R E S E A R C H
Open Access
Basic inequality on a
b
-metric space and
its applications
Tomonari Suzuki
**Correspondence:
[email protected] Department of Basic Sciences, Faculty of Engineering, Kyushu Institute of Technology, Tobata, Kitakyushu, 804-8550, Japan
Abstract
We first prove one of the most basic inequalities on ab-metric space. And then we prove some fixed point theorems. We also consider two similar conditions; one implies the Cauchyness on sequences but the other does not.
MSC: Primary 54E99; secondary 54H25
Keywords: b-metric space; Cauchy sequence; fixed point theorem
1 Introduction
In , Czerwik introduced the following interesting concept.
Definition (Czerwik []) LetXbe a set and letdbe a function fromX×Xinto [,∞). Then (X,d) is said to be ab-metric spaceif the following hold:
(b) d(x,y) = ⇔x=y;
(b) d(x,y) =d(y,x)(symmetry);
(b) There existsK≥satisfyingd(x,z)≤K(d(x,y) +d(y,z))for anyx,y,z∈X
(K-relaxed triangle inequality).
We note that in the case whereK= , everyb-metric space is obviously a metric space. So this concept is a weaker concept than that of a metric space. Conditions (b) and (b) also appear in the definition of metric space. So (b) is a feature of this concept. Therefore it is important to study how to use (b) effectively.
Lemma Let(X,d)be a b-metric space.For n∈Nand(x, . . . ,xn)∈Xn+,
d(x,xn)≤ n–
j=
Kj+d(xj,xj+) +Kn–d(xn–,xn) ()
holds.
Proof Obvious.
Considering the rearrangement inequality, we could tell that () is effective in the case whered(xj+,xj+) is much smaller thand(xj,xj+). Indeed, using (), the following lemma
was proved.
Lemma (Lemma . in []) Let(X,d)be a b-metric space and let{xn}be a sequence in X. Assume that there exists r∈[, /K)satisfying d(xn+,xn+)≤rd(xn,xn+)for any n∈N.
Then{xn}is Cauchy.
In the case whered(xj+,xj+) is not much smaller thand(xj,xj+), how can we use (b)
effectively?
Motivated by this question, in this paper, we first prove some inequality. Using this in-equality, we improve Lemma . In order to understand deeply the mathematical structure of ab-metric space, we give a condition, which does not imply the Cauchyness on se-quences. Finally, we improve some Nadler-type fixed point theorems.
2 Preliminaries
Throughout this paper we denote byNthe set of all positive integers and byRthe set of all real numbers. For an arbitrary setA, we also denote by#Athe cardinal number ofA. For a real numbert, we denote by [t] the maximum integer not exceedingt.
In this section, we give some preliminaries.
Definition Let (X,d) be ab-metric space. Let{xn}be a sequence inXand letAbe a subset ofX.
• {xn}is said toconvergetoxiflimnd(x,xn) = holds.
• {xn}is said to beCauchyiflimnsup{d(xn,xm) :m>n}= holds.
• Xis said to becompleteif every Cauchy sequence converges.
• Ais said to beclosedif for any convergent sequence inA, its limit belongs toA. • Ais said to beboundedifsup{d(x,y) :x,y∈A}<∞holds.
Remark While not everyν-generalized metric space is metrizable [–], everyb-metric space is metrizable. So we note that there is no room for ambiguity in Definition .
Let (X,d) be ab-metric space and letCB(X) be the set of all nonempty, bounded and closed subsets ofX. Forx∈Xand any subsetAofX, we defined(x,A) =inf{d(x,y) :y∈A}. Then theHausdorff metric Hwith respect todis defined by
H(A,B) =maxsupd(u,B) :u∈A,supd(v,A) :v∈B
for allA,B∈CB(X).
3 Basic inequality
In this section, we prove one of the most basic inequalities on ab-metric space.
