ε<dx,y<ε+δimpliesdTx,Ty≤ε,or T is contractive and(3)ε≤dx,y<ε+δimpliesdTx,Ty≤ε.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 x≠y. For more details, see [1] (pages 30-33 and 56-58) and [2]. In 1969, Meir-Keeler [2] proved the following.Theorem 1 ([2], Theorem).
LetX,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 u∈X, and, for each x∈X, limn→∞Tnx=u.For other fixed point results via generalized Meir-Keeler contractions, see [3–5]. Inspired from Meir-Keeler theorem, Ćirić proved the next slightly more general result.Theorem 2 ([1], Theorem 2.5).
LetX,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 u∈X, and, for each x∈X, limn→∞Tnx=u.The following example shows that Ćirić result is a proper generalization of Meir-Keeler theorem.Example 3.
LetX=[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,if0≤x≤1,x=2,⋯,3n-1,⋯(5)Tx=1,ifx=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 [1].Remark 4.
Theorems1 and 2 are true if the self-mapping T:X→X satisfies condition (3). For more details, see [1], pages 30-33.On the other hand, Bakhtin [6] and Czerwik [7] introduced the concept ofb-metric spaces (a generalization of metric spaces) and proved the Banach contraction principle. The definition of ab-metric space is the following.Definition 5 (Bakhtin [6] and Czerwik [7]).
LetX be a nonempty set and let s≥1 be a given real number. A function d:X×X→[0,∞) is said to be ab-metric if and only if, for all x,y,z∈X, the following conditions are satisfied:(b1)
dx,y=0 if and only if x=y.(b2)
dx,y=dy,x.(b3)
dx,z≤sdx,y+dy,z.The tripletX,d,s≥1 is called ab-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 ofb-metric spaces. For more details, see [5, 8–35]. It should be noted that the class ofb-metric spaces is effectively larger than that of standard metric spaces, since ab-metric is a metric when s=1. The following example shows that, in general, ab-metric does not necessarily need to be a metric.Example 6.
LetX,ρ be a metric space and dx,y=ρx,yp where p>1 is a real number. Then d is ab-metric with s=2p-1, but d is not a metric on X.The concepts ofb-convergence,b-completeness,b-Cauchy, andb-closed set inb-metric spaces have been initiated in [6, 7].The following two lemmas are very significant in the class ofb-metric spaces.Lemma 7 ([21], Lemma 3.1).
Letyn be a sequence in ab-metric space X,d,s≥1 such that(6)dyn,yn+1≤λdyn-1,ynfor some λ∈[0,1/s), and each n=1,2,⋯. Then yn is ab-Cauchy sequence in a b-metric space X,d.Lemma 8 ([30], Lemma 2.2).
Letyn be a sequence in ab-metric space X,d,s≥1 such that(7)dyn,yn+1≤λdyn-1,ynfor some λ∈[0,1), and each n=1,2,⋯. Then yn is ab-Cauchy sequence in ab-metric space X,d.Since in general a b-metric is not continuous, we need the following two lemmas.Lemma 9 ([36], Lemma 2.1).
LetX,d,s≥1 be a b-metric space with s≥1. Suppose that xn and yn are b-convergent to x and y, respectively. Then (8)1s2dx,y≤liminfn→∞dxn,yn≤limsupn→∞dxn,yn≤s2dx,y.In particular, if x=y, then we have limn→∞dxn,yn=0. Moreover, for each z∈X, we have (9)1sdx,z≤liminfn→∞dxn,z≤limsupn→∞dxn,z≤s2dx,z.Lemma 10 (see [37]).
