R E S E A R C H
Open Access
Ordered
S
p
-metric spaces and some fixed
point theorems for contractive mappings
with application to periodic boundary value
problems
Z. Mustafa
1*, R.J. Shahkoohi
2, V. Parvaneh
3, Z. Kadelburg
4and M.M.M. Jaradat
1*Correspondence:[email protected] 1Department of Mathematics, Statistics and Physics, Qatar University, Doha, Qatar Full list of author information is available at the end of the article
Abstract
In this paper, we introduce the structure ofSp-metric spaces as a generalization of
bothS-metric andSb-metric spaces. Also, we present the notions ofS-contractive
mappings in the setup of orderedSp-metric spaces and investigate the existence of a
fixed point for such mappings under various contractive conditions. We provide examples to illustrate the results presented herein. An application to periodic boundary value problems is presented.
MSC: Primary 47H10; secondary 54H25
Keywords: Fixed point; Complete metric space; Orderedb-metric space
1 Introduction and preliminaries
There is a large number of generalizations of Banach contraction principle via using dif-ferent forms of contractive conditions in various generalized metric spaces. Some of such generalizations are obtained via contractive conditions expressed by rational terms (see, e.g., [7,9,17,19]).
Ran and Reurings initiated the studying of fixed point results on partially ordered sets in [14]. Further, many researchers have focused on different contractive conditions in com-plete metric spaces endowed with a partial order. For more details, we refer the reader to [11,12].
Parvaneh introduced in [13] the concept of extendedb-metric spaces as follows.
Definition 1.1([13]) LetΞbe a (nonempty) set. A functiond˜:Ξ×Ξ→R+is ap-metric
if there exists a strictly increasing continuous functionΩ: [0,∞)→[0,∞) withΩ–1(t)≤
t≤Ω(t) andΩ–1(0) = 0 =Ω(0) such that, for allζ,η,ω∈Ξ, the following conditions hold: (d1) d(ζ,η) = 0iffζ =η,
(d2) d(ζ,η) =d(η,ζ),
(d3) d(ζ,ω)≤Ω(d(ζ,η) +d(η,ω)).
In this case, the pair (Ξ,d) is called ap-metric space or an extendedb-metric space.
Ab-metric [2] is ap-metric withΩ(t) =stfor some fixeds≥1, while a metric is ap -metric whenΩ(t) =t. We have the following proposition.
Proposition 1.2([13]) Let(Ξ,d)be a metric space,and letd(ζ,η) =ξ(d(ζ,η)),whereξ : [0,∞)→[0,∞)is a strictly increasing continuous function with t≤ξ(t)and0 =ξ(0).In this case,d is a p-metric withΩ(t) =ξ(t).
The above proposition provides several examples ofp-metric spaces.
Example1.3 Let (Ξ,d) be a metric space, and letd(ζ,η) =ed(ζ,η)sec–1(ed(ζ,η)). Thendis a
p-metric withΩ(t) =etsec–1(et).
In [16], Sedghi et al. introduced the notion of anS-metric space as follows.
Definition 1.4([16]) LetΞ be a nonempty set andS:Ξ×Ξ×Ξ →R+be a function
satisfying the following properties:
(S1) S(ζ,η,ω) = 0iffζ=η=ω;
(S2) S(ζ,η,ω)≤S(ζ,ζ,a) +S(η,η,a) +S(ω,ω,a)for allζ,η,ω,a∈Ξ(rectangle inequality).
Then the functionSis called anS-metric onΞ and the pair (Ξ,S) is called anS-metric space.
Example1.5 ([16]) Let R be the real line. ThenS(ζ,η,ω) =|ζ–η|+|ζ–ω|for allζ,η,ω∈R
is anS-metric on R. ThisS-metric on R is called the usualS-metric on R.
Souayaha and Mlaiki in [18], motivated by the concepts ofb-metric andS-metric, in-troduced the concept ofSb-metric spaces, and then they presented some basic properties of such spaces.
The following is the definition of modifiedS-metric spaces, which is a proper general-ization of the notions ofS-metric spaces andSb-metric spaces.
Definition 1.6 LetΞ be a nonempty set andΩ: [0,∞)→[0,∞) be a strictly increasing continuous function such thatt≤Ω(t) for allt> 0 andΩ(0) = 0. Suppose that a mapping S:Ξ×Ξ×Ξ→R+satisfies:
(S1) S(ζ,η,ω) = 0iffζ=η=ω,
(S2) S(ζ,η,ω)≤Ω[S(ζ,ζ,α) +S(η,η,α) +S(ω,ω,α)]for allζ,η,ω,α∈Ξ (rectangle in-equality).
ThenSis called anSp-metric and the pair (Ξ,S) is called anSp-metric space.
Remark1.7 In anSp-metric space, we haveS(ζ,ζ,η)≤Ω[S(η,η,ζ)] for allζ,η∈Ξ. In-deed, putting (ζ,ζ,η,ζ) instead of (ζ,η,ω,α) in (S2) and using (S1), we obtain the previous inequality.
EachS-metric space is anSp-metric space withΩ(t) =tand everySb-metric space with parameters≥1 is anSp-metric space withΩ(t) =st.
Proof For allζ,η,ω,a∈Ξ,
S(ζ,η,ω) =ξS(ζ,η,ω)≤ξsS(ζ,ζ,a) +sS(η,η,a) +sS(ω,ω,a)
≤ξ(sξS(ζ,ζ,a)+sξS(η,η,a) +sξS(ω,ω,a)
=ξsS(ζ,ζ,a) +sS(η,η,a) +sS(ω,ω,a)
=ΩS(ζ,ζ,a) +S(η,η,a) +S(ω,ω,a).
So,Sis anSp-metric withΩ(t) =ξ(st).
The above proposition provides several examples ofSp-metric spaces.