Lemma Let(X,d)be a b-metric space.Define a function f fromNintoN∪ {}by
f(n) = –[–logn]. ()
For n∈Nand(x, . . . ,xn)∈Xn+,
d(x,xn)≤Kf(n) n–
j=
d(xj,xj+) ()
Remark f is as follows:
f() = , f() = , f() =f() = ,
f() =· · ·=f() = , f() =· · ·=f() = , . . . .
Proof We first note that f(n)–<n≤f(n)holds for anyn∈N. It is obvious that () holds
forn= . We assume that () holds fornwithn≤kfor somek∈N∪ {}. Fixn∈Nwith k<n≤k+. Then we notef(n) =k+ ,f(k) =kandf(n– k)≤k. We have
d(x,xn)≤Kd(x,xk) +Kd(xk,xn)
≤KKf(k)
k–
j=
d(xj,xj+) +KKf(n–
k)n–
j=k
d(xj,xj+)
≤Kk+ n–
j=
d(xj,xj+)
=Kf(n) n–
j=
d(xj,xj+).
Thus () holds forn. By induction, we obtain the desired result.
Using Lemma , we give a sufficient condition for the Cauchyness on sequences. The following lemma can be very useful when we prove existence theorems in complete b-metric spaces. In order to demonstrate that, in Section , we improve some fixed point theorems. See [, ] and others.
Lemma Let(X,d)be a b-metric space and let{xn}be a sequence in X.Assume that there exists r∈[, )satisfying
d(xn+,xn+)≤rd(xn,xn+) ()
for any n∈N.Then{xn}is Cauchy.
Remark Compare Lemma with Lemma .
Proof In the case wherer= , the conclusion obviously holds. So we assumer> . We choose∈Nsatisfying
Kr< .
Define a functionf by (). Form,n∈Nwithn<m≤n+ , we have by Lemma
d(xn,xm)≤Kf(m–n) m–
j=n
d(xj,xj+)
≤K m–
j=n
≤K
∞
j=n
rj–d(x,x)
≤KrnC,
where we putC=d(x,x)/(r( –r)). Form,n∈Nwithn+ <m, puttingμ= [(m–n)/],
we have by ()
d(xn,xm)≤ μ–
i=
Ki+d(x
n+i,xn+(i+)) +Kμd(xn+μ,xm)
≤ μ–
i=
Ki++rn+i
C+Kμ+rn+μ C
≤rnC μ
i=
Ki++ri
≤rnC
∞
i=
Ki++ri
=rnC K +
–Kr.
Therefore{xn}is Cauchy.
4 Example, part 1
In this section, we give a typical example of ab-metric space.
Lemma Define a function f by().Then the following hold:
(i) f(n) =f(n) + for anyn∈N.
(ii) f(n+ )∈ {f(n),f(n) + }for anyn∈N. (iii) f is nondecreasing.
Proof Obvious.
Lemma Let K ∈[,∞)and put X=N.Define a function f by().Define a function g fromN∪ {}into[,∞)by
g() = ,
g(n) =n– f(n)Kf(n)+f(n)–nKf(n)–.
()
Then
g(n) =Kg[n/]+gn– [n/], ()
g(n) –g(n– )≥g(n– ) –g(n– ) > , ()
g(n)≤Kg(k) +g(n–k) ()
Remark The functiongis as follows:
g() = , g() = K, g() = K+K,
g() = K, g() = K+ K, g() = K+ K.
Proof We first show (). Fixn∈N. We consider the following two cases:
• n+ = f(n+)–+ ,
• n+ > f(n+)–+ .
We puta= n+ ,b= [(n+ )/] andc=a–b. In the first case, notingf(a)≥f() = , we have
b=f(n+)–+ /= f(n+)–
and
c= f(n+)–+ – f(n+)–= f(n+)–+ .