LetX,d,s≥1 be a b-metric space and xn be a sequence in X such that (10)limn→∞dxn,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)ε≤liminfk→∞dxmk,xnk≤limsupk→∞dxmk,xnk≤εs,εs≤liminfk→∞dxmk,xnk+1≤limsupk→∞dxmk,xnk+1≤εs2,εs≤liminfk→∞dxmk+1,xnk≤limsupk→∞dxmk+1,xnk≤εs2,εs2≤liminfk→∞dxmk+1,xnk+1≤limsupk→∞dxmk+1,xnk+1≤εs3.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 ofb-metric spaces. Also, it is not known that if there are examples such that condition (1) or (2) or (3) holds in the context ofb-metric spaces, but T has no fixed point.Our first result generalizes Lemma 1 of [2]. For some results also see recent paper [38].Lemma 11.
LetX,d,s>1 be ab-completeb-metric space and T:X→X such that condition (1) holds. If Tnx is ab-Cauchy sequence for each x∈X, then T has a unique fixed point, say u∈X and Tnx→u.Proof.
SinceX,d,s>1 isb-complete, each {Tnx} has a limit point, say ηx. Since condition (1) implies the continuity of T, we have(13)Tηx=Tlimn→∞Tnx=limn→∞Tn+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.
LetX,d be a complete b-metric space and let T be a self-mapping on X satisfying the following condition.
Givenε>0 there exists δ>0 such that (14)ε≤dx,y<ε+δimpliessadTx,Ty<ε,where a>0 is given.
ThenT has a unique fixed point, say u∈X, and, for each x∈X,limn→∞Tnx=u.Proof.
It is clear that, for allx,y∈X with dx,y>0, we obtain(15)dTx,Ty≤λdx,y,where λ=1/sa∈[0,1).
Letx0∈X be an arbitrary point. Define the sequence xn by xn+1=Txn for all n≥0. If, for some n, xn=xn+1, then xn is a fixed point of T. From now on, suppose that xn≠xn+1 for all n≥0. From condition (15), we obtain(16)dxn,xn+1≤λdxn-1,xn.Further, according to ([30], Lemma 2.2.), the sequence xn isb-Cauchy in theb-metric space X,d. Byb-completeness of X,d, there exists u∈X such that(17)limn→∞xn=u.Finally, (15) and (17) imply that Tu=u; that is, u is the unique fixed point of T in X.Example 14.
LetX=0,1,2 and define d:X×X→[0,+∞) as follows: dx,x=0, dx,y=dy,x for all x,y∈X, d0,1=1, d0,2=2.2, and d1,2=1.1. Then X,d,22/21>1 is ab-completeb-metric space, but it is not a metric space. Let T:X→X be defined by (18)Tx=0,ifx≠2,1,ifx=2.We shall check that, for all x,y∈X, the contractive condition (15) holds. For this, we distinguish three cases.
(a) x=0,y=1⇒dT0,T1=d0,0=0. Obviously, condition (15) holds.
(b) x=0,y=2⇒dT0,T2=d0,1. Since 22/21ad0,1≤d0,2, i.e., 22/21a·1≤2.2,a>0, which is true, hence, again (15) holds.
(c) x=1,y=2⇒dT1,T2=d0,1=1. Now, we have 22/21a·1≤1.1, i.e., 22/21a≤1.1, 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)ε≤d0,2<ε+δimplies2221<ε,or equivalently (20)ε≤2.2<ε+δimplies2221<ε.Take ε=1/2. Then there exists δ=δ1/2>0 such that 1/2≤2.2<1/2+δ (for example, any δ>17/10). But 22/21<1/2 is false.Now, we give an example supporting Theorem13.Example 15.
LetX=0,1,dx,y=x-y2. Then X,d,2 is a b-complete b-metric space. Let T:X→X be defined as Tx=1/4x2. Taking δ=ε, we get, for x and y satisfying ε≤dx,y<ε+δ=2ε, (21)s·dTx,Ty=2·dTx,Ty=2·116x-y2x+y2≤12x-y2<ε.Hence, all the conditions of Theorem 13 are satisfied. The mapping T has a unique fixed point, which is u=0.LetGs be the class of all mappings g:[0,∞)→[0,1/s) which satisfy the condition: gtn→1/s, whenever tn→0. 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 ofb-metric spaces (see, for instance, [17], where authors use Lemma 1.4.).Theorem 16.