Example1.9 Let (Ξ,S) be anSb-metric space with coefficients≥1. Then:
1. S(ζ,η,ω) =eS(ζ,η,ω)sec–1(eS(ζ,η,ω))is anS
p-metric withΩ(t) =estsec–1(est).
2. S(ζ,η,ω) = [S(ζ,η,ω) + 1]sec–1([S(ζ,η,ω) + 1])is anS
p-metric with Ω(t) = [st+ 1]sec–1([st+ 1]).
3. S(ζ,η,ω) =eS(ζ,η,ω)tan–1(eS(ζ,η,ω)– 1)is anS
p-metric withΩ(t) =esttan–1(est– 1).
4. S(ζ,η,ω) =S(ζ,η,ω)cosh(S(ζ,η,ω))is anSp-metric withΩ(t) =stcosh(st).
5. S(ζ,η,ω) =eS(ζ,η,ω)ln(1 +S(ζ,η,ω))is anS
p-metric withΩ(t) =estln(1 +st).
6. S(ζ,η,ω) =S(ζ,η,ω) +ln(1 +S(ζ,η,ω))is anSp-metric withΩ(t) =st+ln(1 +st). In all the given examples 1–6, it can be checked by routine calculation that the respec-tive functionξsatisfies all the requirements given in Proposition1.8, i.e., it is continuous, strictly increasing,ξ(0) = 0, andξ(t) >tfort> 0.
Definition 1.10 LetΞbe anSp-metric space. A sequence{ζn}inΞis said to be:
(1) Sp-Cauchy if, for eachε> 0, there exists a positive integern0such that, for all
m,n≥n0,S(ζm,ζn,ζn) <ε.
(2) Sp-convergent to a pointζ∈Ξ if, for eachε> 0, there exists a positive integern0 such that, for alln≥n0,S(ζn,ζn,ζ) <ε.
(3) AnSp-metric spaceΞis calledSp-complete if everySp-Cauchy sequence is
Sp-convergent inΞ.
In general, anSp-metric mappingS(ζ,η,ω) with a nontrivial functionΩ need not be jointly continuous in all its variables (see [10]). Thus, in some proofs we will need the following simple lemma about theSp-convergent sequences.
Lemma 1.11 Let(Ξ,S)be an Sp-metric space.
1. Suppose that{ζn}and{ηn}areSp-convergent toζ andη,respectively.Then we have
Ω–1[1 2Ω
–1[S(ζ,η,η)]]
2 ≤lim infn−→∞S(ζn,ηn,ηn)≤lim supn−→∞ S(ζn,ηn,ηn)
≤Ω2ΩS(ζ,η,η).
In particular,ifζ=η,then we havelimn→∞S(ζn,ηn,ηn) = 0.
2. Suppose that{ζn}isSp-convergent toζ andω∈Ξis arbitrary.Then we have
Ω–1[S(ζ,ω,ω)]
2 ≤lim infn−→∞S(ζn,ω,ω)≤lim supn−→∞S(ζn,ω,ω)≤Ω
Proof 1. Using the rectangle inequality in theSp-metric space, it is easy to see that
S(ζ,η,η)≤ΩS(ζ,ζ,ζn) + 2S(η,η,ζn)
≤ΩS(ζ,ζ,ζn) + 2Ω
2S(η,η,ηn) +S(ηn,ζn,ζn)
and
S(ζn,ηn,ηn)≤ΩS(ζn,ζn,ζ) + 2S(ηn,ηn,ζ)
≤ΩS(ζn,ζn,ζ) + 2Ω
2S(ηn,ηn,η) +S(η,ζ,ζ)
.
Taking the lower limit asn→ ∞in the first inequality and the upper limit asn→ ∞in the second inequality, we obtain the desired result.
2. Using the rectangle inequality, we see that
S(ζ,ω,ω)≤ΩS(ζ,ζ,ζn) + 2S(ω,ω,ζn)
and
S(ζn,ω,ω)≤ΩS(ζn,ζn,ζ) + 2S(ω,ω,ζ)
.
LetBdenote the class of all real functionsβ: [0,∞)→[0, 1) satisfying the condition
β(tn)→1 implies that tn→0, asn→ ∞.
In order to generalize the Banach contraction principle, Geraghty proved in 1973 the fol-lowing result.
Theorem 1.12([5]) Let(Ξ,d)be a complete metric space,and let f :Ξ →Ξ be a self-map.Suppose that there existsβ∈Bsuch that
d(fζ,fη)≤βd(ζ,η)d(ζ,η)
holds for allζ,η∈Ξ.Then f has a unique fixed pointω∈Ξand for eachζ∈Ξthe Picard sequence{fnζ}converges toω.
In 2010, Amini-Harandi and Emami [1] characterized the result of Geraghty in the set-ting of a partially ordered complete metric space. In [4], some fixed point theorems for mappings satisfying Geraghty-type contractive conditions were proved in various gener-alized metric spaces. Also, Zabihi and Razani [19] and Shahkoohi and Razani [17] obtained some fixed point results for rational Geraghty contractions inb-metric spaces.
2 Main results
2.1 Fixed point results using Geraghty contractions
Let (Ξ,S) be anSp-metric space with functionΩ, and letBΩ denote the class of all func-tionsβ: [0,∞)→[0,Ω–1(1)) satisfying the following condition:
lim sup
n→∞ β(tn) =Ω
–1(1) implies that t
n→0, asn→ ∞.
Example2.1
(1) LetΞ=RandS(ζ,η,ω) =e|ζ–η|+|η–ω|– 1for allζ,η,ω∈R, withΩ(t) =et– 1. Then,
byβ(t) = (ln2)e–tfort> 0andβ(0)∈[0,ln2), a functionβbelonging toBΩis given.