Hence
f(b) =f(n+ ) – =f(a) –
and
f(c) =f(n+ ) – =f(a) –
hold. We have
g(a) =(n+ ) – f(n+)Kf(a)+f(n+)– n– Kf(a)–
=f(n+)+ – f(n+)Kf(a)+f(n+)– f(n+)–– Kf(a)–
= Kf(a)+ (a– )Kf(a)–
= Kf(c)++bKf(b)++ (c– )Kf(c)
=Kb– f(b)Kf(b)+f(b)–bKf(b)–
+c– f(c)Kf(c)+f(c)–cKf(c)–
=Kg(b) +g(c).
In the second case, noting thatais odd, we have f(a)–>a/ > f(a)–+ . So
f(a)–< f(a)–+ ≤b<a/ <b+ =c≤f(a)–
holds and hencef(b) =f(c) =f(a) – holds. So we have
g(a) =b+ c– f(b)– f(c)Kf(a)+f(b)+ f(c)–b–cKf(a)–
+c– f(c)Kf(c)++f(c)–cKf(c)
=Kg(b) +g(c).
Using Lemma , we have
g(n) =n– f(n)Kf(n)+f(n)– nKf(n)–
=n– f(n)+Kf(n)++f(n)+– nKf(n)
= Kn– f(n)Kf(n)+f(n)–nKf(n)–
=Kg(n) +g(n)
=Kg[n/]+gn– [n/].
We have shown ().
We shall show (). We have
g() –g() = K– ≥ =g() –g() > .
Fixn∈Nwithn≥. In the case wheref(n) =f(n– ), we have
g(n) –g(n– ) = Kf(n)–Kf(n)–.
In the other case, wheref(n) >f(n– ), we haven– = f(n)–andf(n– ) =f(n) – . So
g(n) –g(n– ) = Kf(n)+ (n– )Kf(n)–– (n– )Kf(n–)
= Kf(n)–Kf(n)–.
Sincef is nondecreasing andK≥ holds,{g(n) –g(n– )}is also nondecreasing. Let us prove (). By (), there exists a convex function from [,∞) intoRwhose restric-tion toNisg. So we have
g(n) =Kg[n/]+gn– [n/]
≤Kg[n/] – +gn– [n/] +
≤Kg[n/] – +gn– [n/] +
≤ · · · ≤Kg() +g(n– ).
We have shown ().
By (),gis strictly increasing.
The following example is a typical example of ab-metric space. Indeed, we use this ex-ample in order to make Exex-ample .
Example LetK∈[,∞) and putX=N. Define functionsf andgby () and (), respec-tively. Define a functiondfromX×Xinto [,∞) by
d(x,y) =g|x–y|.
Proof (b) and (b) obviously hold. In order to prove (b), we note
d(,m)≤d(,n) and d(m,n)≤d(,n)
for any,m,n∈Nwith<m<nbecausegis strictly increasing. Fix,m,n∈Nwith<n,
=mandm=n. We consider the following two cases:
(a) m<orn<m, (b) <m<n.
In the case of (a), without loss of generality, we may assumen<m. Then we have
d(,n)≤d(,m)≤d(,m) +d(m,n)≤Kd(,m) +Kd(m,n).
In the case of (b), we have by ()
d(,n)≤Kd(,m) +Kd(m,n).
We have shown (b).
5 Examples, part 2
We have proved that () implies the Cauchyness on sequences. In this section, we give examples of that ∞n=d(xn,xn+) <∞does not imply the Cauchyness on sequences. The
author thinks that such a property is one of characteristics of ab-metric space.
Example LetK∈[,∞) and let{αn}be a sequence in (,∞). Let Xbe a subset of [,∞) satisfyingN⊂Xand#(X∩[,a]) <∞for anya∈[,∞). Define a strictly increasing functionχfromNintoXsatisfyingχ(N) =X. Define a sequence{νj}inNsatisfyingχ(νj) = jfor anyj∈N. Define functionsf andgby () and (), respectively. Define a functiond fromX×Xinto [,∞) by
dχ(m),χ(n)= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
ifm=n,
g(n–m)αk ifνk≤m<n≤νk+for somek∈N,
d(χ(m),j+ ) + ik=–j+d(i,i+ ) +d(k,χ(n))
ifνj≤m<νj+≤νk<n≤νk+for somej,k∈N,
d(χ(n),χ(m)) ifm>n,
for allm,n∈N. Then (X,d) is ab-metric space.