LetX,d,s>1 be a completeb-metric space. Suppose that the mapping T:X→X satisfies the condition (22)dTx,Ty≤gdx,ydx,yfor all x,y∈X and some g∈Gs. Then T has a unique fixed point u∈X and for each x∈X the Picard sequence Tnx converges to u in X,d,s>1.Proof.
Sinceg:[0,∞)→[0,1/s), we get(23)dTx,Ty≤1sdx,y=λdx,y.In view of λ∈0,1, 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 x≠y). 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 [1], pages 56-58).Theorem 17.
LetX,d,s>1 be a compact b-metric space with continuous b-metric d and let T:X→X be a self-mapping. Suppose that the following condition holds: (24)dTx,Ty<dx,yforx≠y.Then T has a unique fixed point, say u∈X, and, for each x∈X,limn→∞Tnx=u.Proof.
Define a functionf: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 u∈X such that(26)fu=du,Tu=minx∈Xdx,Tx.If we assume that u≠Tu, then as T is contractive (dTx,Ty<dx,y whether x≠y), 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.
LetX,d,s>1 be a b-metric space with a continuous b-metric d and let T:X→X be a self-map. Suppose that the following condition holds: (28)dTx,Ty<dx,yforx≠y.If there exists a point x0∈X 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 sequencedTnx0,Tn+1x0. If Tk+1x0=Tkx0 for some k∈N, then Tnx0 for n≥k is a stationary sequence and so Tkx0=u. Thus, Tk+1x0=Tkx0 implies Tu=u. Assume now that Tn+1x0≠Tnx0 for all n∈N. Then as T is contractive, dTnx0,Tn+1x0 is a strictly decreasing sequence of positive reals. Therefore, it converges. Since Tnix0→u as i→∞ and T is continuous, we have(29)limi→∞Tni+1x0=limi→∞TTnix0=Tu;limi→∞Tni+2x0=T2u.Thus (30)limi→∞dTnix0,Tni+1x0=du,Tu,limi→∞dTni+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 considerε-contractive mappings in the context ofb-metric spaces. Namely, we first introduce the following.Definition 20.
A mappingT of ab-metric space X,d,s≥1 into itself is said to be ε-contractive, if and only if there exists ε>0 such that(33)0<dx,y<εimpliesdTx,Ty<dx,y.The following results extend ones from standard metric spaces tob-metric spaces, with a continuousb-metric d.Theorem 21.
LetX,d,s≥1 be a b-metric space with continuous b-metric d and T an ε-contractive self-mapping on X. If, for some x0∈X, the sequence Tnx0 has a convergent subsequence {Tnix0} to u∈X, then u is a periodic point; that is, there exists a positive integer k such that Tku=u.Proof.
SinceTnx0 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)limn→∞dTnx0,Tn+kx0=limi→∞dTnix0,Tni+kx0=limi→∞dTni+1x0,Tni+k+1x0.Since T and theb-metric d are both continuous, we have(35)du,Tku=limi→∞dTnix0,Tni+kx0≤dTpx0,TkTpx0<ε,and(36)dTu,TTku=limi→∞dTni+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 mappingsT of a b-metric space X,d,s≥1 into itself which satisfy the following condition.For everyx,y∈X there exists a positive integer kx,y such that(38)0<dx,yimpliesdTnx,Tny<dx,y∀n≥kx,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 andε-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 x∈X, Tnix→u∈X implies TTnix→Tu.Now, we announce the next result.Theorem 22.
LetX,d,s≥1 be a b-metric space with continuous b-metric d and T:X→X be an eventually contractive and orbitally continuous mapping. If, for some x0∈X, the sequence of iterates Tnx0 has a subsequence {Tnix0} converging to u∈X, then u is the unique fixed point of T and limn→∞Tnx0=u.Proof.