(2) Another example of a function inBΩ may be given byβ(t) =W(1)e–tfort> 0and β(0)∈[0,W(1)), whereS(ζ,η,ω) = (|ζ–η|+|η–ω|)e|ζ–η|+|η–ω|for allζ,η,ω∈R.
Here,Wis the LambertW-function (see, e.g., [3]).
Definition 2.2 Let (Ξ,,S) be an ordered Sp-metric space. A mapping f :Ξ →Ξ is called anSp-Geraghty contraction if there existsβ∈BΩ such that
Ω22S(fζ,fη,fω)≤βM(ζ,η,ω)M(ζ,η,ω) (2.1)
for all mutually comparable elementsζ,η,ω∈Ξ, where
M(ζ,η,ω) =max
S(ζ,η,ω),Ω
–1[S(ζ,fζ,fη)]
2 ,S(η,fη,fω)
.
An orderedSp-metric space (Ξ,,S) is said to have the s.l.c. property if, whenever{ζn} is an increasing sequence inΞ such thatζn→u∈Ξ, one hasζnufor alln∈N.
Theorem 2.3 Let(Ξ,,S)be an ordered Sp-complete Sp-metric space.Let f:Ξ→Ξ be
an increasing mapping with respect tosuch that there exists an elementζ0∈Ξ with
ζ0fζ0.Suppose that f is an Sp-Geraghty contraction.If
(I) f is continuous,or
(II) (Ξ,,S)has the s.l.c.property,
then f has a fixed point.Moreover,the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Proof Putζn=fnζ0. Sinceζ0ζ1 andf is increasing, we obtain by induction that the
sequence{ζn}is increasing w.r.t.. We will show thatlimn→∞S(ζn,ζn+1,ζn+1) = 0.
With-out loss of generality, we may assume thatζn=ζn+1 for alln∈N. Sinceζnandζn+1 are
comparable, then by (2.1) we have
S(ζn,ζn+1,ζn+1) =S(fζn–1,fζn,fζn)≤β
M(ζn–1,ζn,ζn)
M(ζn–1,ζn,ζn), (2.2)
where
M(ζn–1,ζn,ζn)
=max
S(ζn–1,ζn,ζn), 1 2Ω
–1S(ζ
n–1,fζn–1,fζn)
=max
S(ζn–1,ζn,ζn), 1 2Ω
–1S(ζ
n–1,ζn,ζn+1)
,S(ζn,ζn+1,ζn+1)
≤max
S(ζn–1,ζn,ζn),
S(ζn–1,ζn–1,ζn) +S(ζn+1,ζn+1,ζn)
2 ,S(ζn,ζn+1,ζn+1)
=max S(ζn–1,ζn,ζn),S(ζn,ζn+1,ζn+1)
.
Ifmax{S(ζn–1,ζn,ζn),S(ζn,ζn+1,ζn+1)}=S(ζn,ζn+1,ζn+1), then from (2.2) we have
S(ζn,ζn+1,ζn+1)≤β
M(ζn–1,ζn,ζn)S(ζn,ζn+1,ζn+1)
<Ω–1(1)S(ζn,ζn+1,ζn+1)
≤S(ζn,ζn+1,ζn+1),
which is a contradiction.
Hence,max{S(ζn–1,ζn,ζn),S(ζn,ζn+1,ζn+1)}=S(ζn–1,ζn,ζn). So, from (2.2),
S(ζn,ζn+1,ζn+1)≤β
M(ζn–1,ζn,ζn)S(ζn–1,ζn,ζn) <S(ζn–1,ζn,ζn). (2.3)
That is, {S(ζn,ζn+1,ζn+1)} is a decreasing sequence. Then there exists r≥0 such that
limn→∞S(ζn,ζn+1,ζn+1) =r. We will prove thatr= 0. Suppose, on the contrary, thatr> 0.
Then, lettingn→ ∞, from (2.3) we have
r≤ lim
n→∞β
M(ζn–1,ζn,ζn)
r≤Ω–1(1)r,
which implies thatΩ–1(1)≤1≤limn→∞β(M(ζn–1,ζn,ζn))≤Ω–1(1). Now, asβ ∈BΩ, we conclude thatM(ζn–1,ζn,ζn)→0, which yields thatr= 0, a contradiction. Hence, the assumption thatr> 0 is false. That is,
lim
n→∞S(ζn,ζn+1,ζn+1) =nlim→∞S(ζn,ζn,ζn+1) = 0. (2.4)
Now, we prove that the sequence{ζn}is anSp-Cauchy sequence. Suppose the contrary,
i.e., there existsε> 0 for which we can find two subsequences{ζmi}and{ζni}of{ζn}such
thatniis the smallest index for which
ni>mi>i and S(ζmi,ζni,ζni)≥ε. (2.5)
This means that
S(ζmi,ζni–1,ζni–1) <ε. (2.6)
From the rectangular inequality, we get
S(ζmi,ζmi+1,ζni)
≤ΩS(ζmi,ζmi,ζni–1) +S(ζmi+1,ζmi+1,ζni–1) +S(ζni,ζni,ζni–1)
≤ΩS(ζmi,ζmi,ζni–1) +Ω
2S(ζmi+1,ζmi+1,ζmi) +S(ζni–1,ζni–1,ζmi)
+S(ζni,ζni,ζni–1)
Taking the upper limit asi→ ∞and by (2.4), we get
lim sup
i→∞
S(ζmi,ζmi+1,ζni)≤Ω
ε+Ω(ε).
From the rectangular inequality, we get
ε≤S(ζmi,ζni,ζni)≤ΩS(ζmi,ζmi,ζmi+1) +S(ζni,ζni,ζmi+1) +S(ζni,ζni,ζmi+1)
.