Proof It is obvious that (b) and (b) hold. Let us prove (b). Define a functionefrom N×Ninto [,∞) by
e(m,n) =dχ(m),χ(n).
We note
for any,m,n∈Nwith<m<n. Fix,m,n∈Nwith<n,=mandm=n. We consider the following three cases:
(a) m<orn<m,
(b) νk≤<m<n≤νk+for somek∈N,
(c) <m<nandνj≤<νj+≤νk<n≤νk+for somej,k∈N.
In the cases of (a) and (b), we can prove (b) as in the proof of Example . In the case of (c), we further consider the following two cases:
(c-) m≤νj+orνk≤m,
(c-) νj+<m<νk.
In the case of (c-), without loss of generality, we may assumem≤νj+. We have
e(,n) =e(,νj+) +
k–
i=j+
e(νi,νi+) +e(νk,n)
≤Ke(,m) +Ke(m,νj+) +
k–
i=j+
e(νi,νi+) +e(νk,n)
≤Ke(,m) +Ke(m,νj+) +K
k–
i=j+
e(νi,νi+) +Ke(νk,n)
=Ke(,m) +e(m,n).
In the case of (c-), there existsp∈Nsatisfying
νj+≤νp≤m<νp+≤νk.
We have
e(,n) =e(,νj+) +
p–
i=j+
e(νi,νi+)
+e(νp,νp+) +
k–
i=p+
e(νi,νi+) +e(νk,n)
≤e(,νj+) +
p–
i=j+
e(νi,νi+)
+Ke(νp,m) +Ke(m,νp+) +
k–
i=p+
e(νi,νi+) +e(νk,n)
≤Ke(,νj+) +K
p–
i=j+
e(νi,νi+)
+Ke(νp,m) +Ke(m,νp+) +K
k–
i=p+
e(νi,νi+) +Ke(νk,n)
=Ke(,m) +e(m,n).
Example LetK∈(,∞) and define a sequence{αn}in (,∞) byαn= –nK–n. LetXbe a subset of [,∞) satisfyingN⊂Xand#(X∩[n,n+ )) = nfor anyn∈N. Letχ,{ν
j},f,g anddbe as in Example . Then the following hold:
(i) (X,d)is ab-metric space. (ii) ∞n=d(χ(n),χ(n+ )) <∞holds. (iii) {χ(n)}is not Cauchy.
(iv) Xis complete.
Proof We have proved (i) in Example . We have
∞
n=
dχ(n),χ(n+ )=
∞
k=
νk+–
n=νk
dχ(n),χ(n+ )
=
∞
k=
νk+–
n=νk
αk=
∞
k=
kαk=
∞
k=
K–k<∞.
We have
d(n,n+ ) =gnαn= nKnαn=
for anyn∈N. Since the sequence{n}inXis a subsequence of{χ(n)},{χ(n)}is not Cauchy. Let us prove (iv). We noted(n,m) =|n–m|for anym,n∈N. Solimmd(χ(n),χ(m)) =∞ holds for anyn∈N. Let{xn}be a Cauchy sequence inX. Then{xn}is bounded. So, there existsz∈Xsatisfyingxn=zfor sufficiently largen∈N. Thus,{xn}converges toz.
6 Fixed point theorems
Czerwik proved the following fixed point theorem. See page of []. Compare Theo-rem with TheoTheo-rem in [] and TheoTheo-rem . in [].