IfdTn0x0,Tn0+1x0=0 for some n0∈N, then Tnx=Tn0x0 for all n≥n0. Thus, Tn0x0=u and so Tn0+1x0=Tn0x0 implies that Tu=u.
Assume now thatdTn0x0,Tn0+1x0>0 for all n∈N. Consider u=limi→∞Tnix0. Since T is orbitally continuous, for any fixed positive integer p, we have(39)limi→∞dTni+px0,Tni+p+1x0=dTpu,Tp+1u.Assume that Tu≠u. As T is eventually contractive, there exists r∈N such that(40)dTru,TrTu<du,Tu.Hence ε0=1/2du,Tu-dTru,TrTu>0. Since(41)limi→∞dTni+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 n∈N 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+1x0∀n≥k;that is, dTnx0,Tn+1x0<dTq+rx0,Tq+r+1x0 for all n≥k+q+r. So, we obtain that(45)dTnx0,Tn+1x0<12du,Tu+dTru,TrTu∀n≥k+q+r.Hence(46)dTnix0,Tni+1x0<12du,Tu+dTru,TrTu∀ni≥k+q+r.Thus we get(47)du,Tu=limi→∞dTnix0,Tni+1x0≤12du,Tu+dTru,TrTu.Hence(48)du,Tu≤dTru,TrTu,which contradicts the choice of r. Therefore, du,Tu=0; that is, Tu=u.
Now, we show thatlimn∞Tnx0=u. Let ε>0 be arbitrary. Since Tnix0→u as i→∞, there exists a sufficiently large ni=l such that du,Tlx0<ε. If Tnix0≠u, then Tl+1x0=Tu=u and hence Tnx0=u for all n≥l. Assume that du,Tlx0>0. As T is eventually ε-contractive and du,Tlx0<ε, there is k=ku,Tlx0∈N such that(49)dTnu,TnTlx0<du,Tlx0∀n≥k.Since Tu=u, we have(50)du,Tnx0=dTn-lu,Tn-lTlx0<du,Tlu<ε∀n≥k+l.Therefore, limn→∞Tnx0=u.Corollary 23.
IfX,d,s≥1 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 u∈X. Also, for every x∈X,limn→∞Tnx=u.A mappingT of ab-metric space X,d,s≥1 into itself is said to be an eventually ε-contractive mapping if there exists ε>0 such that, for every x,y∈X, there is a positive integer kx,y such that (51)0<dx,y<εimpliesdTnx,Tny<dx,y∀n≥kx,y.For eventually ε-contractive mappings in the context ofb-metric spaces, we announce the next result.Theorem 24.
LetX,d,s≥1 be a b-metric space with a continuous b-metric d and T:X→X be an eventually ε-contractive and orbitally continuous self-map on X,d. If, for some x0∈X, the sequence Tnx0 has a convergent subsequence {Tnix0} to u∈X, 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.
IfX,d,s≥1 is a compact b-metric space with a continuous b-metric d and T:X→X 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 [15]).Theorem 26.
LetX,d,s≥1 be a compact b-metric space with continuous b-metric d and T a continuous self-map on X such that for every x,y∈X there exists a positive integer nx,y such that (52)0<dx,yimpliesdTnx,yx,Tnx,yy<dx,y.Then T has a unique fixed point.Proof.
Definef: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 u∈X. 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.
LetX,d,s≥1 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,y∈X, (54)0<dx,y<εimpliesdTnx,yx,Tnx,yy<dx,y;nx,y∈N.Then the set of periodic points of T is not empty.Proof.
Letx0∈X. 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 u∈X such that fu=minfx:x∈X. Then(56)fu=du,Tku=mindx,Tkx:x∈X≤dTpx0,TkTpx0<ε.Assume that Tku≠u. Then, as du,Tku<ε, by hypothesis there exists n=nu,Tku∈N such that(57)dTnu,TnTku<du,Tku.Thus, fTnu<fu, which contradicts the choice of u. Therefore, Tku=u.



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