Taking the upper limit asi→ ∞and by (2.4), we get
1 2Ω
–1(ε)≤lim sup
i→∞
S(ζmi+1,ζni,ζni).
From the definition ofM(ζ,η,ω) and the above limits,
lim sup
i→∞
M(ζmi,ζni–1,ζni–1)
=lim sup
i→∞
max
S(ζmi,ζni–1,ζni–1),
1 2Ω
–1S(ζ
mi,fζmi,fζni–1)
,
S(ζni–1,fζni–1,fζni–1)
=lim sup
i→∞ max
S(ζmi,ζni–1,ζni–1),
1 2Ω
–1S(ζ
mi,ζmi+1,ζni)
,
S(ζni–1,ζni,ζni)
≤Ω(ε).
Now, since the sequence{ζn}is increasing, we can apply (2.1) and the above inequalities to get
Ω(ε) =Ω2
2·1 2Ω
–1(ε)
≤Ω2
lim sup
i→∞
2S(ζmi+1,ζni,ζni)
≤lim sup
i→∞ β
M(ζmi,ζni–1,ζni–1)
lim sup
i→∞ M(ζmi,ζni–1,ζni–1)
≤Ω(ε)·lim sup
i→∞ β
M(ζmi,ζni–1,ζni–1)
,
which implies thatΩ–1(1)≤1≤lim
n→∞β(M(ζmi,ζni–1,ζni–1))≤Ω–1(1). Now, asβ∈BΩ
we conclude thatM(ζmi,ζni–1,ζni–1)→0, which yields thatS(ζmi,ζni–1,ζni–1)→0.
Conse-quently,
S(ζmi,ζni,ζni)≤ΩS(ζmi,ζmi,ζni–1) + 2S(ζni,ζni,ζni–1)
→0,
We will prove thatuis a fixed point off. First, letf be continuous. Then we have
u= lim
n→∞ζn+1=nlim→∞fζn=fu.
Now, let (II) hold. Using the assumption onΞ, we have thatζnufor alln∈N. Now, by Lemma1.11,
Ω2
2Ω
–1[S(u,fu,fu)]
2
≤Ω2
2lim sup
n→∞
S(ζn+1,fu,fu)
≤lim sup
n→∞ β
M(ζn,u,u)
lim sup
n→∞ M(ζn,u,u),
where
lim
n→∞M(ζn,u,u) =nlim→∞max S(ζn,u,u),S(ζn,ζn+1,fu),S(u,fu,fu)
=S(u,fu,fu).
Therefore, we deduce thatS(u,fu,fu) = 0, sou=fu.
Finally, suppose that the set of fixed points off is well ordered. Assume on the contrary, thatuandvare two fixed points off such thatu=v. Then, by (2.1), we have
S(u,v,v) =S(fu,fv,fv)≤βM(u,v,v)M(u,v,v) =βS(u,v,v)S(u,v,v)
<Ω–1(1)S(u,v,v)
becauseM(u,v,v) =S(u,v,v). So, we getS(u,v,v) <Ω–1(1)S(u,v,v), a contradiction. Hence,
u=v, andf has a unique fixed point. Conversely, iff has a unique fixed point, then the set
of fixed points off is trivially well ordered.
2.2 Fixed point results via comparison functions
LetΨ be the family of all nondecreasing functionsψ: [0,∞)→[0,∞) such that
lim
n→∞ψ n(t) = 0
for allt> 0. It is easy to see that the following holds.
Lemma 2.4(see, e.g., [15]) Ifψ∈Ψ,then the following are satisfied:
(a) ψ(t) <tfor allt> 0; (b) ψ(0) = 0.
Definition 2.5 Let (Ξ,,S) be an ordered Sp-metric space. A mapping f :Ξ →Ξ is called anSp-ψ-contraction if there existsψ∈Ψ such that
Ω22S(fζ,fη,fω)≤ψM(ζ,η,ω) (2.7)
for all mutually comparable elementsζ,η,ω∈Ξ, where
M(ζ,η,ω) =max
S(ζ,η,ω),1 2Ω
–1S(ζ,fζ,fη),S(η,fη,fω)
Theorem 2.6 Let(Ξ,,S)be an ordered Sp-complete Sp-metric space.Let f:Ξ→Ξ be
an increasing mapping with respect tosuch that there exists an elementζ0∈Ξ with
ζ0fζ0.Suppose that f is an Sp-ψ-contractive mapping.If
(I) f is continuous,or
(II) (Ξ,,S)has the s.l.c.property,
then f has a fixed point.Moreover,the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Proof Putζn=fnζ0.
Step 1. We will show thatlimn→∞S(ζn,ζn+1,ζn+1) = 0. We assume thatζn=ζn+1 for all
n∈N(otherwise there is nothing to prove). As in the proof of Theorem2.3, we have that the sequence{ζn}is increasing. Hence, by (2.7) we have
S(ζn,ζn+1,ζn+1) =S(fζn–1,fζn,fζn)≤Ω(2S
f(ζn–1,fζn,fζn)
≤ψM(ζn–1,ζn,ζn)
≤ψS(ζn–1,ζn,ζn)
<S(ζn–1,ζn,ζn) (2.8)
because
M(ζn–1,ζn,ζn+1)≤max S(ζn–1,ζn,ζn),S(ζn,ζn+1,ζn+1)
,
and it is easy to see that max{S(ζn–1,ζn,ζn),S(ζn,ζn+1,ζn+1)} =S(ζn–1,ζn,ζn). So, from (2.8) we conclude that{S(ζn,ζn+1,ζn+1)}is decreasing. Then there existsr≥0 such that
limn→∞S(ζn,ζn+1,ζn+2) =r. It is easy to verify that
r= lim
n→∞S(ζn–1,ζn,ζn) = 0.