Theorem ([]) Let(X,d)be a complete b-metric space and let T be a mapping from X intoCB(X).Assume that there exists r∈[, /K)satisfying
H(Tx,Ty)≤rd(x,y) ()
for all x,y∈X.Then there exists z∈X satisfying z∈Tz.
Using Lemma , we improve Theorem . We begin with a generalization of Theorem in [].
Theorem Let(X,d)be a complete b-metric space and let T be a mapping from X into
CB(X).Assume there existsε> satisfying the following:
(i) There existsr∈[, )such that()holds for allx,y∈Xwithd(x,y) <ε. (ii) There existsx∈Xsuch thatd(x,Tx) <ε.
Then there exists z∈X satisfying z∈Tz.
Proof Putq:= ( +r)/∈(, ). Then we have the following:
In particular, puttingu=y, we obtain the following:
• Forx∈Xandy∈Txwithd(x,y) <ε, there existsv∈Tysatisfyingd(y,v)≤qd(x,y).
Therefore we can choose a sequence{un}inXsatisfying
d(u,Tu)≤d(u,u) <ε,
un+∈Tun,
d(un+,un+)≤qd(un,un+)
forn∈N. By Lemma ,{un}is Cauchy. SinceXis complete,{un}converges to some point z∈X. Since
d(z,Tz)≤ lim
n→∞K
d(z,un+) +d(un+,Tz)
=K lim
n→∞d(un+,Tz)≤Knlim→∞H(Tun,Tz)
≤Kq lim
n→∞d(un,z) =
holds andTzis closed, we obtainz∈Tz.
As a direct consequence, we obtain a generalization of Nadler’s fixed point theorem [].
Corollary Let(X,d)be a complete b-metric space and let T be a mapping from X into
CB(X).Assume that there exists r∈[, )such that()holds for all x,y∈X.Then there
exists z∈X satisfying z∈Tz.
Using Theorem , we can prove a generalization of Mizoguchi and Takahashi’s fixed point theorem []. See also [, –] and others.
Corollary Let(X,d)be a complete b-metric space and let T be a mapping from X into
CB(X).Assume that there exists a functionαfrom[,∞)into[, )satisfying
H(Tx,Ty)≤αd(x,y)d(x,y)
for all x,y∈X and
lim sup
s→t+
α(s) <
for all t∈[,∞).Then there exists z∈X satisfying z∈Tz.
Proof Sincelim sups→+α(s) < , we can chooseε> andr∈[, ) satisfyingα(t)≤rfor anyt∈[,ε). Thus (i) of Theorem holds. So we only have to prove (ii) of Theorem . Define a functionβ from [,∞) into (, ) byβ(t) = (α(t) + )/ fort∈[,∞). As in the proof of Theorem , we can choose a sequence{un}inXsatisfying
un+∈Tun and d(un+,un+)≤β
d(un,un+)
forn∈N. Sinceβ(t) < for anyt∈[,∞),{d(un,un+)}is a nonincreasing sequence in
[,∞). So{d(un,un+)}converges to someτ ∈[,∞). Sincelim sups→τ+β(s) < andβ(τ) <
, there exist q∈[, ) andδ> satisfyingβ(s)≤q for alls∈[τ,τ +δ]. Chooseν∈N
satisfyingd(uν,uν+)≤τ+δ. Then, for anyn∈Nwithn≥ν, we have
d(un+,un+)≤β
d(un,un+)
d(un,un+)≤qd(un,un+),
which implieslimnd(un,un+) = . Therefored(un,un+)≤d(un,Tun) <εholds for
suffi-ciently largen∈N. We have shown (ii) of Theorem .
Acknowledgements
The author is supported in part by JSPS KAKENHI Grant Number 16K05207 from Japan Society for the Promotion of Science. The author is grateful to the referees for letting the author know article [10].
Competing interests
The author declares that he has no competing interests.
Authors’ contributions
The author read and approved the final manuscript.
Publisher’s Note
Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
Received: 21 June 2017 Accepted: 26 September 2017
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