Step 2. Similarly as in the proof of Theorem2.3, if{ζn}were not anSp-Cauchy sequence, then there would existε> 0 for which we can find two subsequences{ζmi}and{ζni}of{ζn}
such that (2.5) and (2.6) hold. Then we would have
lim sup
i→∞
M(ζmi,ζni–1,ζni–1)≤Ω(ε).
Now, from (2.7) and the mentioned inequalities, we have
Ω(ε) =Ω2
2·1 2Ω
–1ε
≤Ω2
lim sup
i→∞
2S(ζmi+1,ζni,ζni)
≤lim sup
i→∞
ψM(ζmi,ζni–1,ζni–1)
<Ω(ε),
which is a contradiction. Hence,{ζn}is aSp-Cauchy sequence andSp-completeness ofζ yields that{ζn}Sp-converges to a pointu∈Ξ.
Step 3.uis a fixed point off. This step is proved as in the proof of Theorem2.3with
If in the above theorem we takeψ(t) =sinhtandS(ζ,η,ω) =sinh(Sb(ζ,η,ω)), then we have the following corollary in the framework ofSbmetric spaces.
Corollary 2.7 Let(Ξ,,Sb)be an ordered Sb-complete Sb-metric space with coefficient
s> 1.Let f :Ξ→Ξ be an increasing mapping with respect tosuch that there exists an elementζ0∈Ξwithζ0fζ0.Suppose that
sinhs·sinhs·2sinhSb(fζ,fη,fω)
≤sinhM(ζ,η,ω)
for all mutually comparable elementsζ,η,ω∈Ξ,where
M(ζ,η,ω) =max
sinhSb(ζ,η,ω)
,Sb(ζ,fζ,fη) 2s ,sinh
Sb(η,fη,fω)
.
If
(I) f is continuous,or
(II) (Ξ,,Sb)enjoys the s.l.c.property,
then f has a fixed point.
2.3 Fixed point results related toJS-contractions
Jleli et al. [8] introduced the classΘ0consisting of all functionsθ: (0,∞)→(1,∞)
satis-fying the following conditions:
(θ1) θis nondecreasing;
(θ2) for each sequence{tn} ⊆(0,∞),limn→∞θ(tn) = 1if and only iflimn→∞tn= 0;
(θ3) there existr∈(0, 1)and∈(0,∞]such thatlimt→0+ θ(t)–1 tr =; (θ4) θis continuous.
They proved the following result.
Theorem 2.8([8, Corollary 2.1]) Let(Ξ,d)be a complete metric space,and let T:Ξ→Ξ
be a given mapping.Suppose that there existθ∈Θ0and k∈(0, 1)such that
ζ,η∈Ξ, d(Tx,Ty)= 0 ⇒ θd(Tx,Tη)≤θd(ζ,η)k.
Then T has a unique fixed point.
From now on, we denote byΘthe set of all functionsθ: [0,∞)→[1,∞) satisfying the following conditions:
θ1. θis a continuous strictly increasing function;
θ2. for each sequence{tn} ⊆(0,∞),limn→∞θ(tn) = 1if and only iflimn→∞tn= 0.
Remark2.9 ([6]) It is clear thatf(t) =etdoes not belong toΘ
0, but it belongs toΘ. Other
examples aref(t) =cosht,f(t) =1+2coshcoshtt,f(t) = 1 +ln(1 +t),f(t) =2+22+lnln(1+(1+tt)),f(t) =etet, and
f(t) = 2etet
1+etet for allt> 0.
Definition 2.10 Let (Ξ,,S) be an orderedSp-metric space. A mappingf :Ξ →Ξ is called anSprational-JS-contraction if
for all mutually comparable elementsζ,η,ω∈ζ, whereθ∈Θ,k∈[0, 1) and
M(ζ,η,ω) =max
S(ζ,η,ω), S(ζ,ζ,fζ)S(η,η,fη) 1 +S(ζ,η,η) +S(ζ,ω,ω),
S(η,η,fη)S(ω,ω,fω) 1 +S(η,fω,fω) +S(η,ζ,ζ)
.
Theorem 2.11 Let(Ξ,,S)be an ordered Sp-complete Sp-metric space.Let f :Ξ→Ξ
be an increasing mapping with respect tosuch that there exists an elementζ0∈Ξ with
ζ0fζ0.Suppose that f is an Sp-rational JS-contractive mapping.If
(I) f is continuous,or
(II) (Ξ,,S)enjoys the s.l.c.property,
then f has a fixed point.Moreover,the set of fixed points of f is well ordered if and only if f has one and only one fixed point.
Proof Putζn=fnζ0.
Step 1. We will show thatlimn→∞S(ζn,ζn+1,ζn+1) = 0. Without loss of generality, we may
assume thatζn=ζn+1for alln∈N. Sinceζn–1ζnfor eachn∈N, then by (2.9) we have
θS(ζn,ζn+1,ζn+1)
≤θΩ2S(ζn,ζn+1,ζn+1)
=θS(fζn–1,fζn,fζn)
≤θM(ζn–1,ζn,ζn) k
≤θS(ζn–1,ζn,ζn) k
(2.10)
because
M(ζn–1,ζn,ζn) =max
S(ζn–1,ζn,ζn),
S(ζn–1,ζn–1,fζn–1)S(ζn,ζn,fζn) 1 +S(ζn–1,ζn,ζn) +S(ζn–1,ζn,ζn) S(ζn,ζn,fζn)S(ζn,ζn,fζn)
1 +S(ζn,fζn,fζn) +S(ζn,ζn–1,ζn–1)
=max
S(ζn–1,ζn,ζn),
S(ζn–1,ζn–1,ζn)S(ζn,ζn,ζn+1)
1 +S(ζn–1,ζn,ζn) +S(ζn–1,ζn,ζn) S(ζn,ζn,ζn+1)S(ζn,ζn,ζn+1)
1 +S(ζn,ζn+1,ζn+1) +S(ζn,ζn–1,ζn–1)
≤max S(ζn–1,ζn,ζn),S(ζn,ζn,ζn+1)
.
From (2.10) we deduce that
θS(ζn,ζn+1,ζn+1)
≤θS(ζn–1,ζn,ζn) k
.
Therefore,
1≤θS(ζn,ζn+1,ζn+1)
≤θS(ζn–1,ζn,ζn)
k≤ · · · ≤
θS(ζ0,ζ1,ζ1)
kn
. (2.11)
Taking the limit asn→ ∞in (2.11), we have
lim
n→∞θS(ζn,ζn+1,ζn+1)
= 1,
and sinceθ∈Θ, we obtainlimn→∞S(ζn,ζn+1,ζn+1) = 0. Therefore, we have
lim
Step 2. Similarly as in the proof of Theorem2.3, if{ζn}were not anSp-Cauchy sequence, then there would existε> 0 for which we can find two subsequences{ζmi}and{ζni}of{ζn}
such that (2.5) and (2.6) hold. Hence,
S(ζmi,ζmi,ζni–1) <ε.
From the rectangular inequality, we get
ε≤S(ζmi,ζni,ζni)≤ΩS(ζmi,ζmi,ζmi+1) + 2S(ζmi+1,ζni,ζni)
.
By taking the upper limit asi→ ∞, we get
1 2Ω
–1(ε)≤lim sup
i→∞
S(ζmi+1,ζni,ζni).
From the definition ofM(ζ,η,ω) and the above limits,
lim sup
i→∞
M(ζmi,ζni–1,ζni–1) =lim sup
i→∞
max
S(ζmi,ζni–1,ζni–1),
S(ζmi,ζmi,fζmi)S(ζni–1,ζni–1,fζni–1)
1 +S(ζmi,ζni–1,ζni–1) +S(ζmi,ζni–1,ζni–1)
S(ζni–1,ζni–1,fζni–1)S(ζni–1,ζni–1,fζni–1)
1 +S(ζni–1,fζni–1,fζni–1) +S(ζni–1,ζmi,ζmi)
≤ε.
Now, from (2.9) and the above inequalities, we have
θ
Ω
2·1 2Ω
–1(ε)
≤lim sup
i→∞ θ
Ω2S(ζmi+1,ζni,ζni)
≤lim sup
i→∞
θM(ζmi,ζni–1,ζni–1)
k
≤θ(ε)k,
which implies thatε= 0, a contradiction. So, we conclude that{ζn}is anSp-Cauchy se-quence. BySp-completeness ofΞ, it follows that{ζn}Sp-converges to a pointu∈Ξ.
Step 3.uis a fixed point off.
Whenf is continuous, the proof is straightforward.
Now, let (II) hold. Using the assumption onΞ, we haveζnu. By Lemma1.11,
θ
Ω
2·Ω
–1[S(u,u,fu)]
2
≤lim sup
n→∞ θ
Ω2S(ζn+1,ζn+1,fu)
≤lim sup
n→∞ θ
M(ζn,ζn,u) k
where
lim
n→∞M(ζn,ζn,u) =nlim→∞max
S(ζn,ζn,u),
S(ζn,ζn,fζn)S(ζn,ζn,fζn) 1 +S(ζn,ζn,ζn) +S(ζn,u,u)
,
S(ζn,ζn,fζn)S(u,u,fu) 1 +S(ζn,fu,fu) +S(ζn,ζn,ζn)
= 0.
Therefore, we deduce thatS(u,fu,fu) = 0, sou=fu.
Finally, suppose that the set of fixed points off is well ordered. Assume, on the contrary, thatuandvare two fixed points off such thatu=v. Then, by (2.9), we have
θS(u,v,v)=θS(fu,fv,fv)≤θM(u,v,v)k=θS(u,v,v)k.
So, we getS(u,v,v) = 0, a contradiction. Henceu=v, andf has a unique fixed point.
If in the above theorem we takeθ(t) = 2etet
1+etet andS(ζ,η,ω) =e
Sb(ζ,η,ω)– 1, then we have
the following corollary in the framework ofSbmetric spaces.
Corollary 2.12 Let(Ξ,,Sb)be an ordered Sb-complete Sb-metric space with coefficient
s> 1.Let f :Ξ→Ξ be an increasing mapping with respect tosuch that there exists an elementζ0∈Ξwithζ0fζ0.Suppose that
2e[es·[2eSb(fζ,fη,fω)–1]–1]ees·[2eSb
(fζ,fη,fω)–1]
–1
1 +e[es·[2eSb(fζ,fη,fω)–1]–1]ees·[2eSb(fζ,fη,fω)–1]–1 ≤
2eM(ζ,η,ω)eM(ζ,η,ω) 1 +eM(ζ,η,ω)eM(ζ,η,ω)
for all mutually comparable elementsζ,η,ω∈Ξ,where
M(ζ,η,ω) =max
eSb(ζ,η,ω)– 1, [e
Sb(ζ,ζ,fζ)– 1][eSb(η,η,fη)– 1]
1 +eSb(ζ,η,η)– 1 +eSb(ζ,ζ,fη)– 1,
[eSb(η,η,fη)– 1][eSb(ω,ω,fω)– 1]
1 +eSb(η,fω,fω)– 1 +eSb(η,ζ,ζ)– 1
.
If
(I) f is continuous,or
(II) (Ξ,,Sb)enjoys the s.l.c.property,
then f has a fixed point.
3 Examples
Example3.1 LetΞ={0,12,13,15}be equipped with the partial ordergiven as
:= (0, 0), 0,1 2 , 0,1 3 , 0,1 5 , 1 2, 1 2 , 1 3, 1 3 , 1 5, 1 5 .
Define a metricdonΞ by
d(ζ,η) = ⎧ ⎨ ⎩
0, ifζ=η,
and letS(ζ,η,ω) =sinh[d(ζ,ω) +d(η,ω)]. It is easy to see that (Ξ,S) is anSp-completeSp -metric space. Indeed, in the same way as the usualS-metric is formed in Example1.5, an
S-metricS(ζ,η,ω) =d(ζ,ω) +d(η,ω) is formed, and using the functionξ(t) =sinht, as in Proposition1.8, one obtains theSp-metricS.
Define a self-mapf by
f =
0 12 13 15 0 1
5 0 0
.
We see thatf is an increasing mapping w.r.t. and (Ξ,,S) enjoys the s.l.c. property. Also, 0f0.
Defineθ : [0,∞)→[1,∞) byθ(t) =cosh(t) and takek=23. One can easily check that
f is anSp-rationalJS-contractive mapping. Indeed, as a sample, we check some cases as follows:
1. (ζ,η,ω) = (0, 0,1 2). Then
θΩ2S(fζ,fη,fω)=cosh
sinh2sinh2
f0 +f1
2
=cosh
sinh2sinh2
0 +1 5
≈1.451≤1.465≈ 3
cosh
sinh2
0 +1 2
2
= 3
θM(ζ,η)2.
2. (ζ,η,ω) = (0,12,12). Then
θΩ2S(fζ,fη,fω)=cosh
sinh sinh2
f0 +f1
2
=cosh
sinh sinh2
0 +1 5
≈1.087≤1.091≈ 3
cosh
sinh
0 +1
2 2
= 3
θM(ζ,η)2.
3. (ζ,η,ω) = (0, 0,13). Then
θΩ2S(fζ,fη,fω)=cosh
sinh sinh2
f0 +f1
3
=coshsinh sinh2(0 + 0)
= 0≤1.172≈ 3
cosh
sinh2
0 +1 3
2
= 3
4. (ζ,η,ω) = (0,15,15). Then
θΩ2S(fζ,fη,fω)=cosh
sinh2sinh
f0 +f1
5
=coshsinh sinh2(0 + 0)
= 0≤1.0135≈ 3
cosh
sinh
0 +1
5 2
= 3
θM(ζ,η)2.
The rest of the cases can be checked similarly. Thus, all the conditions of Theorem2.11
are satisfied and hencef has a fixed point. Indeed, 0 is the fixed point off.
Example3.2 LetΞ= [0, 1.5] be equipped with theSp-metric
S(ζ,η,ω) =e|ζ–ω|+|η–ω|– 1
for allζ,η,ω∈Ξ, whereΩ(t) =et– 1 (the mappingSis obtained from the usualS-metric (Example1.5) using the functionξ(t) =et– 1). Define a relationonΞbyζηiffη≤ζ, a mappingf:Ξ→Ξby
fζ=ζ 8e
–ζ2
and a function β by β(t) = 12 < 0.882≈Ω–1(1). For all mutually comparable elements
ζ,η,ω∈Ξ, we have
S(fζ,fη,fω) =e|fζ–fω|+|fη–fω|– 1 =e|ζ8e–ζ2–ω8e–ω2|+|η 8e
–η2
–ω8e–ω2| – 1
≤e18|ζ–ω|+18|η–ω|– 1≤1 8
e|ζ–ω|+|η–ω|– 1
=1
8S(ζ,η,ω).
So,
Ω22Ssp(fζ,fη,fω)≤Ω2
1
4S(ζ,η,ω)
=e(e
1
4S(ζ,η,ω)–1)– 1≤e21(e12S(ζ,η,ω)–1)– 1
≤1
2S(ζ,η,ω) =β
M(ζ,η,ω)S(ζ,η,ω).
Therefore, it follows from Theorem2.3thatf has a fixed point (which isu= 0).
Example3.3 LetΞ= [0, 3] be equipped with theSp-metric
as in the previous example. Define a relationonΞbyζ ηiffη≤ζ, a mappingf :Ξ→ Ξ by
fζ=ln(ζ+ 12),
and a functionψbyψ(t) =14(et– 1). For all mutually comparable elementsζ,η,ω∈Ξ, by the mean value theorem, we have
S(fζ,fη,fω) =e|ln(ζ+12)–ln(ω+12)|+|ln(η+12)–ln(ω+12)|– 1
=e|lnζω+12+12|+|ln
η+12
ω+12|– 1 =ζ+ 12 ω+ 12·
η–ω ω+ 12
– 1
≤ 1 12
|ζ–ω| · |η–ω|– 1≤ 1 12
e|ζ–ω|+|η–ω|– 1
= 1
12S(ζ,η,ω).
So,
Ω22S(fζ,fη,fω)≤Ω2
1
6S(ζ,η,ω)
=e(e
1 6S(ζ,η,ω)–1)
– 1
≤e14S(ζ,η,ω)– 1 =ψS(ζ,η,ω).
Therefore, from Theorem2.6,f has a fixed point.
Example3.4 LetS:Ξ×Ξ×Ξ→R+be defined onΞ= [0, 20] by
S(ζ,η,ω) =e13(|ζ–η|+|η–ω|+|ω–ζ|)– 1
for allζ,η,ω∈Ξ (the mappingSis obtained from theS-metric
S(ζ,η,ω) = 1 3
|ζ–η|+|η–ω|+|ω–ζ|
using the functionξ(t) =et– 1). Then (Ξ,S) is anS
p-completeSp-metric space withΩ(t) =
et– 1. Defineθ∈Θbyθ(t) =etet. LetΞ be endowed with the standard order≤, and let
f :Ξ →Ξ be defined byfζ =arctan(32ζ ). It is easy to see thatf is an ordered increasing and continuous self-map onζ and 0≤f0. For any mutually comparableζ,η,ω∈Ξ, we have
S(fζ,fη,fω) =e13(|fζ–fη|+|fη–fω|+|fω–fζ|)– 1
=e13(|arctan
ζ
32–arctan
η
32|+|arctan
η
32–arctan32ω|+|arctan32ω–arctan
ζ
32|)– 1
≤e13(|32ζ–32η|+|32η–32ω|+|32ω–32ζ|)– 1≤ 1 32
e13(|ζ–η|+|η–ω|+|ω–ζ|)– 1
= 1
So,
Ω2S(fζ,fη,fω)=e2S(fζ,fη,fω)– 1≤e161S(ζ,η,ω)– 1 ≤1
8S(ζ,η,ω).
Therefore,
θΩ2S(fζ,fη,fω)=eΩ[2S(fζ,fη,fω)]eΩ[2S(fζ,fη,fω)]≤e18S(ζ,η,ω)e 1 8S(ζ,η,ω)
≤eS(ζ,η,ω)eS(ζ,η,ω)
1 √
2 =θS(ζ,η,ω) 1 √ 2.
Thus, (2.9) is satisfied withk=√1
2. Hence, all the conditions of Theorem2.11are satisfied.
We see that 0 is the unique fixed point off.
4 Application
In this section we present an application of Theorem2.6. This application is inspired by [12].
LetΞ=C(I) be the set of all real continuous functions onI= [0,T]. Obviously, this set with thep-metric given by
d(ζ,η) =sinh
max
t∈I
ζ(t) –η(t)
for allζ,η∈Ξ is ap-completep-metric space withΩ(t) =sinht(in the sense of [13]). ThenS(ζ,η,ω) =d(ζ,ω) +d(η,ω) is anSp-metric onΞ assinhtis super-additive. (Ξ,S) can also be equipped with the order given by
ζ η iff ζ(t)≤η(t) for allt∈I.
Moreover, as in [12], it can be proved that (C(I),,S) has the s.l.c. property. Consider the first-order periodic boundary value problem
⎧ ⎨ ⎩
ζ(t) =f(t,ζ(t)),
ζ(0) =ζ(T), (4.1)
where t∈I andf :I×R→Ris a continuous function. A lower solution for (4.1) is a functionα∈C1(I) such that
⎧ ⎨ ⎩
α(t)≤f(t,α(t)),
α(0)≤α(T),
wheret∈I. Assume that there existsλ> 0 such that, for allζ,η∈Ξandt∈I, we have
ft,ζ(t)+λζ(t) –ft,η(t)–λη(t)≤λ
We will show that the existence of a lower solution for (4.1) provides the existence of a unique solution of (4.1). Problem (4.1) can be rewritten as
⎧ ⎨ ⎩
ζ(t) +λζ(t) =f(t,ζ(t)) +λζ(t)≡F(t,ζ(t)),
ζ(0) =ζ(T).
It is well known that this problem is equivalent to the integral equation
ζ(t) = T
0
G(t,s)Fs,ζ(s)ds,
whereGis the Green’s function given as
G(t,s) = ⎧ ⎨ ⎩
eλ(T+s–t)
eλT–1 , 0≤s≤t≤T,
eλ(s–t)
eλT–1, 0≤t≤s≤T.
Easy calculation gives
T
0
G(t,s)ds=1 λ.
Now define an operatorH:C(I)→C(I) as
Hζ(t) = T
0
G(t,s)Fs,ζ(s)ds. (4.3)
It is easy to see that the mappingHis increasing w.r.t.. Note that ifu∈C1(I) is a fixed point ofH, thenuis a solution of (4.1).
Letζ,ω∈Ξbe comparable. Then we have
d(Hx,Hz) =sinh
max
t∈I Hx(t) –Hz(t)
=sinh
max
t∈I T
0
G(t,s)Fs,ζ(s)ds– T
0
G(t,s)Fs,ω(s)ds
≤sinh
max
t∈I T
0
G(t,s)·Fs,ζ(s)–Fs,ω(s)ds
≤sinh
1 λmaxs∈I
λ
8ζ(s) –ω(s)
=sinh
1 8sinh
–1d(ζ,ω)
≤1 8d(ζ,ω).
Therefore,
S(Hx,Hy,Hz) =d(Hx,Hz) +d(Hy,Hz)≤1 8
d(ζ,ω) +d(η,ω)=1
So,
Ω22S(Hx,Hy,Hz)≤Ω2
1
4S(ζ,η,ω)
=sinh
sinh
1
4S(ζ,η,ω)
=ψS(ζ,η,ω)≤ψM(ζ,η,ω),
where
M(ζ,η) =max
S(ζ,η,ω),sinh
–1(S(ζ,fζ,fη))
2 ,S(η,fη,fω)
,
andψ(t) =sinh(sinh(t/4)). Finally, if there exists a lower solutionαof4.1, the hypotheses of Theorem2.6are satisfied. Therefore, there exists a fixed pointζˆ∈C(I) ofH, which is a solution of the given boundary value problem.
Acknowledgements
The publication of this article was funded by the Qatar National Library.
Funding
The publication of this article funded by the Qatar National Library.
Abbreviations
Not applicable.
Availability of data and materials
Data sharing not applicable to this article as no datasets were generated or analysed during the current study.
Competing interests
The authors declare that they have no conflict of interests.
Authors’ contributions
All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, Statistics and Physics, Qatar University, Doha, Qatar.2Department of Mathematics, Aliabad Katool Branch, Islamic Azad University, Aliabad katool, Iran.3Department of Mathematics, Gilan-E-Gharb Branch, Islamic Azad University, Gilan-E-Gharb, Iran.4Faculty of Mathematics, University of Belgrade, Beograd, Serbia.
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Received: 7 April 2019 Accepted: 23 October 2019
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